Miscellaneous Exercise

Integrate the functions in Exercises 1 to 24.

${"type":"$","font":{"family":"Arial","color":"#000000","size":12},"code":"$1.\\,\\frac{1}{x-x^{3}}$","id":"4-0-0-0-0","ts":1602667091155,"cs":"SQ5AKSc7fplWDIvuHQy1hg==","size":{"width":61,"height":25}}$

Solun:- Let f(x) = ${"id":"10","type":"$","code":"$\\frac{1}{x-x^{3}}$","font":{"color":"#000000","family":"Arial","size":12},"ts":1602667129550,"cs":"KiFBBDuQB8blM3NxShl/sA==","size":{"width":41,"height":25}}$

Integrate f(x):-

${"type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{1}{x-x^{3}}.dx}\\\\\n{I}&={\\int_{}^{}\\frac{1}{x\\left(1-x^{2}\\right)}.dx}\\\\\n{I}&={\\int_{}^{}\\frac{1}{x\\left(1-x\\right)\\left(1+x\\right)}.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0","ts":1602671337662,"cs":"K+MC63ucgZCj8qSC9gLO6g==","size":{"width":185,"height":118}}$

By Partial Fraction:-

${"font":{"color":"#000000","size":12,"family":"Arial"},"code":"$\\frac{1}{x\\left(1-x\\right)\\left(1+x\\right)}=\\frac{A}{x}+\\frac{B}{1-x}+\\frac{C}{1+x}$","type":"$","id":"1-0-0-0-0-0","ts":1602671433105,"cs":"2wRvFNLcgeKjtJcw8+ee0A==","size":{"width":260,"height":28}}$

⇒ 1 = A(1 - x2) + B(x)(1 + x) + C(x)(1 - x)......(1)

Put x = 0 in eq. 1:-

⇒ 1 = A(1)

⇒ A = 1

Put x = 1 in eq. 1:-

⇒ 1 = B(1)(2)

⇒ B = 1/2

Put x = - 1 in eq. 1:-

⇒ 1 = C(-1)(2)

⇒ C = - 1/2

${"font":{"family":"Arial","color":"#000000","size":10},"type":"align*","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-1-0-0-0-0-0","code":"\\begin{align*}\n{I}&={\\int_{}^{}\\left[\\frac{1}{x}+\\frac{1}{2}\\frac{1}{1-x}-\\frac{1}{2}\\frac{1}{1+x}\\right].dx}\\\\\n{I}&={\\int_{}^{}\\frac{1}{x}.dx+\\frac{1}{2}\\int_{}^{}\\frac{1}{1-x}.dx-\\frac{1}{2}\\int_{}^{}\\frac{1}{1+x}.dx}\t\n\\end{align*}","ts":1602672701274,"cs":"R2Kh+plt3PFVxS+NqezL6w==","size":{"width":336,"height":77}}$

We know that:-

${"id":"5-0-0-0","type":"$","font":{"color":"#222222","family":"Arial","size":10},"code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}x+C$","ts":1602672753650,"cs":"nCWTyeeuna6EW91VW2bTKA==","size":{"width":136,"height":18}}$

${"type":"align*","code":"\\begin{align*}\n{I}&={\\log_{}x-\\frac{1}{2}\\log_{}\\left|1-x\\right|-\\frac{1}{2}\\log_{}\\left|1+x\\right|}\\\\\n{I}&={\\log_{}x-\\frac{1}{2}\\left[\\log_{}\\left|1-x\\right|+\\log_{}\\left|1+x\\right|\\right]}\t\n\\end{align*}","id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-0-0-0-0","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1602673783731,"cs":"4Cu60G1GTYUWhwkGGwK1EA==","size":{"width":268,"height":68}}$

We know that:-

⇒ log m - log n = log (m/n)

⇒ log mn = n.log m

⇒ log m + log n = log (mn)

${"id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-1","code":"\\begin{align*}\n{I}&={\\log_{}x-\\frac{1}{2}\\log_{}\\left|\\left(1-x\\right)\\left(1+x\\right)\\right|+C}\\\\\n{I}&={\\frac{2}{2}\\log_{}x-\\frac{1}{2}\\log_{}\\left|\\left(1-x^{2}\\right)\\right|+C}\\\\\n{I}&={\\frac{1}{2}\\log_{}x^{2}-\\frac{1}{2}\\log_{}\\left|\\left(1-x^{2}\\right)\\right|+C}\\\\\n{I}&={\\frac{1}{2}\\left[\\log_{}x^{2}-\\log_{}\\left|\\left(1-x^{2}\\right)\\right|\\right]+C}\\\\\n{I}&={\\frac{1}{2}\\log_{}\\left|\\frac{x^{2}}{1-x^{2}}\\right|+C}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","ts":1602674197340,"cs":"Y5lyzduDZmbRTrsCshDU9g==","size":{"width":256,"height":186}}$

${"code":"$2.\\,\\frac{1}{{\\sqrt[]{x+a}}+{\\sqrt[]{x+b}}}$","font":{"family":"Arial","size":12,"color":"#000000"},"type":"$","id":"4-0-0-0-1-0","ts":1602674346575,"cs":"BXfmE4OBGMd2a8rcA9u+AA==","size":{"width":116,"height":29}}$

Solun:- Let f(x) = ${"code":"$\\frac{1}{{\\sqrt[]{x+a}}+{\\sqrt[]{x+b}}}$","type":"$","id":"7","font":{"color":"#000000","size":12,"family":"Arial"},"ts":1602674363529,"cs":"2jWOi8zR+tx1XeaSo29/Gw==","size":{"width":96,"height":29}}$

Integrate f(x):-

${"code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{x+a}}+{\\sqrt[]{x+b}}}.dx}\\\\\n{I}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{x+a}}+{\\sqrt[]{x+b}}}\\times\\frac{{\\sqrt[]{x+a}}-{\\sqrt[]{x+b}}}{{\\sqrt[]{x+a}}-{\\sqrt[]{x+b}}}.dx}\\\\\n{I}&={\\int_{}^{}\\frac{{\\sqrt[]{x+a}}-{\\sqrt[]{x+b}}}{\\left(x+a\\right)-\\left(x+b\\right)}.dx}\\\\\n{I}&={\\int_{}^{}\\frac{{\\sqrt[]{x+a}}-{\\sqrt[]{x+b}}}{a-b}.dx}\\\\\n{I}&={\\frac{1}{a-b}\\int_{}^{}\\left[{\\sqrt[]{x+a}}-{\\sqrt[]{x+b}}\\right].dx}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"type":"align*","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-0","ts":1602674643138,"cs":"LIxySkRri+tfMIUZUkT7aA==","size":{"width":342,"height":216}}$

We know that:-

${"type":"$","code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+C$","font":{"family":"Arial","size":10,"color":"#222222"},"id":"6-0","ts":1602674707519,"cs":"lu5Rv9pte2N7xB3fYyG8Gg==","size":{"width":136,"height":21}}$

${"id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-1-0","code":"\\begin{align*}\n{I}&={\\frac{1}{a-b}\\left[\\frac{\\left(x+a\\right)^{\\frac{3}{2}}}{\\frac{3}{2}}-\\frac{\\left(x+b\\right)^{\\frac{3}{2}}}{\\frac{3}{2}}\\right]+C}\\\\\n{I}&={\\frac{1}{a-b}\\left[\\frac{2\\left(x+a\\right)^{\\frac{3}{2}}}{3}-\\frac{2\\left(x+b\\right)^{\\frac{3}{2}}}{3}\\right]+C}\\\\\n{I}&={\\frac{2}{3\\left(a-b\\right)}\\left[\\left(x+a\\right)^{\\frac{3}{2}}-\\left(x+b\\right)^{\\frac{3}{2}}\\right]+C}\t\n\\end{align*}","type":"align*","font":{"family":"Arial","color":"#000000","size":10},"ts":1602675047836,"cs":"v/3TuLLG0Vd59tHEaUhRNw==","size":{"width":285,"height":141}}$

${"code":"$3.\\,\\frac{1}{x{\\sqrt[]{ax-x^{2}}}}$","type":"$","font":{"size":12,"color":"#000000","family":"Arial"},"id":"4-0-0-0-1-1-0","ts":1602675254639,"cs":"8dRVeNfxMbwP1KzwV60vyw==","size":{"width":89,"height":29}}$

Solun:- Let f(x) = ${"id":"8-0","type":"$","font":{"size":12,"color":"#000000","family":"Arial"},"code":"$\\frac{1}{x{\\sqrt[]{ax-x^{2}}}}$","ts":1602675337817,"cs":"Wmm8WkI1g6OwKlBbcl0HpA==","size":{"width":70,"height":29}}$

Integrate f(x):-

${"font":{"family":"Arial","color":"#000000","size":10},"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0","code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{1}{x{\\sqrt[]{ax-x^{2}}}}.dx}\t\n\\end{align*}","type":"align*","ts":1602675576351,"cs":"b54ZEM7qjjIFpT6Uk4exkA==","size":{"width":154,"height":37}}$

Let x = a/t

Differentiate w.r.t. to t:-

⇒ dx = - a/t2.dt

${"code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{t}{a{\\sqrt[]{a.\\frac{a}{t}-\\left(\\frac{a}{t}\\right)^{2}}}}.\\left(\\frac{-a}{t^{2}}\\right).dt}\\\\\n{I}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{a.\\frac{a}{t}-\\left(\\frac{a}{t}\\right)^{2}}}}.\\left(\\frac{-1}{t}\\right).dt}\\\\\n{I}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{\\frac{a^{2}}{t}-\\frac{a^{2}}{t^{2}}}}}.\\left(\\frac{-1}{t}\\right).dt}\\\\\n{I}&={\\int_{}^{}\\frac{1}{a{\\sqrt[]{\\frac{1}{t}-\\frac{1}{t^{2}}}}}.\\left(\\frac{-1}{t}\\right).dt}\\\\\n{I}&={\\int_{}^{}\\frac{1}{a{\\sqrt[]{\\frac{t-1}{t^{2}}}}}.\\left(\\frac{-1}{t}\\right).dt}\\\\\n{I}&={\\frac{-1}{a}\\int_{}^{}\\frac{1}{{\\sqrt[]{t-1}}}.dt}\\\\\n{I}&={\\frac{-1}{a}\\int_{}^{}\\left(t-1\\right)^{\\frac{-1}{2}}.dt}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-1-0-0","font":{"family":"Arial","color":"#000000","size":10},"type":"align*","ts":1602676817057,"cs":"rCzoU6hTvDCSO0yvgEBzhA==","size":{"width":240,"height":360}}$

We know that:-

${"type":"$","code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+C$","font":{"family":"Arial","size":10,"color":"#222222"},"id":"6-1","ts":1602674707519,"cs":"AFhuO8euY5Glk0PZ0HlRMg==","size":{"width":136,"height":21}}$

${"font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{I}&={\\frac{-1}{a}\\left[\\frac{\\left(t-1\\right)^{\\frac{1}{2}}}{\\frac{1}{2}}\\right]+C}\\\\\n{I}&={\\frac{-1}{a}\\left[2\\left(t-1\\right)^{\\frac{1}{2}}\\right]+C}\\\\\n{I}&={\\frac{-2}{a}{\\sqrt[]{t-1}}+C}\t\n\\end{align*}","type":"align*","id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-1-1-0","ts":1602677131812,"cs":"QZ2Lwl+vFAwCV3k1WoKqow==","size":{"width":168,"height":121}}$

Put the value of t:-

${"code":"\\begin{align*}\n{I}&={\\frac{-2}{a}{\\sqrt[]{\\frac{a}{x}-1}}+C}\\\\\n{I}&={\\frac{-2}{a}{\\sqrt[]{\\frac{a-x}{x}}}+C}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#222222"},"id":"9-0","type":"align*","ts":1602677345856,"cs":"1HXJN/t0HkiRzjN4u+MJfw==","size":{"width":148,"height":82}}$

${"code":"$4.\\,\\frac{1}{x^{2}\\left(x^{4}+1\\right)^{\\frac{3}{4}}}$","font":{"color":"#000000","family":"Arial","size":12},"type":"$","id":"4-0-0-0-1-1-1","ts":1602753085821,"cs":"rggeOdbx+7WjcqsXx4ZsGw==","size":{"width":96,"height":36}}$

Solun:- Let f(x) = ${"id":"8-1-0","code":"$\\frac{1}{x^{2}\\left(x^{4}+1\\right)^{\\frac{3}{4}}}$","font":{"size":12,"color":"#000000","family":"Arial"},"type":"$","ts":1602753131517,"cs":"Y33zV1dS2Pj8BfK+8apMsA==","size":{"width":76,"height":36}}$

Integrate f(x):-

${"font":{"size":10,"color":"#000000","family":"Arial"},"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-1-0-0","type":"align*","code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{1}{x^{2}\\left(x^{4}+1\\right)^{\\frac{3}{4}}}.dx}\\\\\n{I}&={\\int_{}^{}\\frac{1}{x^{2}\\times \\left(x^{4}\\right)^{\\frac{3}{4}}\\times\\left(1+\\frac{1}{x^{4}}\\right)^{\\frac{3}{4}}}.dx}\\\\\n{I}&={\\int_{}^{}\\frac{1}{x^{2}\\times x^{3}\\times\\left(1+\\frac{1}{x^{4}}\\right)^{\\frac{3}{4}}}.dx}\\\\\n{I}&={\\int_{}^{}\\frac{1}{x^{5}\\times\\left(1+\\frac{1}{x^{4}}\\right)^{\\frac{3}{4}}}.dx}\t\n\\end{align*}","ts":1602753274759,"cs":"r2PftGVMFUF+8UTYATkr8A==","size":{"width":240,"height":194}}$

Let 1 + 1/x4 = t4

Differentiate w.r.t. to t:-

⇒ (0 + (-4/x5)).dx = 4t3.dt

⇒ 1/x5.dx = (-4t3/4).dt

⇒ 1/x5.dx = -t3.dt

${"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-1-1","font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{-t^{3}.dt}{\\left(t^{4}\\right)^{\\frac{3}{4}}}}\\\\\n{I}&={\\int_{}^{}\\frac{-t^{3}.dt}{t^{3}}}\\\\\n{I}&={\\int_{}^{}-1.dt}\\\\\n{I}&={-\\int_{}^{}1.dt}\\\\\n{I}&={-t+C}\t\n\\end{align*}","ts":1602754198289,"cs":"UGphEfaizKjUUmLk3hGJqQ==","size":{"width":100,"height":188}}$

Put the value of t:-

${"font":{"color":"#222222","size":10,"family":"Arial"},"code":"\\begin{align*}\n{I}&={-\\left(1+\\frac{1}{x^{4}}\\right)^{\\frac{1}{4}}+C}\t\n\\end{align*}","id":"9-1-0","type":"align*","ts":1602754183226,"cs":"nyHErGyvDQjC01TBNpabsA==","size":{"width":156,"height":42}}$

${"font":{"family":"Arial","color":"#000000","size":12},"type":"$","code":"$5.\\,\\frac{1}{x^{\\frac{1}{2}}+x^{\\frac{1}{3}}}$","id":"4-0-0-0-1-1-2-0","ts":1602754378905,"cs":"oGkTu9sZlNQx15B1sBPLUw==","size":{"width":76,"height":32}}$

Solun:- Let f(x) = ${"type":"$","code":"$\\frac{1}{x^{\\frac{1}{2}}+x^{\\frac{1}{3}}}$","id":"8-1-1","font":{"color":"#000000","family":"Arial","size":12},"ts":1602754409945,"cs":"H4dIkRcDseKJt7/1tPB15A==","size":{"width":56,"height":32}}$

Integrate f(x):-

${"code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{1}{x^{\\frac{1}{2}}+x^{\\frac{1}{3}}}.dx}\t\n\\end{align*}","type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-1-0-1-0","ts":1602754432472,"cs":"Zr1+cKAOtPV45p4iBmTdnA==","size":{"width":136,"height":37}}$

Let x = t6

Differentiate w.r.t. to t:-

⇒ dx = 6t5.dt

${"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-1-0-1-1-0","font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{6t^{5}.dt}{\\left(t^{6}\\right)^{\\frac{1}{2}}+\\left(t^{6}\\right)^{\\frac{1}{3}}}}\\\\\n{I}&={6\\int_{}^{}\\frac{t^{5}.dt}{t^{3}+t^{2}}}\\\\\n{I}&={6\\int_{}^{}\\frac{t^{5}.dt}{t^{2}\\left(t+1\\right)}}\\\\\n{I}&={6\\int_{}^{}\\frac{t^{3}.dt}{t+1}}\t\n\\end{align*}","type":"align*","ts":1602756478508,"cs":"PBimUjQQhTBjZ6Vc7vuJDg==","size":{"width":142,"height":173}}$

On Dividing:-

${"font":{"color":"#000000","size":10,"family":"Arial"},"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-1-0-1-1-1-0","type":"align*","code":"\\begin{align*}\n{I}&={6\\int_{}^{}\\left[\\left(t^{2}-t+1\\right)-\\frac{1}{t+1}\\right].dt}\\\\\n{I}&={6\\left[\\frac{t^{3}}{3}-\\frac{t^{2}}{2}+t-\\log_{}\\left(t+1\\right)\\right]+C}\\\\\n{I}&={2t^{3}-3t^{2}+6t-6\\log_{}\\left(t+1\\right)+C}\t\n\\end{align*}","ts":1603108384715,"cs":"dcFPvlo2MFx3UgkrelH+9A==","size":{"width":256,"height":104}}$

Put the value of t:-

${"font":{"size":10,"color":"#000000","family":"Arial"},"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-1-0-1-1-1-1","code":"\\begin{align*}\n{I}&={2x^{\\frac{1}{2}}-3x^{\\frac{1}{3}}+6x^{\\frac{1}{6}}-6\\log_{}\\left(x^{\\frac{1}{6}}+1\\right)+C}\t\n\\end{align*}","type":"align*","ts":1603108403503,"cs":"AbmDMgOptBO1rKrpv1vkhA==","size":{"width":297,"height":28}}$

${"id":"4-0-0-0-1-1-2-1-0","font":{"family":"Arial","color":"#000000","size":12},"type":"$","code":"$6.\\,\\frac{5x}{\\left(x+1\\right)\\left(x^{2}+9\\right)}$","ts":1602757474021,"cs":"DvfJJbibAgYYZYEyCHLJug==","size":{"width":109,"height":28}}$

Solun:- Let f(x) = ${"type":"$","font":{"color":"#000000","family":"Arial","size":12},"code":"$\\frac{5x}{\\left(x+1\\right)\\left(x^{2}+9\\right)}$","id":"11-0","ts":1602757934834,"cs":"g/o7letu4hW/lUJsWL86VA==","size":{"width":89,"height":28}}$

Integrate f(x):-

${"type":"align*","code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{5x}{\\left(x+1\\right)\\left(x^{2}+9\\right)}.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-0","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1602758156037,"cs":"FG/DNtyh1EN1CQux98NC5A==","size":{"width":184,"height":36}}$

By Partial Fraction:-

${"id":"1-0-1","code":"$\\frac{5x}{\\left(x+1\\right)\\left(x^{2}+9\\right)}=\\frac{A}{x+1}+\\frac{Bx+C}{x^{2}+9}$","font":{"size":12,"family":"Arial","color":"#000000"},"type":"$","ts":1602758191312,"cs":"7UNrWmSk4tpZNkHbpmMbjQ==","size":{"width":228,"height":29}}$

⇒ 5x = A(x2 + 9) + (Bx + C)(x + 1).....(1)

Put x = -1 in eq. 1:-

⇒ -5 = A(10)

⇒ A = -1/2

Put x = 0 in eq. 1:-

⇒ 0 = A(9) + C(1)

⇒ 0 =(-1/2)(9) + C

⇒ 0 =(-9/2) + C

⇒ C = 9/2

Put x = 1 in eq. 1:-

⇒ 5 = A(10) + (B + C)(2)

⇒ 5 = (-1/2)(10) + (B + 9/2)(2)

⇒ 5 = -5 + 2B + 9

⇒ 1 = 2B

⇒ B = 1/2

${"font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","code":"\\begin{align*}\n{I}&={\\int_{}^{}\\left[\\frac{-1}{2}\\frac{1}{x+1}+\\frac{\\frac{1}{2}x+\\frac{9}{2}}{x^{2}+9}\\right].dx}\\\\\n{I}&={\\frac{-1}{2}\\int_{}^{}\\frac{1}{x+1}.dx+\\frac{1}{2}\\int_{}^{}\\frac{x}{x^{2}+9}.dx+\\frac{9}{2}\\int_{}^{}\\frac{1}{x^{2}+\\left(3\\right)^{2}}.dx}\\\\\n{I}&={\\frac{-1}{2}\\int_{}^{}\\frac{1}{x+1}.dx+\\frac{1}{4}\\int_{}^{}\\frac{2x}{x^{2}+9}.dx+\\frac{9}{2}\\int_{}^{}\\frac{1}{x^{2}+\\left(3\\right)^{2}}.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-1-0-1","ts":1602759505743,"cs":"+3JT5AwnE+Ug1GZtQCzatg==","size":{"width":425,"height":136}}$

We know that:-

${"id":"5-1-0-0","type":"$","font":{"color":"#222222","family":"Arial","size":10},"code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}x+C$","ts":1602672753650,"cs":"s+RxJtRHo5Mo+KWRlxcgCA==","size":{"width":136,"height":18}}$

${"code":"$\\int_{}^{}\\frac{f^{\\prime }\\left(x\\right)}{f\\left(x\\right)}.dx=\\log_{}\\left(f\\left(x\\right)\\right)+C$","id":"12","type":"$","font":{"size":10,"color":"#222222","family":"Arial"},"ts":1603108880295,"cs":"fHb9gZxzqxpJk7Z3/Q8sOQ==","size":{"width":188,"height":24}}$

${"font":{"family":"Arial","color":"#222222","size":10},"code":"$\\int_{}^{}\\frac{1}{x^{2}+a^{2}}.dx=\\frac{1}{a}\\tan^{-1}\\left(\\frac{x}{a}\\right)+C$","id":"5-1-1-0","type":"$","ts":1602759703882,"cs":"N6cOEyLnL9ZIZuutWfrfOw==","size":{"width":208,"height":20}}$

${"font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-0-1-0","code":"\\begin{align*}\n{I}&={-\\frac{1}{2}\\log_{}\\left|x+1\\right|+\\frac{1}{4}\\log_{}\\left|x^{2}+9\\right|+\\frac{9}{2}\\times\\frac{1}{3}\\tan^{-1}\\left(\\frac{x}{3}\\right)+C}\\\\\n{I}&={-\\frac{1}{2}\\log_{}\\left|x+1\\right|+\\frac{1}{4}\\log_{}\\left|x^{2}+9\\right|+\\frac{3}{2}\\tan^{-1}\\left(\\frac{x}{3}\\right)+C}\t\n\\end{align*}","ts":1602759872826,"cs":"78Ta4kgyZ2WBF0CsVgdM9Q==","size":{"width":412,"height":69}}$

${"code":"$7.\\,\\frac{\\sin x}{\\sin\\left(x-a\\right)}$","type":"$","id":"4-0-0-0-1-1-2-1-1-0-0","font":{"family":"Arial","color":"#000000","size":12},"ts":1602760002864,"cs":"fH1Hjahmb90gP0mykbLcpw==","size":{"width":84,"height":28}}$

Solun:- Let f(x) = ${"font":{"size":12,"color":"#000000","family":"Arial"},"code":"$\\frac{\\sin x}{\\sin\\left(x-a\\right)}$","id":"11-1-0-0","type":"$","ts":1602760032351,"cs":"lOxgA1rkY+MeDnt6ruGs4g==","size":{"width":64,"height":28}}$

Integrate f(x):-

${"font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{\\sin x}{\\sin\\left(x-a\\right)}.dx}\t\n\\end{align*}","type":"align*","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-0","ts":1602760048771,"cs":"fEZFisBwnIVFzINcZAh/UQ==","size":{"width":150,"height":36}}$

Let x - a = t

Differentiate w.r.t. to t:-

⇒ dx = dt

${"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-1-0-0","type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{\\sin \\left(t+a\\right)}{\\sin t}.dt}\\\\\n{I}&={\\int_{}^{}\\frac{\\sin t.\\cos a+\\cos t.\\sin a}{\\sin t}.dt}\\\\\n{I}&={\\int_{}^{}\\frac{\\sin t.\\cos a}{\\sin t}.dt+\\int_{}^{}\\frac{\\cos t.\\sin a}{\\sin t}.dt}\\\\\n{I}&={\\int_{}^{}\\cos a.dt+\\int_{}^{}\\cot t.\\sin a.dt}\\\\\n{I}&={\\cos a\\int_{}^{}1.dt+\\sin a\\int_{}^{}\\cot t.dt}\t\n\\end{align*}","ts":1603111256163,"cs":"l6fgVtTd97KVX8sJ9OpA4g==","size":{"width":276,"height":198}}$

We know that:-

${"font":{"family":"Arial","size":10,"color":"#222222"},"type":"$","id":"5-1-0-1-0-0-0","code":"$\\int_{}^{}\\cot x.dx=\\log_{}\\left(\\sin x\\right)+C$","ts":1602760385689,"cs":"v/YbNuhsDZmENvCkbsxQNg==","size":{"width":190,"height":17}}$

${"id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-0-1-1-0-0-0-0","code":"\\begin{align*}\n{I}&={t.\\cos a+\\sin a.\\log_{}\\left|\\sin t\\right|+C}\t\n\\end{align*}","type":"align*","font":{"color":"#000000","family":"Arial","size":10},"ts":1602760542108,"cs":"4U4cnkbeD1KjumkN4Wrq5Q==","size":{"width":222,"height":16}}$

Put the value of t:-

${"id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-0-1-1-1-0-0","font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{I}&={\\left(x-a\\right).\\cos a+\\sin a.\\log_{}\\left|\\sin \\left(x-a\\right)\\right|+C}\\\\\n{I}&={x.\\cos a-a\\cos a+\\sin a.\\log_{}\\left|\\sin \\left(x-a\\right)\\right|+C}\\\\\n{I}&={x.\\cos a+\\sin a.\\log_{}\\left|\\sin \\left(x-a\\right)\\right|+C}\t\n\\end{align*}","type":"align*","ts":1602760691760,"cs":"9mEMZ6qNw6BinsWIu71ENg==","size":{"width":332,"height":56}}$

${"code":"$8.\\,\\frac{e^{5\\log_{}x}-e^{4\\log_{}x}}{e^{3\\log_{}x}-e^{2\\log_{}x}}$","font":{"color":"#000000","family":"Arial","size":12},"type":"$","id":"4-0-0-0-1-1-2-1-1-1","ts":1603111408227,"cs":"OE3/rdZbYKfpzztHlHiKOA==","size":{"width":109,"height":29}}$

Solun:- Let f(x) = ${"type":"$","code":"$\\frac{e^{5\\log_{}x}-e^{4\\log_{}x}}{e^{3\\log_{}x}-e^{2\\log_{}x}}$","font":{"size":12,"family":"Arial","color":"#000000"},"id":"11-1-1","ts":1602761096176,"cs":"O6GNOb42J5BNYEKgemlM6g==","size":{"width":90,"height":29}}$

Integrate f(x):-

${"font":{"size":10,"color":"#000000","family":"Arial"},"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-1","type":"align*","code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{e^{5\\log_{}x}-e^{4\\log_{}x}}{e^{3\\log_{}x}-e^{2\\log_{}x}}.dx}\\\\\n{I}&={\\int_{}^{}\\frac{e^{\\log_{}x^{5}}-e^{\\log_{}x^{4}}}{e^{\\log_{}x^{3}}-e^{\\log_{}x^{2}}}.dx}\\\\\n{I}&={\\int_{}^{}\\frac{x^{5}-x^{4}}{x^{3}-x^{2}}.dx}\\\\\n{I}&={\\int_{}^{}\\frac{x^{4}\\left(x-1\\right)}{x^{2}\\left(x-1\\right)}.dx}\\\\\n{I}&={\\int_{}^{}x^{2}.dx}\t\n\\end{align*}","ts":1602761347902,"cs":"rXtQkPDSM+1egfVLNxrZqg==","size":{"width":176,"height":210}}$

We know that:-

${"id":"5-1-0-1-1","type":"$","font":{"color":"#222222","family":"Arial","size":10},"code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+C$","ts":1602761535164,"cs":"AgCv1SCkw+bRVzl5Ls78Bw==","size":{"width":136,"height":21}}$

${"type":"align*","code":"\\begin{align*}\n{I}&={\\frac{x^{3}}{3}+C}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-0-1-1-0-1","ts":1602761681417,"cs":"1u8hWqa3dTlCgHoCYxbw3w==","size":{"width":82,"height":34}}$

${"id":"4-0-0-0-1-1-2-1-1-0-1-2-0","code":"$9.\\,\\frac{\\cos x}{{\\sqrt[]{4-\\sin^{2}x}}}$","type":"$","font":{"size":12,"family":"Arial","color":"#000000"},"ts":1603111661431,"cs":"1E5Lu9yjnsqjHss/bdkWSA==","size":{"width":94,"height":28}}$

Solun:- Let f(x) = ${"id":"11-1-0-1-1-1-0-0","type":"$","code":"$\\frac{\\cos x}{{\\sqrt[]{4-\\sin^{2}x}}}$","font":{"size":12,"color":"#000000","family":"Arial"},"ts":1603111843286,"cs":"1C0QpOUOoz6LwwgA1tq7xw==","size":{"width":76,"height":28}}$

Integrate f(x):-

${"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-0-0","type":"align*","font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{\\cos x}{{\\sqrt[]{4-\\sin^{2}x}}}.dx}\t\n\\end{align*}","ts":1603111890870,"cs":"okrofyJWUEpl3T5Aa5kwvA==","size":{"width":160,"height":40}}$

Let sin x = t

Differentiate w.r.t. to t:-

⇒ (cos x).dx = dt

${"font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-1-0-1-0-0-0","code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{4-t^{2}}}}.dt}\\\\\n{I}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{\\left(2\\right)^{2}-t^{2}}}}.dt}\t\n\\end{align*}","ts":1603111973453,"cs":"Kasz4M2XrxuR/0e/Te9b8A==","size":{"width":150,"height":92}}$

We know that:-

${"type":"$","id":"5-1-0-1-0-0-1-0-0-0","code":"$\\int_{}^{}\\frac{1}{{\\sqrt[]{a^{2}-x^{2}}}}.dx=\\sin^{-1}\\left(\\frac{x}{a}\\right)+C$","font":{"color":"#222222","family":"Arial","size":10},"ts":1603112014093,"cs":"tAQvSL76P1h5Ww7vZBJmXQ==","size":{"width":202,"height":22}}$

${"font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-0-1-1-0-0-0-1-0-0-0-0","code":"\\begin{align*}\n{I}&={\\sin^{-1}\\left(\\frac{t}{2}\\right)+C}\t\n\\end{align*}","ts":1603112064169,"cs":"P5SETXk6Cgoi3bRs5pXajQ==","size":{"width":136,"height":37}}$

Put the value of t:-

${"font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{I}&={\\sin^{-1}\\left(\\frac{\\sin x}{2}\\right)+C}\t\n\\end{align*}","id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-0-1-1-0-0-0-1-0-0-0-1","type":"align*","ts":1603112099503,"cs":"v1q9TqtWBzlKqzxfzURQJA==","size":{"width":158,"height":37}}$

${"code":"$10.\\,\\frac{\\sin^{8}x-\\cos ^{8}x}{1-2\\sin^{2}x\\cos ^{2}x}$","type":"$","id":"4-0-0-0-1-1-2-1-1-0-1-0","font":{"family":"Arial","color":"#000000","size":12},"ts":1602762466322,"cs":"O4QMIZ/dMUGCr7+Rg6gSOg==","size":{"width":138,"height":30}}$

Solun:- Let f(x) = ${"type":"$","id":"11-1-0-1-0","font":{"size":12,"family":"Arial","color":"#000000"},"code":"$\\frac{\\sin^{8}x-\\cos ^{8}x}{1-2\\sin^{2}x\\cos ^{2}x}$","ts":1602762501293,"cs":"/A4MrGd5ulLUoCkmPrnqhQ==","size":{"width":108,"height":30}}$

Integrate f(x):-

${"type":"align*","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-0","code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{\\sin^{8}x-\\cos ^{8}x}{1-2\\sin^{2}x\\cos ^{2}x}.dx}\\\\\n{I}&={\\int_{}^{}\\frac{\\left(\\sin^{4}x-\\cos ^{4}x\\right)\\left(\\sin^{4}x+\\cos ^{4}x\\right)}{\\sin^{2}x+\\cos^{2}x-2\\sin^{2}x\\cos ^{2}x}.dx}\\\\\n{I}&={\\int_{}^{}\\frac{\\left(\\sin^{2}x-\\cos ^{2}x\\right)\\left(\\sin^{2}x+\\cos ^{2}x\\right)\\left(\\sin^{4}x+\\cos ^{4}x\\right)}{\\sin^{2}x+\\cos^{2}x-\\sin^{2}x\\cos ^{2}x-\\sin^{2}x\\cos ^{2}x}.dx}\\\\\n{I}&={\\int_{}^{}\\frac{\\left(\\sin^{2}x-\\cos ^{2}x\\right)\\left(\\sin^{4}x+\\cos ^{4}x\\right)}{\\sin^{2}x-\\sin^{2}x\\cos ^{2}x-\\sin^{2}x\\cos ^{2}x+\\cos^{2}x}.dx}\\\\\n{I}&={\\int_{}^{}\\frac{\\left(\\sin^{2}x-\\cos ^{2}x\\right)\\left(\\sin^{4}x+\\cos ^{4}x\\right)}{\\sin^{2}x\\left(1-\\cos ^{2}x\\right)+\\cos ^{2}x\\left(1-\\sin^{2}x\\right)}.dx}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1602762861153,"cs":"onlhHy2szxxnXmwNmhTfAA==","size":{"width":412,"height":225}}$

${"type":"align*","font":{"color":"#000000","family":"Arial","size":10},"id":"13-0-0","code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{\\left(\\sin^{2}x-\\cos ^{2}x\\right)\\left(\\sin^{4}x+\\cos ^{4}x\\right)}{\\sin^{2}x.\\sin^{2}x+\\cos ^{2}x.\\cos ^{2}x}.dx}\\\\\n{I}&={\\int_{}^{}\\frac{\\left(\\sin^{2}x-\\cos ^{2}x\\right)\\left(\\sin^{4}x+\\cos ^{4}x\\right)}{\\sin^{4}x+\\cos ^{4}x}.dx}\\\\\n{I}&={\\int_{}^{}\\left(\\sin^{2}x-\\cos ^{2}x\\right).dx}\t\n\\end{align*}","ts":1602762900718,"cs":"Y2O31Pwy/mG+Z4XmJrdXMA==","size":{"width":300,"height":128}}$

We know that:-

⇒ cos2x - sin2x = cos 2x

${"code":"\\begin{align*}\n{I}&={-\\int_{}^{}\\left(\\cos ^{2}x-\\sin^{2}x\\right).dx}\\\\\n{I}&={-\\int_{}^{}\\cos2x.dx}\t\n\\end{align*}","id":"13-1","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","ts":1602764767079,"cs":"cABlFTyisS2cQIaYCqebjA==","size":{"width":196,"height":76}}$

We know that:-

${"id":"5-1-0-1-0-1-0","type":"$","font":{"color":"#222222","family":"Arial","size":10},"code":"$\\int_{}^{}\\cos x.dx=\\sin x+C$","ts":1602764793027,"cs":"brZEzQt+Upt4/g66b9eaig==","size":{"width":154,"height":17}}$

${"font":{"family":"Arial","color":"#000000","size":10},"id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-0-1-1-0-0-1-0-0","type":"align*","code":"\\begin{align*}\n{I}&={\\frac{-\\sin2x}{2}+C}\t\n\\end{align*}","ts":1602764850611,"cs":"g1Ls+IYIG5Np+fjk1D/aeA==","size":{"width":120,"height":32}}$

${"id":"4-0-0-0-1-1-2-1-1-0-1-1","code":"$11.\\,\\frac{1}{\\cos\\left(x+a\\right)\\cos\\left(x+b\\right)}$","type":"$","font":{"family":"Arial","color":"#000000","size":12},"ts":1602840813698,"cs":"oxqWv8n1qsb3rFYrVHQLyg==","size":{"width":153,"height":28}}$

Solun:- Let f(x) = ${"code":"$\\frac{1}{\\cos\\left(x+a\\right)\\cos\\left(x+b\\right)}$","type":"$","font":{"family":"Arial","color":"#000000","size":12},"id":"11-1-0-1-1-0","ts":1602840827026,"cs":"cJIJ4N7Dvw4bJpa6e3l6PA==","size":{"width":124,"height":28}}$

Integrate f(x):-

${"type":"align*","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-0","code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{1}{\\cos\\left(x+a\\right)\\cos\\left(x+b\\right)}.dx}\\\\\n{I}&={\\frac{1}{\\sin\\left(a-b\\right)}\\int_{}^{}\\frac{\\sin\\left(a-b\\right)}{\\cos\\left(x+a\\right)\\cos\\left(x+b\\right)}.dx}\\\\\n{I}&={\\frac{1}{\\sin\\left(a-b\\right)}\\int_{}^{}\\frac{\\sin\\left(x-x+a-b\\right)}{\\cos\\left(x+a\\right)\\cos\\left(x+b\\right)}.dx}\\\\\n{I}&={\\frac{1}{\\sin\\left(a-b\\right)}\\int_{}^{}\\frac{\\sin\\left(x+a-b-x\\right)}{\\cos\\left(x+a\\right)\\cos\\left(x+b\\right)}.dx}\\\\\n{I}&={\\frac{1}{\\sin\\left(a-b\\right)}\\int_{}^{}\\frac{\\sin\\left[\\left(x+a\\right)-\\left(x+b\\right)\\right]}{\\cos\\left(x+a\\right)\\cos\\left(x+b\\right)}.dx}\\\\\n{I}&={\\frac{1}{\\sin\\left(a-b\\right)}\\int_{}^{}\\frac{\\sin\\left(x+a\\right)\\cos\\left(x+b\\right)-\\cos\\left(x+a\\right)\\sin\\left(x+b\\right)}{\\cos\\left(x+a\\right)\\cos\\left(x+b\\right)}.dx}\\\\\n{I}&={\\frac{1}{\\sin\\left(a-b\\right)}\\int_{}^{}\\left[\\tan\\left(x+a\\right)-\\tan\\left(x+b\\right)\\right].dx}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1602841708677,"cs":"F9JXqoD8hXj52bAkDnsScQ==","size":{"width":468,"height":293}}$

We know that:-

${"type":"$","font":{"size":10,"family":"Arial","color":"#222222"},"code":"$\\int_{}^{}\\tan x.dx=-\\log_{}\\left|\\cos x\\right|+C$","id":"5-1-0-1-0-1-1","ts":1602842458397,"cs":"8wPRq/ZErmejRbZcec67fg==","size":{"width":205,"height":17}}$

${"id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-0","type":"align*","code":"\\begin{align*}\n{I}&={\\frac{1}{\\sin\\left(a-b\\right)}\\left[-\\log_{}\\left|\\cos\\left(x+a\\right)\\right|+\\log_{}\\left|\\cos\\left(x+b\\right)\\right|\\right]+C}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"ts":1602842541114,"cs":"OdI7cWDlyg1+dj93nteG1A==","size":{"width":388,"height":36}}$

${"font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1","code":"\\begin{align*}\n{I}&={\\frac{1}{\\sin\\left(a-b\\right)}\\log_{}\\left|\\frac{\\cos\\left(x+b\\right)}{\\cos\\left(x+a\\right)}\\right|+C}\t\n\\end{align*}","ts":1602842587435,"cs":"8e7BvshC7y2vYvdabZPY0w==","size":{"width":250,"height":37}}$

${"font":{"color":"#000000","family":"Arial","size":12},"code":"$12.\\,\\frac{x^{3}}{{\\sqrt[]{1-x^{8}}}}$","id":"4-0-0-0-1-1-2-1-1-0-1-2-0","type":"$","ts":1602842692471,"cs":"n0bZZZqzRi4lNzDf147tFw==","size":{"width":82,"height":32}}$

Solun:- Let f(x) = ${"id":"11-1-0-1-1-1-0-1","code":"$\\frac{x^{3}}{{\\sqrt[]{1-x^{8}}}}$","font":{"color":"#000000","size":12,"family":"Arial"},"type":"$","ts":1602842716066,"cs":"uEHShgwEIyKQ42wri4IGAg==","size":{"width":53,"height":32}}$

Integrate f(x):-

${"type":"align*","font":{"family":"Arial","color":"#000000","size":10},"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-0-1","code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{x^{3}}{{\\sqrt[]{1-x^{8}}}}.dx}\\\\\n{I}&={\\int_{}^{}\\frac{x^{3}}{{\\sqrt[]{1-\\left(x^{4}\\right)^{2}}}}.dx}\t\n\\end{align*}","ts":1602842770807,"cs":"aT/iRfeWqh4TuJIJiR4eXg==","size":{"width":156,"height":97}}$

Let x4 = t

Differentiate w.r.t. to t:-

⇒ 4x3.dx = dt

⇒ x3.dx = (1/4).dt

${"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-1-0-1-0-0-1","font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{1-t^{2}}}}.\\frac{dt}{4}}\\\\\n{I}&={\\frac{1}{4}\\int_{}^{}\\frac{1}{{\\sqrt[]{1-t^{2}}}}.dt}\t\n\\end{align*}","type":"align*","ts":1602842973625,"cs":"2fUga7K5ZufRyVtLktd8KA==","size":{"width":146,"height":80}}$

We know that:-

${"font":{"color":"#222222","size":10,"family":"Arial"},"type":"$","code":"$\\int_{}^{}\\frac{1}{{\\sqrt[]{1-x^{2}}}}.dx=\\sin^{-1}x+C$","id":"5-1-0-1-0-0-1-0-0-1","ts":1602843023149,"cs":"RRF5tOlHdbM5ni4iBcllcw==","size":{"width":178,"height":22}}$

${"code":"\\begin{align*}\n{I}&={\\frac{1}{4}\\sin^{-1}\\left(t\\right)+C}\t\n\\end{align*}","id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-0-1-1-0-0-0-1-0-0-1","type":"align*","font":{"color":"#000000","family":"Arial","size":10},"ts":1602843098380,"cs":"OrBUgSRI8DxHSmCbHqbZvA==","size":{"width":130,"height":32}}$

Put the value of t:-

${"code":"\\begin{align*}\n{I}&={\\frac{1}{4}\\sin^{-1}\\left(x^{4}\\right)+C}\t\n\\end{align*}","id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-0-1-1-1-0-1-0-0-1","type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1602843147319,"cs":"oTQ9ppTebkgCx1yXHBk0fQ==","size":{"width":142,"height":32}}$

${"id":"4-0-0-0-1-1-2-1-1-0-1-2-1-0","code":"$13.\\,\\frac{e^{x}}{\\left(1+e^{x}\\right)\\left(2+e^{x}\\right)}$","font":{"family":"Arial","color":"#000000","size":12},"type":"$","ts":1602843206642,"cs":"pO/y59K/P0qXTYZ5VWmzCg==","size":{"width":124,"height":28}}$

Solun:- Let f(x) = ${"font":{"color":"#000000","size":12,"family":"Arial"},"code":"$\\frac{e^{x}}{\\left(1+e^{x}\\right)\\left(2+e^{x}\\right)}$","id":"11-1-0-1-1-1-1-0","type":"$","ts":1602843218197,"cs":"0GGGXmz628G5bG7wex0R0w==","size":{"width":94,"height":28}}$

Integrate f(x):-

${"font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-0","code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{e^{x}}{\\left(1+e^{x}\\right)\\left(2+e^{x}\\right)}.dx}\t\n\\end{align*}","ts":1602843241174,"cs":"T9kTh8NGl6VNtLOQT2mV0A==","size":{"width":188,"height":36}}$

Let ex = t

Differentiate w.r.t. to t:-

⇒ ex.dx = dt

${"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-1-0-1-1","type":"align*","code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{1}{\\left(1+t\\right)\\left(2+t\\right)}.dt}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"ts":1602843317935,"cs":"ZShyQdyWPVuBhoDaNfUYGQ==","size":{"width":166,"height":36}}$

By Partial Fraction:-

${"font":{"family":"Arial","color":"#000000","size":12},"type":"$","id":"1-0-0-1-0","code":"$\\frac{1}{\\left(1+t\\right)\\left(2+t\\right)}=\\frac{A}{\\left(1+t\\right)}+\\frac{B}{\\left(2+t\\right)}$","ts":1602843492608,"cs":"S3asYfZcfPs2GKrgssWpqg==","size":{"width":217,"height":28}}$

⇒ 1 = A(2 + t) + B(1 + t)....(1)

Put t = - 1 in eq. 1:-

⇒ 1 = A(1)

⇒ A = 1

Put t = - 2 in eq. 1:-

⇒ 1 = B(-1)

⇒ B = - 1

${"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-1-0-0-1-0","code":"\\begin{align*}\n{I}&={\\int_{}^{}\\left[\\frac{1}{1+t}-\\frac{1}{2+t}\\right].dx}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","ts":1602843767963,"cs":"gxWZBI1FJU25VKIKcEjs2Q==","size":{"width":185,"height":37}}$

We know that:-

${"type":"$","id":"5-1-0-1-0-0-1-1-0","code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}x+C$","font":{"color":"#222222","family":"Arial","size":10},"ts":1602843795367,"cs":"cccpw+bfex4Xgmv1GX6i3g==","size":{"width":136,"height":18}}$

${"id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-0-1-1-0-0-0-1-1-0","code":"\\begin{align*}\n{I}&={\\log_{}\\left(1+t\\right)-\\log_{}\\left(2+t\\right)+C}\t\n\\end{align*}","type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1602843832218,"cs":"akuhIGaXfH5PxTrMSSmZNQ==","size":{"width":217,"height":16}}$

Put the value of t:-

${"id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-0-1-1-1-0-1-1-0","code":"\\begin{align*}\n{I}&={\\log_{}\\left(1+e^{x}\\right)-\\log_{}\\left(2+e^{x}\\right)+C}\\\\\n{I}&={\\log_{}\\left|\\frac{1+e^{x}}{2+e^{x}}\\right|+C}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","ts":1602843892469,"cs":"OfiGSDMaIi3Sv7Fzb6wmNg==","size":{"width":236,"height":54}}$

${"code":"$14.\\,\\frac{1}{\\left(x^{2}+1\\right)\\left(x^{2}+4\\right)}$","type":"$","id":"4-0-0-0-1-1-2-1-1-0-1-2-1-1-0","font":{"family":"Arial","size":12,"color":"#000000"},"ts":1602844002415,"cs":"06Lb+CXW9moghwPYCSMJ5Q==","size":{"width":125,"height":28}}$

Solun:- Let f(x) = ${"code":"$\\frac{1}{\\left(x^{2}+1\\right)\\left(x^{2}+4\\right)}$","font":{"family":"Arial","size":12,"color":"#000000"},"id":"11-1-0-1-1-1-1-1-0","type":"$","ts":1602844021548,"cs":"NOZcjNCTK4rUdvSRglPJJw==","size":{"width":96,"height":28}}$

Integrate f(x):-

${"font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{1}{\\left(x^{2}+1\\right)\\left(x^{2}+4\\right)}.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-0","type":"align*","ts":1602844041109,"cs":"ByzJ2GWSBUS6nfhDxj2waQ==","size":{"width":189,"height":36}}$

Put x2 = t

${"code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{1}{\\left(t+1\\right)\\left(t+4\\right)}.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-1","font":{"color":"#000000","family":"Arial","size":10},"type":"align*","ts":1602844090372,"cs":"5cUHULpl9IwHUTYj6xKq0w==","size":{"width":170,"height":36}}$

By Partial Fraction:-

${"code":"$\\frac{1}{\\left(t+1\\right)\\left(t+4\\right)}=\\frac{A}{\\left(t+1\\right)}+\\frac{B}{\\left(t+4\\right)}$","id":"1-0-0-1-1","font":{"color":"#000000","size":12,"family":"Arial"},"type":"$","ts":1602844189703,"cs":"Mjp+HcGWkZm5/RK3Ui2YDw==","size":{"width":217,"height":28}}$

⇒ 1 = A(t + 4) + B(t + 1)....(1)

Put t = - 1 in eq. 1:-

⇒ 1 = A(3)

⇒ A = 1/3

Put t = - 4 in eq. 1:-

⇒ 1 = B(-3)

⇒ B = - 1/3

${"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-1-0-0-1-1-0","font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{I}&={\\int_{}^{}\\left[\\frac{1}{3}\\frac{1}{t+1}-\\frac{1}{3}\\frac{1}{t+4}\\right].dx}\t\n\\end{align*}","type":"align*","ts":1602844464826,"cs":"LVANDc3U9PfXTghH4kiXdg==","size":{"width":216,"height":37}}$

${"type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{I}&={\\int_{}^{}\\left[\\frac{1}{3}\\frac{1}{x^{2}+1}-\\frac{1}{3}\\frac{1}{x^{2}+4}\\right].dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-1-0-0-1-1-1","ts":1602844704576,"cs":"PtqoiUdqleKtCpDe+6IQCQ==","size":{"width":236,"height":37}}$

We know that:-

${"code":"$\\int_{}^{}\\frac{1}{x^{2}+a^{2}}.dx=\\frac{1}{a}\\tan^{-1}\\left(\\frac{x}{a}\\right)+C$","type":"$","id":"5-1-0-1-0-0-1-1-1","font":{"family":"Arial","color":"#222222","size":10},"ts":1602844789588,"cs":"j5VJ4eSNm7tWtaL20QM43Q==","size":{"width":208,"height":20}}$

${"code":"\\begin{align*}\n{I}&={\\frac{1}{3}\\left[\\frac{1}{1}\\tan^{-1}\\left(\\frac{x}{1}\\right)-\\frac{1}{2}\\tan^{-1}\\left(\\frac{x}{2}\\right)\\right]+C}\\\\\n{I}&={\\frac{1}{3}\\left[\\tan^{-1}x-\\frac{1}{2}\\tan^{-1}\\left(\\frac{x}{2}\\right)\\right]+C}\\\\\n{I}&={\\frac{1}{3}\\tan^{-1}x-\\frac{1}{6}\\tan^{-1}\\left(\\frac{x}{2}\\right)+C}\t\n\\end{align*}","type":"align*","id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-0-1-1-0-0-0-1-1-1","font":{"color":"#000000","family":"Arial","size":10},"ts":1602844964368,"cs":"HTvMiV8GGMzTs+haN1aHBw==","size":{"width":292,"height":117}}$

${"id":"4-0-0-0-1-1-2-1-1-0-1-2-1-1-1-0","code":"$15.\\,\\cos^{3}x.e^{\\log_{}\\sin x}$","type":"$","font":{"size":11,"family":"Arial","color":"#000000"},"ts":1603197116356,"cs":"w4b8deQeeT6vwv6JE6G8Bw==","size":{"width":136,"height":16}}$

Solun:- Let f(x) = cos3x.elog sinx

Integrate f(x):-

${"font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{I}&={\\int_{}^{}\\cos^{3}x.e^{\\log_{}\\sin x}.dx}\\\\\n{I}&={\\int_{}^{}\\cos^{3}x.\\sin x.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-0","type":"align*","ts":1602845301992,"cs":"kPU+a8YVAJHq8dPDLmS8vw==","size":{"width":165,"height":76}}$

Let cos x = t

Differentiate w.r.t. to t:-

⇒ - sin x.dx = dt

⇒ sin x.dx = - dt

${"type":"align*","code":"\\begin{align*}\n{I}&={-\\int_{}^{}t^{3}.dt}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-1-0-1-0-1-0","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1602845585580,"cs":"ERca/XHP25uoV21YoK7z2A==","size":{"width":96,"height":36}}$

We know that:-

${"type":"$","id":"5-1-0-1-0-0-1-0-1-0","code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+C$","font":{"color":"#222222","size":10,"family":"Arial"},"ts":1602845619687,"cs":"5UawUj2WmhiUBpbG8BlQ3Q==","size":{"width":136,"height":21}}$

${"id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-0-1-1-0-0-0-1-0-1-0","type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{I}&={\\frac{-t^{4}}{4}+C}\t\n\\end{align*}","ts":1602845665899,"cs":"Ego2NSAn6PXWaBxJOoMmlg==","size":{"width":92,"height":34}}$

Put the value of t:-

${"id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-0-1-1-1-0-1-0-1-0","type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{I}&={\\frac{-1}{4}\\cos^{4}x+C}\t\n\\end{align*}","ts":1602845694174,"cs":"vY85BopDtWY/OMCb+bwBxA==","size":{"width":126,"height":32}}$

${"font":{"size":10,"color":"#000000","family":"Arial"},"type":"$","code":"$16.\\,e^{3\\log_{}x}.\\left(x^{4}+1\\right)^{-1}$","id":"4-0-0-0-1-1-2-1-1-0-1-2-1-1-1-1-0","ts":1602845806178,"cs":"wj6HeTyhmI2IzS6rSmNVlA==","size":{"width":140,"height":21}}$

Solun:- Let f(x) = ${"type":"$","code":"$e^{3\\log_{}x}.\\left(x^{4}+1\\right)^{-1}$","id":"14-0","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1602845880952,"cs":"885mlEybe3g5cBXk0+YleQ==","size":{"width":117,"height":21}}$

Integrate f(x):-

${"type":"align*","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-0","code":"\\begin{align*}\n{I}&={\\int_{}^{}e^{3\\log_{}x}.\\left(x^{4}+1\\right)^{-1}.dx}\\\\\n{I}&={\\int_{}^{}\\frac{e^{\\log_{}x^{3}}}{\\left(x^{4}+1\\right)}.dx}\\\\\n{I}&={\\int_{}^{}\\frac{x^{3}}{\\left(x^{4}+1\\right)}.dx}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"ts":1602845942981,"cs":"c1icl8Ce35uEfSX1mi0Zcg==","size":{"width":190,"height":125}}$

Let x4 + 1 = t

Differentiate w.r.t. to t:-

⇒ 4x3.dx = dt

⇒ x3.dx = (1/4).dt

${"font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{1}{t}.\\frac{dt}{4}}\\\\\n{I}&={\\frac{1}{4}\\int_{}^{}\\frac{1}{t}.dt}\t\n\\end{align*}","type":"align*","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-1-0-1-0-1-1-0","ts":1602846165506,"cs":"HJJM+FdvK6+WvprLNY9KMA==","size":{"width":101,"height":76}}$

We know that:-

${"font":{"size":10,"family":"Arial","color":"#222222"},"code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}x+C$","type":"$","id":"5-1-0-1-0-0-1-0-1-1-0","ts":1602846192268,"cs":"zGLNqIDW/uvoX3lUVbovhg==","size":{"width":136,"height":18}}$

${"type":"align*","code":"\\begin{align*}\n{I}&={\\frac{1}{4}\\log_{}t+C}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-0-1-1-0-0-0-1-0-1-1-0","ts":1602846218678,"cs":"IE9WTHuZ82w2/k6ktoWpxw==","size":{"width":104,"height":32}}$

Put the value of t:-

${"font":{"family":"Arial","color":"#000000","size":10},"type":"align*","id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-0-1-1-1-0-1-0-1-1-0","code":"\\begin{align*}\n{I}&={\\frac{1}{4}\\log_{}\\left(x^{4}+1\\right)+C}\t\n\\end{align*}","ts":1602846257638,"cs":"Bvx+uAAc0yyEAgNYtScy7Q==","size":{"width":156,"height":32}}$

${"font":{"color":"#000000","size":10,"family":"Arial"},"code":"$17.\\,f^{\\prime }\\left(ax+b\\right)\\left[f\\left(ax+b\\right)\\right]^{n}$","id":"4-0-0-0-1-1-2-1-1-0-1-2-1-1-1-1-1-0","type":"$","ts":1602846354911,"cs":"ARXAy/8devHeFAmX/WVPGg==","size":{"width":172,"height":16}}$

Solun:- Let f(x) = ${"code":"$f^{\\prime }\\left(ax+b\\right)\\left[f\\left(ax+b\\right)\\right]^{n}$","type":"$","font":{"size":10,"color":"#000000","family":"Arial"},"id":"14-1-0","ts":1602846376172,"cs":"jr6oYcyH+FMG9Q5mdnzoNg==","size":{"width":149,"height":16}}$

Integrate f(x):-

${"code":"\\begin{align*}\n{I}&={\\int_{}^{}f^{\\prime }\\left(ax+b\\right)\\left[f\\left(ax+b\\right)\\right]^{n}.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-0","font":{"color":"#000000","family":"Arial","size":10},"type":"align*","ts":1602846412492,"cs":"IlxRZQ8gJp3or/YAhPjZPw==","size":{"width":222,"height":36}}$

Let f(ax + b) = t

Differentiate w.r.t. to t:-

⇒ a.f’(ax + b).dx = dt

⇒ f’(ax + b).dx = (1/a).dt

${"code":"\\begin{align*}\n{I}&={\\frac{1}{a}\\int_{}^{}t^{n}.dt}\t\n\\end{align*}","type":"align*","font":{"color":"#000000","family":"Arial","size":10},"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-1-0-1-0-1-1-1-0","ts":1602846674805,"cs":"2eNoJV7aYMr7b1JKmqsLbA==","size":{"width":100,"height":36}}$

We know that:-

${"code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+C$","font":{"color":"#222222","size":10,"family":"Arial"},"type":"$","id":"5-1-0-1-0-0-1-0-1-1-1-0","ts":1602846540610,"cs":"ub8YvMTF2xi7aCZUloJv8g==","size":{"width":136,"height":21}}$

${"font":{"color":"#000000","size":10,"family":"Arial"},"id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-0-1-1-0-0-0-1-0-1-1-1-0","type":"align*","code":"\\begin{align*}\n{I}&={\\frac{1}{a}.\\frac{t^{n+1}}{n+1}+C}\t\n\\end{align*}","ts":1602846701317,"cs":"PPPAvtUiAZXnB2LzDZgkoQ==","size":{"width":126,"height":36}}$

Put the value of t:-

${"code":"\\begin{align*}\n{I}&={\\frac{\\left[f\\left(ax+b\\right)\\right]^{n+1}}{a\\left(n+1\\right)}+C}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#000000"},"id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-0-1-1-1-0-1-0-1-1-1-0","type":"align*","ts":1602846809054,"cs":"/o/e65mDgbI4hzMSaWbjug==","size":{"width":162,"height":40}}$

${"font":{"color":"#000000","family":"Arial","size":10},"code":"$18.\\,\\frac{1}{{\\sqrt[]{\\sin^{3}x.\\sin\\left(x+\\alpha\\right)}}}$","id":"4-0-0-0-1-1-2-1-1-0-1-2-1-1-1-1-1-1-0","type":"$","ts":1602846984708,"cs":"bUs9nsT3bEtxnKh1gw/RvQ==","size":{"width":116,"height":32}}$

Solun:- Let f(x) = ${"code":"$\\frac{1}{{\\sqrt[]{\\sin^{3}x.\\sin\\left(x+\\alpha\\right)}}}$","id":"14-1-1-0","type":"$","font":{"color":"#000000","family":"Arial","size":10},"ts":1602847022143,"cs":"gQwauyvezPs3gR00d+vH7A==","size":{"width":92,"height":32}}$

Integrate f(x):-

${"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-1-0","type":"align*","code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{\\sin^{3}x.\\sin\\left(x+\\alpha\\right)}}}.dx}\\\\\n{I}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{\\sin^{3}x.\\left[\\sin x.\\cos\\alpha+\\cos x.\\sin\\alpha\\right]}}}.dx}\\\\\n{I}&={\\int_{}^{}\\frac{1}{{\\sqrt[]{\\sin^{4}x.\\cos\\alpha+\\sin^{3}x.\\cos x.\\sin\\alpha}}}.dx}\\\\\n{I}&={\\int_{}^{}\\frac{1}{\\sin^{2}x{\\sqrt[]{\\cos\\alpha+\\frac{\\cos x}{\\sin x}.\\sin\\alpha}}}.dx}\\\\\n{I}&={\\int_{}^{}\\frac{\\cos ec^{2}x}{{\\sqrt[]{\\cos\\alpha+\\cot x.\\sin\\alpha}}}.dx}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1602847240513,"cs":"PjMdQJhnhB1qpFLJNtXF2w==","size":{"width":312,"height":241}}$

Let cos α + cot x.sin α = t

Differentiate w.r.t. to t:-

⇒ - sin α.cosec2x.dx = dt

⇒ cosec2x.dx = (-1/sin α).dt

${"code":"\\begin{align*}\n{I}&={\\frac{-1}{\\sin\\alpha}\\int_{}^{}\\frac{dt}{{\\sqrt[]{t}}}}\\\\\n{I}&={\\frac{-1}{\\sin\\alpha}\\int_{}^{}t^{\\frac{-1}{2}}.dt}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-1-0-1-0-1-1-1-1-0","type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1602848013326,"cs":"GtmAgfgDaCG1scMpZ4ZfAQ==","size":{"width":132,"height":77}}$

We know that:-

${"code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+C$","font":{"color":"#222222","size":10,"family":"Arial"},"type":"$","id":"5-1-0-1-0-0-1-0-1-1-1-1-0","ts":1602846540610,"cs":"jGhoObNYoPlL9fqdoz2+jQ==","size":{"width":136,"height":21}}$

${"font":{"family":"Arial","color":"#000000","size":10},"type":"align*","code":"\\begin{align*}\n{I}&={\\frac{-1}{\\sin\\alpha}\\times\\frac{t^{\\frac{1}{2}}}{\\frac{1}{2}}+C}\\\\\n{I}&={\\frac{-2t^{\\frac{1}{2}}}{\\sin\\alpha}+C}\\\\\n{I}&={\\frac{-2{\\sqrt[]{t}}}{\\sin\\alpha}+C}\t\n\\end{align*}","id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-0-1-1-0-0-0-1-0-1-1-1-1-0-0","ts":1602848083799,"cs":"7CzeYrvGD3z3axcxIUuKmQ==","size":{"width":140,"height":128}}$

Put the value of t:-

${"id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-0-1-1-1-0-1-0-1-1-1-1-0","type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{I}&={\\frac{-2}{\\sin\\alpha}{\\sqrt[]{\\cos\\alpha+\\cot x.\\sin\\alpha}}+C}\\\\\n{I}&={\\frac{-2}{\\sin\\alpha}{\\sqrt[]{\\cos\\alpha+\\frac{\\cos x}{\\sin x}.\\sin\\alpha}}+C}\\\\\n{I}&={\\frac{-2}{\\sin\\alpha}{\\sqrt[]{\\frac{\\sin x.\\cos\\alpha+\\cos x.\\sin\\alpha}{\\sin x}}}+C}\\\\\n{I}&={\\frac{-2}{\\sin\\alpha}{\\sqrt[]{\\frac{\\sin\\left(x+\\alpha\\right)}{\\sin x}}}+C}\t\n\\end{align*}","ts":1602848161184,"cs":"Q5hxn0OBEql3d0CJT5q8hw==","size":{"width":285,"height":162}}$

${"font":{"size":11,"family":"Arial","color":"#000000"},"code":"$19.\\,\\frac{\\sin^{-1}{\\sqrt[]{x}}-\\cos^{-1}{\\sqrt[]{x}}}{\\sin^{-1}{\\sqrt[]{x}}+\\cos^{-1}{\\sqrt[]{x}}}$","id":"4-0-0-0-1-1-2-1-1-0-1-2-1-1-1-1-1-1-1-0","type":"$","ts":1602848281153,"cs":"KOPCSB7BSYz4pvP2+FYoCQ==","size":{"width":141,"height":30}}$

Solun:- Let f(x) = ${"type":"$","font":{"color":"#000000","family":"Arial","size":10},"id":"14-1-1-1-0","code":"$\\frac{\\sin^{-1}{\\sqrt[]{x}}-\\cos^{-1}{\\sqrt[]{x}}}{\\sin^{-1}{\\sqrt[]{x}}+\\cos^{-1}{\\sqrt[]{x}}}$","ts":1602848297282,"cs":"GrTEpHTw58U9e2H2zl/hlg==","size":{"width":100,"height":26}}$

Integrate f(x):-

${"code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{\\sin^{-1}{\\sqrt[]{x}}-\\cos^{-1}{\\sqrt[]{x}}}{\\sin^{-1}{\\sqrt[]{x}}+\\cos^{-1}{\\sqrt[]{x}}}.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-1-1-0-0","font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","ts":1602848318455,"cs":"lIrI6M2gebQC4cPcjQvvGQ==","size":{"width":220,"height":40}}$

We know that:-

sin-1x + cos-1x = π/2

${"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-1-1-1-0","code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{\\left(\\sin^{-1}{\\sqrt[]{x}}-\\cos^{-1}{\\sqrt[]{x}}\\right)}{\\frac{\\Pi}{2}}.dx}\\\\\n{I}&={\\frac{2}{\\Pi}\\int_{}^{}\\left[\\sin^{-1}{\\sqrt[]{x}}-\\left(\\frac{\\Pi}{2}-\\sin^{-1}{\\sqrt[]{x}}\\right)\\right].dx}\\\\\n{I}&={\\frac{2}{\\Pi}\\int_{}^{}\\left[2\\sin^{-1}{\\sqrt[]{x}}-\\frac{\\Pi}{2}\\right].dx}\\\\\n{I}&={\\int_{}^{}\\left[\\frac{4}{\\Pi}\\sin^{-1}{\\sqrt[]{x}}-1\\right].dx}\\\\\n{I}&={\\frac{4}{\\Pi}\\int_{}^{}\\sin^{-1}{\\sqrt[]{x}}.dx-\\int_{}^{}1.dx}\t\n\\end{align*}","type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1602850033414,"cs":"7rENYBpM5tUF7SYT38M7XQ==","size":{"width":308,"height":213}}$

Let sin-1√x = t

⇒ √x = sin t

⇒ x = sin2 t

Differentiate w.r.t. to t:-

⇒ dx = 2sint.cost.dt

⇒ dx = sin 2t.dt

${"type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-1-0-1-0-1-1-1-1-1","code":"\\begin{align*}\n{I}&={\\frac{4}{\\Pi}\\int_{}^{}t.\\sin 2t.dt-\\int_{}^{}1.dx}\t\n\\end{align*}","ts":1602850135763,"cs":"KlUCpabkt+J7RzBDtY0Y7Q==","size":{"width":208,"height":36}}$

We know that:-

${"id":"5-1-0-1-0-0-1-0-1-1-1-1-1-0","type":"$","code":"$\\int_{}^{}\\left(u.v\\right).dx=u\\int_{}^{}v.dx-\\int_{}^{}\\left(\\diff{u}{x}\\int_{}^{}v.dx\\right).dx+C$","font":{"color":"#222222","size":10,"family":"Arial"},"ts":1602850217892,"cs":"5hn9765no0qGiXd10GqUeg==","size":{"width":320,"height":20}}$

${"code":"\\begin{align*}\n{I}&={\\frac{4}{\\Pi}\\left[t\\int_{}^{}\\sin 2t.dt-\\int_{}^{}\\left(\\diff{t}{t}\\int_{}^{}\\sin 2t.dt\\right).dt\\right]-x+C}\\\\\n{I}&={\\frac{4}{\\Pi}\\left[-\\frac{t\\cos 2t}{2}-\\frac{1}{2}\\int_{}^{}-\\cos 2t.dt\\right]-x+C}\\\\\n{I}&={\\frac{4}{\\Pi}\\left[-\\frac{t\\cos 2t}{2}+\\frac{1}{2}\\int_{}^{}\\cos 2t.dt\\right]-x+C}\\\\\n{I}&={\\frac{4}{\\Pi}\\left[-\\frac{t\\cos 2t}{2}+\\frac{1}{4}\\sin 2t\\right]-x+C}\t\n\\end{align*}","type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-0-1-1-0-0-0-1-0-1-1-1-1-0-1-0","ts":1602852877370,"cs":"DdO94n/lW+9GLKCLZy3Dgg==","size":{"width":389,"height":165}}$

Put the value of t:-

⇒ cos t = √(1-x)

⇒ sin 2t = 2√(x-x2)

⇒ cos 2t = 1 - 2x

${"id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-0-1-1-1-0-1-0-1-1-1-1-1-0","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","code":"\\begin{align*}\n{I}&={\\frac{4}{\\Pi}\\left[\\frac{-\\left(1-2x\\right)}{2}.\\sin^{-1}{\\sqrt[]{x}}+\\frac{1}{4}\\times2{\\sqrt[]{x-x^{2}}}\\right]-x+C}\\\\\n{I}&={\\frac{2\\left(2x-1\\right)}{\\Pi}.\\sin^{-1}{\\sqrt[]{x}}+\\frac{2{\\sqrt[]{x-x^{2}}}}{\\Pi}-x+C}\t\n\\end{align*}","ts":1602912014694,"cs":"4zU8NFPSYjprdGbUZ7mLAw==","size":{"width":388,"height":80}}$

${"type":"$","id":"4-0-0-0-1-1-2-1-1-0-1-2-1-1-1-1-1-1-1-1-0","font":{"size":11,"color":"#000000","family":"Arial"},"code":"$20.\\,{\\sqrt[]{\\frac{1-{\\sqrt[]{x}}}{1+{\\sqrt[]{x}}}}}$","ts":1602912188980,"cs":"dzL/Elu8VWaLBaFlnlIX4A==","size":{"width":86,"height":44}}$

Solun:- Let f(x) = ${"id":"14-1-1-1-1-0","font":{"size":10,"family":"Arial","color":"#000000"},"type":"$","code":"${\\sqrt[]{\\frac{1-{\\sqrt[]{x}}}{1+{\\sqrt[]{x}}}}}$","ts":1602913400528,"cs":"tNS2WNzw7FeDurkmNWO+Zg==","size":{"width":52,"height":38}}$

Integrate f(x):-

${"code":"\\begin{align*}\n{I}&={\\int_{}^{}{\\sqrt[]{\\frac{1-{\\sqrt[]{x}}}{1+{\\sqrt[]{x}}}}}.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-1-1-0-1-0-0","font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","ts":1602914336106,"cs":"D6oShWRSymHGPCQggh7rQw==","size":{"width":145,"height":48}}$

Let x = cos2y

Differentiate w.r.t. to y:-

⇒ dx = 2cosy.(-siny).dy

⇒ dx = - sin 2y.dy

${"font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-1-1-0-1-1-0","code":"\\begin{align*}\n{I}&={-\\int_{}^{}{\\sqrt[]{\\frac{1-\\cos y}{1+\\cos y}}}.\\sin2y.dy}\t\n\\end{align*}","ts":1602914798172,"cs":"oPgOV4lTHuvXS4SBZ+H6UQ==","size":{"width":212,"height":48}}$

We know that:-

⇒ cos 2y = 2cos2y - 1

⇒ cos 2y = 1 - 2sin2y

${"type":"align*","code":"\\begin{align*}\n{I}&={-\\int_{}^{}{\\sqrt[]{\\frac{2\\sin^{2}\\frac{y}{2}}{2\\cos^{2}\\frac{y}{2}}}}.\\sin2y.dy}\\\\\n{I}&={-\\int_{}^{}\\tan \\frac{y}{2}.\\sin2y.dy}\\\\\n{I}&={-\\int_{}^{}\\frac{\\sin\\frac{y}{2}}{\\cos\\frac{y}{2}}.\\left(2\\sin y.\\cos y\\right).dy}\\\\\n{I}&={-2\\int_{}^{}\\frac{\\sin\\frac{y}{2}}{\\cos\\frac{y}{2}}.\\left(\\sin y\\right).\\cos y.dy}\\\\\n{I}&={-2\\int_{}^{}\\frac{\\sin\\frac{y}{2}}{\\cos\\frac{y}{2}}.\\left(2\\sin \\frac{y}{2}\\cos\\frac{y}{2}\\right).\\cos y.dy}\\\\\n{I}&={-2\\int_{}^{}\\sin\\frac{y}{2}.\\left(2\\sin \\frac{y}{2}\\right).\\cos y.dy}\\\\\n{I}&={-4\\int_{}^{}\\sin^{2}\\frac{y}{2}.\\cos y.dy}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-1-1-0-1-2","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1602916982394,"cs":"jGpzbSNl1Km7aerj26xCyg==","size":{"width":288,"height":312}}$

${"type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{I}&={-4\\int_{}^{}\\frac{\\left(1-\\cos y\\right)}{2}.\\cos y.dy}\\\\\n{I}&={-2\\int_{}^{}\\left(\\cos y-\\cos ^{2}y\\right).dy}\\\\\n{I}&={-2\\int_{}^{}\\cos y+2\\int_{}^{}\\cos ^{2}y.dy}\\\\\n{I}&={-2\\int_{}^{}\\cos y+2\\int_{}^{}\\frac{1+\\cos2y}{2}.dy}\\\\\n{I}&={-2\\int_{}^{}\\cos y+\\int_{}^{}1+\\cos2y.dy}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-1-1-1-1","ts":1602917021908,"cs":"Jm/PhNsSdnC2Ufm+sDksKQ==","size":{"width":248,"height":198}}$

We know that:-

${"font":{"size":10,"color":"#222222","family":"Arial"},"code":"$\\int_{}^{}\\cos x.dx=\\sin x+C$","id":"5-1-0-1-0-0-1-0-1-1-1-1-1-1-0-0","type":"$","ts":1603184692816,"cs":"QLvusRFKzmuQ+OnWsAAJdQ==","size":{"width":154,"height":17}}$

${"type":"align*","id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-0-1-1-0-0-0-1-0-1-1-1-1-0-1-1-0-0","code":"\\begin{align*}\n{I}&={-2\\sin y+y+\\frac{\\sin2y}{2}+C}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1602917208927,"cs":"eTupryBLLh2uuZHEMZnrtQ==","size":{"width":204,"height":32}}$

Put the value of y:-

⇒ cos y = √x

⇒ sin y = √(1-x)

⇒ sin 2y = 2√(x - x2)

${"type":"align*","font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{I}&={-2{\\sqrt[]{1-x}}+\\cos^{-1}{\\sqrt[]{x}}+\\frac{1}{2}\\times2{\\sqrt[]{x-x^{2}}}+C}\\\\\n{I}&={-2{\\sqrt[]{1-x}}+\\cos^{-1}{\\sqrt[]{x}}+{\\sqrt[]{x-x^{2}}}+C}\t\n\\end{align*}","id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-0-1-1-1-0-1-0-1-1-1-1-1-1-0","ts":1602917656553,"cs":"4k0Nx2joTO6b2/D4g2Njwg==","size":{"width":332,"height":57}}$

${"font":{"family":"Arial","size":11,"color":"#000000"},"code":"$21.\\,\\frac{2+\\sin2x}{1+\\cos2x}.e^{x}$","id":"4-0-0-0-1-1-2-1-1-0-1-2-1-1-1-1-1-1-1-1-1-0","type":"$","ts":1602918483162,"cs":"e+CDnGnLhsyS9VzCeb9cRg==","size":{"width":108,"height":22}}$

Solun:- Let f(x) = ${"font":{"size":10,"color":"#000000","family":"Arial"},"id":"14-1-1-1-1-1-0","code":"$\\frac{2+\\sin2x}{1+\\cos2x}.e^{x}$","type":"$","ts":1602918515289,"cs":"LT/nYc3aJZsiP1CrC86fUg==","size":{"width":72,"height":20}}$

Integrate f(x):-

${"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-1-1-0-1-0-1-0","font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{2+2\\sin x.\\cos x}{2\\cos^{2}x}.e^{x}.dx}\\\\\n{I}&={\\int_{}^{}\\left(\\sec^{2}x+\\tan x\\right).e^{x}.dx}\t\n\\end{align*}","type":"align*","ts":1602918724529,"cs":"3pP9ClU575iKxgOpiHCblQ==","size":{"width":209,"height":76}}$

We know that:-

${"font":{"family":"Arial","color":"#222222","size":10},"type":"$","code":"$\\int_{}^{}e^{x}\\left(f\\left(x\\right)+f^{\\prime }\\left(x\\right)\\right).dx=e^{x}.f\\left(x\\right)+C$","id":"5-1-0-1-0-0-1-0-1-1-1-1-1-1-1","ts":1602918768734,"cs":"YfrUNFy+X4VD+ICJd1BrrA==","size":{"width":253,"height":17}}$

Let f(x) = tan x

⇒ f’(x) = sec2x

${"code":"\\begin{align*}\n{I}&={e^{x}.\\tan x+C}\t\n\\end{align*}","id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-0-1-1-0-0-0-1-0-1-1-1-1-0-1-1-1","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","ts":1602918832221,"cs":"qur3M80zUbcHLC6CcTJMbA==","size":{"width":116,"height":16}}$

${"font":{"color":"#000000","family":"Arial","size":11},"code":"$22.\\,\\frac{x^{2}+x+1}{\\left(x+1\\right)^{2}\\left(x+2\\right)}$","type":"$","id":"4-0-0-0-1-1-2-1-1-0-1-2-1-1-1-1-1-1-1-1-1-1-0","ts":1602918974775,"cs":"vBftlqiyueSUDpzKZ/bPuQ==","size":{"width":106,"height":29}}$

Solun:- Let f(x) = ${"id":"14-1-1-1-1-1-1-0","font":{"size":10,"color":"#000000","family":"Arial"},"type":"$","code":"$\\frac{x^{2}+x+1}{\\left(x+1\\right)^{2}\\left(x+2\\right)}$","ts":1602918994871,"cs":"xw9bavBUKzM5Mb+/rvFfLQ==","size":{"width":69,"height":26}}$

Integrate f(x):-

${"font":{"color":"#000000","size":10,"family":"Arial"},"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-1-1-0-1-0-1-1-0","code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{x^{2}+x+1}{\\left(x+1\\right)^{2}\\left(x+2\\right)}.dx}\t\n\\end{align*}","type":"align*","ts":1602919115333,"cs":"BbEhp2hXd8US4TEG2CLU+Q==","size":{"width":184,"height":41}}$

By Partial Fraction:-

${"id":"1-0-0-0-1","font":{"size":12,"family":"Arial","color":"#000000"},"code":"$\\frac{x^{2}+x+1}{\\left(x+1\\right)^{2}\\left(x+2\\right)}=\\frac{A}{\\left(x+1\\right)}+\\frac{B}{\\left(x+1\\right)^{2}}+\\frac{C}{\\left(x+2\\right)}$","type":"$","ts":1602919261954,"cs":"I2S6Rti6YIm63BYtMEmLzw==","size":{"width":313,"height":33}}$

⇒ x2+x+1 = A(x + 1)(x + 2) + B(x + 2) + C(x + 1)2......(1)

Put x = -1 in eq. 1:-

⇒ 1 = B(1)

⇒ B = 1

Put x = -2 in eq. 1:-

⇒ 3 = C(1)

⇒ C = 3

Put x = 0 in eq. 1:-

⇒ 1 = 2A + 2B + C

⇒ 1 = 2A + 2 + 3

⇒ 2A = - 4

⇒ A = -2

${"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-1-0-0-0-1","font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{I}&={\\int_{}^{}\\left[\\frac{-2}{\\left(x+1\\right)}+\\frac{1}{\\left(x+1\\right)^{2}}+\\frac{3}{\\left(x+2\\right)}\\right].dx}\\\\\n{I}&={-2\\int_{}^{}\\frac{1}{\\left(x+1\\right)}.dx+\\int_{}^{}\\frac{1}{\\left(x+1\\right)^{2}}.dx+3\\int_{}^{}\\frac{1}{\\left(x+2\\right)}.dx}\t\n\\end{align*}","type":"align*","ts":1602920745438,"cs":"cW8/CeGY/IR+eSBkU4R+Hg==","size":{"width":405,"height":92}}$

We know that:-

${"id":"5-0-1","type":"$","font":{"color":"#222222","family":"Arial","size":10},"code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}x+C$","ts":1602672753650,"cs":"cnrPd6D+Y+6Cknmb6/Ea9g==","size":{"width":136,"height":18}}$

${"id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-0-0-1","type":"align*","code":"\\begin{align*}\n{I}&={-2\\log_{}\\left|x+1\\right|-\\frac{1}{x+1}+3\\log_{}\\left|x+2\\right|+C}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1602920855341,"cs":"8bQyx0TS+OSMwOX7iBKhew==","size":{"width":313,"height":32}}$

${"font":{"color":"#000000","size":11,"family":"Arial"},"id":"4-0-0-0-1-1-2-1-1-0-1-2-1-1-1-1-1-1-1-1-1-1-1-0","code":"$23.\\,\\tan^{-1}{\\sqrt[]{\\frac{1-x}{1+x}}}$","type":"$","ts":1602923316297,"cs":"uHc1XfqM1vNRsEWFChB26Q==","size":{"width":125,"height":33}}$

Solun:- Let f(x) = ${"id":"14-1-1-1-1-1-1-1-0","font":{"color":"#000000","family":"Arial","size":10},"code":"$\\tan^{-1}{\\sqrt[]{\\frac{1-x}{1+x}}}$","type":"$","ts":1602923340677,"cs":"7iibeDJQ6odiTBdRMIb2rg==","size":{"width":84,"height":28}}$

Integrate f(x):-

${"type":"align*","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-1-1-0-1-0-1-1-1-0","code":"\\begin{align*}\n{I}&={\\int_{}^{}\\tan^{-1}{\\sqrt[]{\\frac{1-x}{1+x}}}.dx}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"ts":1602923370939,"cs":"ICyO2yNXazVKxHZJVZsNWQ==","size":{"width":172,"height":40}}$

Let x = cos y

Differentiate w.r.t. to y:-

⇒ dx = -siny.dy

${"type":"align*","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-1-1-0-1-1-1-0","font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{I}&={-\\int_{}^{}\\tan^{-1}\\left({\\sqrt[]{\\frac{1-\\cos y}{1+\\cos y}}}\\right).\\sin y.dy}\\\\\n{I}&={-\\int_{}^{}\\tan^{-1}\\left({\\sqrt[]{\\frac{2\\sin^{2}\\frac{y}{2}}{2\\cos ^{2}\\frac{y}{2}}}}\\right).\\sin y.dy}\\\\\n{I}&={-\\int_{}^{}\\tan^{-1}\\left(\\tan\\frac{y}{2}\\right).\\sin y.dy}\t\n\\end{align*}","ts":1602926888377,"cs":"/ndwllGpy/DYaK06pnLqOQ==","size":{"width":273,"height":152}}$

${"code":"\\begin{align*}\n{I}&={-\\int_{}^{}\\frac{y}{2}.\\sin y.dy}\\\\\n{I}&={-\\frac{1}{2}\\int_{}^{}y.\\sin y.dy}\t\n\\end{align*}","font":{"color":"#222222","family":"Arial","size":10},"type":"align*","id":"16","ts":1602924304207,"cs":"E/aC9wTZFDMliusOGlHpYw==","size":{"width":145,"height":76}}$

We know that:-

${"font":{"size":10,"color":"#222222","family":"Arial"},"id":"5-1-0-1-0-0-1-0-1-1-1-1-1-1-0-1-0","type":"$","code":"$\\int_{}^{}\\left(u.v\\right).dx=u\\int_{}^{}v.dx-\\int_{}^{}\\left[\\diff{u}{x}\\int_{}^{}v.dx\\right].dx+C$","ts":1602925083002,"cs":"jsXxaUTJueBIShsPd00O/w==","size":{"width":318,"height":20}}$

According to ILATE:- u = y and v = sin y

${"font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{I}&={\\frac{-1}{2}\\left[y\\int_{}^{}\\sin y.dy-\\int_{}^{}\\left[\\diff{y}{y}\\int_{}^{}\\sin y.dy\\right].dy\\right]+C}\t\n\\end{align*}","type":"align*","id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-0-1-1-0-0-0-1-0-1-1-1-1-0-1-1-0-1-0-0","ts":1602926907512,"cs":"a2nNOa2/n8Vqg2A9NVkJ5Q==","size":{"width":361,"height":37}}$

We know that:-

${"font":{"family":"Arial","size":10,"color":"#222222"},"id":"17","code":"$\\int_{}^{}\\sin x.dx=-\\cos x+C$","type":"$","ts":1602925370969,"cs":"KcXDnn//nnDQzvu3cvqqQw==","size":{"width":169,"height":17}}$

${"code":"$\\int_{}^{}\\cos x.dx=\\sin x+C$","id":"17","type":"$","font":{"size":10,"family":"Arial","color":"#222222"},"ts":1602925531176,"cs":"KDQDHpi3UD1JWeEU4gteQg==","size":{"width":154,"height":17}}$

${"code":"\\begin{align*}\n{I}&={\\frac{-1}{2}\\left[-y\\cos y-\\int_{}^{}-\\cos y.dy\\right]+C}\\\\\n{I}&={\\frac{-1}{2}\\left[-y\\cos y+\\int_{}^{}\\cos y.dy\\right]+C}\\\\\n{I}&={\\frac{-1}{2}\\left[-y\\cos y+\\sin y\\right]+C}\t\n\\end{align*}","id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-0-1-1-0-0-0-1-0-1-1-1-1-0-1-1-0-1-1","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","ts":1602926933666,"cs":"p9zXB4/kQ+8l/ovFqS/QqQ==","size":{"width":266,"height":117}}$

Put the value of y:-

⇒ cos y = x

⇒ sin y = √(1-x2)

${"code":"\\begin{align*}\n{I}&={\\frac{-1}{2}\\left[-x\\cos^{-1}x+{\\sqrt[]{1-x^{2}}}\\right]+C}\\\\\n{I}&={\\frac{1}{2}\\left[x\\cos^{-1}x-{\\sqrt[]{1-x^{2}}}\\right]+C}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"type":"align*","id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-0-1-1-1-0-1-0-1-1-1-1-1-1-1-0","ts":1602926966264,"cs":"P8VnXVL3eB1zr8/sYpZgnA==","size":{"width":252,"height":68}}$

${"font":{"size":11,"color":"#000000","family":"Arial"},"code":"$24.\\,\\frac{{\\sqrt[]{x^{2}+1}}\\left[\\log_{}\\left(x^{2}+1\\right)-2\\log_{}x\\right]}{x^{4}}$","type":"$","id":"4-0-0-0-1-1-2-1-1-0-1-2-1-1-1-1-1-1-1-1-1-1-1-1-0","ts":1602927295945,"cs":"xNgEeX5C6wQPk3D6e7SOrw==","size":{"width":186,"height":29}}$

Solun:- Let f(x) = ${"id":"14-1-1-1-1-1-1-1-1-0","type":"$","code":"$\\frac{{\\sqrt[]{x^{2}+1}}\\left[\\log_{}\\left(x^{2}+1\\right)-2\\log_{}x\\right]}{x^{4}}$","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1602927348015,"cs":"QD9duYMmXUGq1nRgi/atxg==","size":{"width":140,"height":25}}$

Integrate f(x):-

${"font":{"size":10,"family":"Arial","color":"#000000"},"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-1-1-0-1-0-1-1-1-1-0","code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{{\\sqrt[]{x^{2}+1}}\\left[\\log_{}\\left(x^{2}+1\\right)-2\\log_{}x\\right]}{x^{4}}.dx}\\\\\n{I}&={\\int_{}^{}\\frac{{\\sqrt[]{x^{2}+1}}}{x^{4}}\\left[\\log_{}\\left(x^{2}+1\\right)-\\log_{}x^{2}\\right].dx}\\\\\n{I}&={\\int_{}^{}\\frac{{\\sqrt[]{x^{2}+1}}}{x^{4}}\\log_{}\\left|\\frac{x^{2}+1}{x^{2}}\\right|.dx}\\\\\n{I}&={\\int_{}^{}\\frac{1}{x^{3}}\\times\\frac{{\\sqrt[]{x^{2}+1}}}{x}\\log_{}\\left|1+\\frac{1}{x^{2}}\\right|.dx}\\\\\n{I}&={\\int_{}^{}\\frac{1}{x^{3}}\\times{\\sqrt[]{\\frac{x^{2}+1}{x^{2}}}}\\log_{}\\left|1+\\frac{1}{x^{2}}\\right|.dx}\t\n\\end{align*}","type":"align*","ts":1602927586389,"cs":"gQp8rkSjnWUEtnoKngn8lQ==","size":{"width":292,"height":224}}$

${"code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{1}{x^{3}}\\times{\\sqrt[]{1+\\frac{1}{x^{2}}}}\\log_{}\\left|1+\\frac{1}{x^{2}}\\right|.dx}\t\n\\end{align*}","id":"18","font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","ts":1602927611877,"cs":"BKH67LD6v1udFv1aiJ2FRw==","size":{"width":266,"height":40}}$

Let 1+1/x2 = t

Differentiate w.r.t. to t:-

⇒ -2/x3.dx = dt

⇒ 1/x3.dx = (-1/2).dt

${"code":"\\begin{align*}\n{I}&={-\\int_{}^{}{\\sqrt[]{t}}.\\log_{}\\left|t\\right|.\\frac{dt}{2}}\\\\\n{I}&={\\frac{-1}{2}\\int_{}^{}{\\sqrt[]{t}}.\\log_{}\\left|t\\right|.dt}\t\n\\end{align*}","type":"align*","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-1-1-0-1-1-1-1","font":{"color":"#000000","family":"Arial","size":10},"ts":1602928206068,"cs":"jUAVP0CziYkOjQvFcc7JNg==","size":{"width":164,"height":76}}$

We know that:-

${"font":{"size":10,"color":"#222222","family":"Arial"},"id":"5-1-0-1-0-0-1-0-1-1-1-1-1-1-0-1-1","type":"$","code":"$\\int_{}^{}\\left(u.v\\right).dx=u\\int_{}^{}v.dx-\\int_{}^{}\\left[\\diff{u}{x}\\int_{}^{}v.dx\\right].dx+C$","ts":1602925083002,"cs":"KLMpD+jbMaN7wqRUsOFpZw==","size":{"width":318,"height":20}}$

According to ILATE:- u = log t and v = √t

${"type":"align*","code":"\\begin{align*}\n{I}&={\\frac{-1}{2}\\left[\\log_{}t.\\int_{}^{}{\\sqrt[]{t}}.dt-\\int_{}^{}\\left[\\diff{\\left(\\log_{}t\\right)}{t}\\int_{}^{}{\\sqrt[]{t}}.dt\\right].dt\\right]+C}\\\\\n{I}&={\\frac{-1}{2}\\left[\\frac{t^{\\frac{3}{2}}}{\\frac{3}{2}}.\\log_{}t-\\int_{}^{}\\left[\\frac{1}{t}.\\frac{t^{\\frac{3}{2}}}{\\frac{3}{2}}\\right].dt\\right]+C}\\\\\n{I}&={\\frac{-1}{2}\\left[\\frac{2t^{\\frac{3}{2}}}{3}.\\log_{}t-\\frac{2}{3}\\int_{}^{}\\left[t^{\\frac{1}{2}}\\right].dt\\right]+C}\\\\\n{I}&={\\frac{-1}{2}\\left[\\frac{2t^{\\frac{3}{2}}}{3}.\\log_{}t-\\frac{2}{3}\\frac{t^{\\frac{3}{2}}}{\\frac{3}{2}}\\right]+C}\\\\\n{I}&={\\frac{-1}{2}\\left[\\frac{2t^{\\frac{3}{2}}}{3}.\\log_{}t-\\frac{4}{9}t^{\\frac{3}{2}}\\right]+C}\\\\\n{I}&={-t^{\\frac{3}{2}}\\left[\\frac{1}{3}.\\log_{}t-\\frac{2}{9}\\right]+C}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-0-1-1-0-0-0-1-0-1-1-1-1-0-1-1-0-1-0-1","ts":1603186576916,"cs":"Y+IaMi1tsEfOw+YssuW4CQ==","size":{"width":393,"height":289}}$

Put the value of t:-

${"type":"align*","code":"\\begin{align*}\n{I}&={-{\\left(1+\\frac{1}{x^{2}}\\right)}^{\\frac{3}{2}}\\left[\\frac{1}{3}.\\log_{}\\left|1+\\frac{1}{x^{2}}\\right|-\\frac{2}{9}\\right]+C}\\\\\n{I}&={-\\frac{1}{3}\\left(1+\\frac{1}{x^{2}}\\right)^{\\frac{3}{2}}\\left[\\log_{}\\left|1+\\frac{1}{x^{2}}\\right|-\\frac{2}{3}\\right]+C}\t\n\\end{align*}","id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-0-1-1-1-0-1-0-1-1-1-1-1-1-1-1-0","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1603186641123,"cs":"QRyMvW8ROANyuS7y18E/fw==","size":{"width":312,"height":90}}$

Evaluate the definite integrals in Exercises 25 to 33.

${"id":"4-0-0-0-1-1-2-1-1-0-1-2-1-1-1-1-1-1-1-1-1-1-1-1-1-0","font":{"size":11,"family":"Arial","color":"#000000"},"code":"$25.\\,\\int_{\\frac{\\Pi}{2}}^{\\Pi}e^{x}\\left(\\frac{1-\\sin x}{1-\\cos x}\\right).dx$","type":"$","ts":1602930005097,"cs":"baOfSnub5imjJd7C1uv8AA==","size":{"width":166,"height":28}}$

Solun:- Let f(x) = ${"id":"14-1-1-1-1-1-1-1-1-1-0","code":"$e^{x}\\left(\\frac{1-\\sin x}{1-\\cos x}\\right)$","font":{"size":10,"color":"#000000","family":"Arial"},"type":"$","ts":1602930107426,"cs":"lJROjLvfA/f25cfOOyszrw==","size":{"width":73,"height":20}}$

Integrate f(x):-

${"type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-1-1-0-1-0-1-1-1-1-1-0","code":"\\begin{align*}\n{I}&={\\int_{\\frac{\\Pi}{2}}^{\\Pi}e^{x}\\left(\\frac{1-\\sin x}{1-\\cos x}\\right).dx....\\left(1\\right)}\t\n\\end{align*}","ts":1602931108647,"cs":"E7/JpfhUvA3lvsUNhGioKg==","size":{"width":234,"height":42}}$

${"code":"\\begin{align*}\n{I}&={\\int_{\\frac{\\Pi}{2}}^{\\Pi}e^{x}\\left(\\frac{1-2\\sin \\frac{x}{2}\\cos\\frac{x}{2}}{2\\sin^{2}\\frac{x}{2}}\\right).dx}\\\\\n{I}&={\\int_{\\frac{\\Pi}{2}}^{\\Pi}e^{x}\\left(\\frac{1}{2}\\cos ec^{2}\\frac{x}{2}-\\cot\\frac{x}{2}\\right).dx}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-1-1-0-1-0-1-1-1-1-1-1","ts":1602931920733,"cs":"lY5EJk3FkSeUpqASBhmd6A==","size":{"width":253,"height":94}}$

We know that:-

${"id":"21-0","code":"$\\int_{}^{}e^{x}\\left[f\\left(x\\right)+f^{\\prime }\\left(x\\right)\\right].dx=e^{x}.f\\left(x\\right)+C$","type":"$","font":{"family":"Arial","color":"#000000","size":10},"ts":1602932003494,"cs":"IhJXLkHc1BhxZtHECJpIeg==","size":{"width":249,"height":17}}$

Let f(x) = - cot (x/2)

Then

${"type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"id":"22","code":"\\begin{align*}\n{f^{\\prime }\\left(x\\right)}&={-\\left[\\frac{-1}{2}\\cos ec^{2}\\frac{x}{2}\\right]}\\\\\n{f^{\\prime }\\left(x\\right)}&={\\frac{1}{2}\\cos ec^{2}\\frac{x}{2}}\t\n\\end{align*}","ts":1602932116714,"cs":"LWy80Ip5WLb0VtZ68yxAUQ==","size":{"width":172,"height":74}}$

${"id":"20-0","font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{I}&={-\\left[e^{x}.\\cot\\frac{x}{2}\\right]_{\\frac{\\Pi}{2}}^{\\Pi}}\\\\\n{I}&={-\\left[e^{\\Pi}.\\cot\\frac{\\Pi}{2}-e^{\\frac{\\Pi}{2}}.\\cot\\frac{\\Pi}{4}\\right]}\\\\\n{I}&={-\\left[e^{\\Pi}\\times0-e^{\\frac{\\Pi}{2}}\\times1\\right]}\\\\\n{I}&={e^{\\frac{\\Pi}{2}}}\t\n\\end{align*}","type":"align*","ts":1603187038943,"cs":"ftI39zyji7WS9mFdFb628w==","size":{"width":212,"height":137}}$

${"font":{"size":11,"family":"Arial","color":"#000000"},"type":"$","id":"4-0-0-0-1-1-2-1-1-0-1-2-1-1-1-1-1-1-1-1-1-1-1-1-1-1-0","code":"$26.\\,\\int_{0}^{\\frac{\\Pi}{4}}\\frac{\\sin x\\cos x}{\\cos ^{4}x+ \\sin^{4}x}.dx$","ts":1602934766650,"cs":"Ay9hswRkH1WptDcB69JiqA==","size":{"width":164,"height":30}}$

Solun:- Let f(x) = ${"id":"14-1-1-1-1-1-1-1-1-1-1-0","type":"$","code":"$\\frac{\\sin x\\cos x}{\\cos ^{4}x+ \\sin^{4}x}$","font":{"color":"#000000","family":"Arial","size":10},"ts":1602934791756,"cs":"StrjbypsH/u1q5pOpR3zZQ==","size":{"width":69,"height":21}}$

Integrate f(x):-

${"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-1-1-0-1-0-1-1-1-1-1-2-0","type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{I}&={\\int_{0}^{\\frac{\\Pi}{4}}\\frac{\\sin x\\cos x}{\\cos ^{4}x+ \\sin^{4}x}.dx}\\\\\n{I}&={\\int_{0}^{\\frac{\\Pi}{4}}\\frac{\\frac{\\sin x\\cos x}{\\cos ^{4}x}}{\\frac{\\cos ^{4}x}{\\cos ^{4}x}+ \\frac{\\sin^{4}x}{\\cos ^{4}x}}.dx}\\\\\n{I}&={\\int_{0}^{\\frac{\\Pi}{4}}\\frac{\\tan x.\\sec^{2}x}{1+ \\tan^{4}x}.dx}\\\\\n{I}&={\\int_{0}^{\\frac{\\Pi}{4}}\\frac{\\tan x.\\sec^{2}x}{1+ \\left(\\tan^{2}x\\right)^{2}}.dx}\t\n\\end{align*}","ts":1602935035295,"cs":"UPaq57KikGoFZKpKlVZEHg==","size":{"width":186,"height":194}}$

Let tan2x = t…....(1)

Differentiate w.r.t. to t:-

⇒ 2tan x.sec2x.dx = dt

⇒ tan x.sec2x.dx = (1/2).dt

Limits Change:-

Put x = 0 in eq. 1:-

⇒ t = 0

Put x = π/4 in eq. 1:-

⇒ t = 1

${"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-1-1-0-1-0-1-1-1-1-1-2-0","type":"align*","font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{I}&={\\int_{0}^{1}\\frac{1}{1+ t^{2}}.\\frac{dt}{2}}\\\\\n{I}&={\\frac{1}{2}\\int_{0}^{1}\\frac{1}{1+ t^{2}}.dt}\t\n\\end{align*}","ts":1603187436461,"cs":"0UgqRhd/hF7jCt57Dl8bnQ==","size":{"width":138,"height":82}}$

We know that:-

${"id":"21-1","type":"$","font":{"family":"Arial","size":10,"color":"#000000"},"code":"$\\int_{}^{}\\frac{1}{1+x^{2}}.dx=\\tan^{-1}x+C$","ts":1603187477544,"cs":"Hw+L9MB6LxPMx5i0sMFxVQ==","size":{"width":172,"height":20}}$

${"type":"align*","id":"20-1","code":"\\begin{align*}\n{I}&={\\frac{1}{2}\\left[\\tan^{-1}t\\right]_{0}^{1}}\\\\\n{I}&={\\frac{1}{2}\\left[\\tan^{-1}1-\\tan^{-1}0\\right]}\\\\\n{I}&={\\frac{1}{2}\\times\\frac{\\Pi}{4}}\\\\\n{I}&={\\frac{\\Pi}{8}}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1603187713795,"cs":"uLqQ7EqjD8H4qat1Jqf0kA==","size":{"width":173,"height":144}}$

${"type":"$","id":"4-0-0-0-1-1-2-1-1-0-1-2-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1","font":{"color":"#000000","size":11,"family":"Arial"},"code":"$27.\\,\\int_{0}^{\\frac{\\Pi}{2}}\\frac{\\cos ^{2}x}{\\cos ^{2}x+ 4\\sin^{2}x}.dx$","ts":1603010265990,"cs":"cdeoQMFz8GlqXHdY9aGk7Q==","size":{"width":173,"height":30}}$

Solun:- Let f(x) = ${"font":{"color":"#000000","family":"Arial","size":10},"code":"$\\frac{\\cos ^{2}x}{\\cos ^{2}x+ 4\\sin^{2}x}$","id":"14-1-1-1-1-1-1-1-1-1-1-1-0","type":"$","ts":1603010291701,"cs":"BycAhpZrYOwCYJ9p/+0ihw==","size":{"width":76,"height":24}}$

Integrate f(x):-

${"font":{"family":"Arial","size":9.260959500653215,"color":"#000000"},"type":"align*","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-1-1-0-1-0-1-1-1-1-1-2-1-0","code":"\\begin{align*}\n{I}&={\\int_{0}^{\\frac{\\Pi}{2}}\\frac{\\cos ^{2}x}{\\cos ^{2}x+ 4\\sin^{2}x}.dx}\\\\\n{I}&={\\int_{0}^{\\frac{\\Pi}{2}}\\frac{\\cos ^{2}x}{\\cos ^{2}x+ 4\\left(1-\\cos^{2}x\\right)}.dx}\\\\\n{I}&={\\int_{0}^{\\frac{\\Pi}{2}}\\frac{\\cos ^{2}x}{\\cos ^{2}x+ 4-4\\cos^{2}x}.dx}\\\\\n{I}&={\\frac{-1}{3}\\int_{0}^{\\frac{\\Pi}{2}}\\frac{-3\\cos ^{2}x}{4-3\\cos^{2}x}.dx}\\\\\n{I}&={\\frac{-1}{3}\\int_{0}^{\\frac{\\Pi}{2}}\\frac{4-3\\cos ^{2}x-4}{4-3\\cos^{2}x}.dx}\\\\\n{I}&={\\frac{-1}{3}\\left[\\int_{0}^{\\frac{\\Pi}{2}}\\frac{4-3\\cos ^{2}x}{4-3\\cos^{2}x}.dx-\\int_{0}^{\\frac{\\Pi}{2}}\\frac{\\frac{4}{\\cos^{2}x}}{\\frac{4}{\\cos^{2}x}-\\frac{3\\cos^{2}x}{\\cos^{2}x}}.dx\\right]}\\\\\n{I}&={\\frac{-1}{3}\\left[\\int_{0}^{\\frac{\\Pi}{2}}1.dx-4\\int_{0}^{\\frac{\\Pi}{2}}\\frac{\\sec^{2}x}{4\\sec^{2}x-3}.dx\\right]}\t\n\\end{align*}","ts":1603011773290,"cs":"BTBFPRJ4P8j4YqyMsB47xw==","size":{"width":390,"height":318}}$

${"code":"\\begin{align*}\n{I}&={\\frac{-1}{3}\\left[\\int_{0}^{\\frac{\\Pi}{2}}1.dx-4\\int_{0}^{\\frac{\\Pi}{2}}\\frac{\\sec^{2}x}{4\\left(1+\\tan^{2}x\\right)-3}.dx\\right]}\\\\\n{I}&={\\frac{-1}{3}\\left[\\int_{0}^{\\frac{\\Pi}{2}}1.dx-4\\int_{0}^{\\frac{\\Pi}{2}}\\frac{\\sec^{2}x}{4\\tan^{2}x+1}.dx\\right]}\t\n\\end{align*}","id":"25-0","font":{"family":"Arial","color":"#000000","size":10},"type":"align*","ts":1603011849725,"cs":"QKR7fWfueOiKYfVgZeVBQA==","size":{"width":344,"height":100}}$

Let tan x = t…....(1)

Differentiate w.r.t. to t:-

⇒ sec2x.dx = dt

Limits change:-

Put x = 0 in eq. 1:-

⇒ t = 0

Put x = π/2 in eq. 1:-

⇒ t = ∞

${"font":{"color":"#000000","family":"Arial","size":10},"id":"24-0","code":"\\begin{align*}\n{I}&={\\frac{-1}{3}\\left[\\int_{0}^{\\frac{\\Pi}{2}}1.dx-4\\int_{0}^{\\infty}\\frac{1}{4t^{2}+1}.dt\\right]}\\\\\n{I}&={\\frac{-1}{3}\\left[\\int_{0}^{\\frac{\\Pi}{2}}1.dx-\\frac{4}{4}\\int_{0}^{\\infty}\\frac{1}{t^{2}+\\frac{1}{4}}.dt\\right]}\\\\\n{I}&={\\frac{-1}{3}\\left[\\int_{0}^{\\frac{\\Pi}{2}}1.dx-\\int_{0}^{\\infty}\\frac{1}{t^{2}+\\left(\\frac{1}{2}\\right)^{2}}.dt\\right]}\t\n\\end{align*}","type":"align*","ts":1603012632616,"cs":"lOdHpf2WmwkPsLatBTxGMQ==","size":{"width":277,"height":153}}$

We know that:-

${"type":"$","code":"$\\int_{}^{}\\frac{1}{x^{2}+a^{2}}.dx=\\frac{1}{a}\\tan^{-1}\\left(\\frac{x}{a}\\right)+C$","font":{"color":"#222222","family":"Arial","size":10},"id":"26-0","ts":1603012792140,"cs":"eeIbRz1YQ+C4CJuA3VvJxQ==","size":{"width":208,"height":20}}$

${"code":"\\begin{align*}\n{I}&={\\frac{-1}{3}\\left[\\left[x\\right]_{0}^{\\frac{\\Pi}{2}}-\\left[\\frac{1}{\\frac{1}{2}}\\tan^{-1}\\left(\\frac{t}{\\frac{1}{2}}\\right)\\right]_{0}^{\\infty}\\right]}\\\\\n{I}&={\\frac{-1}{3}\\left[\\left[x\\right]_{0}^{\\frac{\\Pi}{2}}-\\left[2\\tan^{-1}\\left(2t\\right)\\right]_{0}^{\\infty}\\right]}\\\\\n{I}&={\\frac{-1}{3}\\left[\\left[x\\right]_{0}^{\\frac{\\Pi}{2}}-2\\left[\\tan^{-1}\\left(2t\\right)\\right]_{0}^{\\infty}\\right]}\\\\\n{I}&={\\frac{-1}{3}\\left[\\left[\\frac{\\Pi}{2}-0\\right]-2\\left[\\tan^{-1}\\left(\\infty\\right)-\\tan^{-1}\\left(0\\right)\\right]\\right]}\\\\\n{I}&={\\frac{-1}{3}\\left[\\frac{\\Pi}{2}-2\\left[\\frac{\\Pi}{2}-0\\right]\\right]}\\\\\n{I}&={\\frac{-1}{3}\\left[\\frac{\\Pi}{2}-\\Pi\\right]}\\\\\n{I}&={\\frac{\\Pi}{6}}\t\n\\end{align*}","id":"27-0","type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1603013529439,"cs":"BgDYCyO+5FoNW4EldLzfMg==","size":{"width":324,"height":288}}$

${"font":{"color":"#000000","family":"Arial","size":11},"type":"$","code":"$28.\\,\\int_{\\frac{\\Pi}{6}}^{\\frac{\\Pi}{3}}\\frac{\\sin x+\\cos x}{{\\sqrt[]{\\sin2x}}}.dx$","id":"4-0-0-0-1-1-2-1-1-0-1-2-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-0","ts":1603014197720,"cs":"ns0ur3PKrluzHhHD7mKbtg==","size":{"width":154,"height":34}}$

Solun:- Let f(x) = ${"code":"$\\frac{\\sin x+\\cos x}{{\\sqrt[]{\\sin2x}}}$","id":"14-1-1-1-1-1-1-1-1-1-1-1-1-0","type":"$","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1603014222153,"cs":"DzgdkbFLQRPl4MO08zpj4A==","size":{"width":60,"height":22}}$

Integrate f(x):-

${"type":"align*","code":"\\begin{align*}\n{I}&={\\int_{\\frac{\\Pi}{6}}^{\\frac{\\Pi}{3}}\\frac{\\sin x+\\cos x}{{\\sqrt[]{\\sin2x}}}.dx}\\\\\n{I}&={\\int_{\\frac{\\Pi}{6}}^{\\frac{\\Pi}{3}}\\frac{\\sin x+\\cos x}{{\\sqrt[]{-\\left(-\\sin2x\\right)}}}.dx}\\\\\n{I}&={\\int_{\\frac{\\Pi}{6}}^{\\frac{\\Pi}{3}}\\frac{\\sin x+\\cos x}{{\\sqrt[]{-\\left(1-\\sin2x-1\\right)}}}.dx}\\\\\n{I}&={\\int_{\\frac{\\Pi}{6}}^{\\frac{\\Pi}{3}}\\frac{\\sin x+\\cos x}{{\\sqrt[]{-\\left[\\sin^{2}x+\\cos^{2}x-2\\sin x.\\cos x-1\\right]}}}.dx}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"id":"53","ts":1603188266456,"cs":"IJaucTbWRDFEMpvoVvkH6A==","size":{"width":356,"height":208}}$

${"font":{"color":"#000000","family":"Arial","size":9.260959500653215},"type":"align*","code":"\\begin{align*}\n{I}&={\\int_{\\frac{\\Pi}{6}}^{\\frac{\\Pi}{3}}\\frac{\\sin x+\\cos x}{{\\sqrt[]{-\\left[\\left(\\sin x-\\cos x\\right)^{2}-1\\right]}}}.dx}\\\\\n{I}&={\\int_{\\frac{\\Pi}{6}}^{\\frac{\\Pi}{3}}\\frac{\\sin x+\\cos x}{{\\sqrt[]{1-\\left(\\sin x-\\cos x\\right)^{2}}}}.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-1-1-0-1-0-1-1-1-1-1-2-1-1-0","ts":1603188346077,"cs":"Z2Dc2Xb54ZxXU9OpPFwizg==","size":{"width":264,"height":125}}$

Let sin x - cos x = t…....(1)

Differentiate w.r.t. to t:-

⇒ (cos x + sin x).dx = dt

Limits change:-

Put x = π/6 in eq. 1:-

⇒ t = (1 - √3)/2

Put x = π/3 in eq. 1:-

⇒ t = (√3 - 1)/2

${"code":"\\begin{align*}\n{I}&={\\int_{\\left(\\frac{1-{\\sqrt[]{3}}}{2}\\right)}^{\\left(\\frac{{\\sqrt[]{3}}-1}{2}\\right)}\\frac{1}{{\\sqrt[]{1-t^{2}}}}.dt}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","id":"24-1-0","ts":1603015525967,"cs":"Lpd3l56SWEyYc9165Cg/9g==","size":{"width":168,"height":52}}$

We know that:-

${"type":"$","code":"$\\int_{}^{}\\frac{1}{{\\sqrt[]{1-x^{2}}}}.dx=\\sin^{-1}x+C$","id":"26-1-0-0","font":{"size":10,"color":"#222222","family":"Arial"},"ts":1603015559734,"cs":"qGhCaHsW9VPgqopFRgaI5g==","size":{"width":178,"height":22}}$

${"id":"27-1-0-0","type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{I}&={\\left[\\sin^{-1}\\left(t\\right)\\right]_{\\frac{1-{\\sqrt[]{3}}}{2}}^{\\frac{{\\sqrt[]{3}}-1}{2}}}\\\\\n{I}&={\\left[\\sin^{-1}\\left(\\frac{{\\sqrt[]{3}}-1}{2}\\right)-\\sin^{-1}\\left(\\frac{1-{\\sqrt[]{3}}}{2}\\right)\\right]}\\\\\n{I}&={\\left[\\sin^{-1}\\left(\\frac{{\\sqrt[]{3}}-1}{2}\\right)-\\sin^{-1}\\left(\\frac{-\\left({\\sqrt[]{3}}-1\\right)}{2}\\right)\\right]}\t\n\\end{align*}","ts":1603015819986,"cs":"HGgXODQn1f8fCvg9K20iMA==","size":{"width":338,"height":156}}$

We know that:-

sin-1(-x) = -sin-1(x)

${"font":{"size":10,"family":"Arial","color":"#222222"},"code":"\\begin{align*}\n{I}&={\\left[\\sin^{-1}\\left(\\frac{{\\sqrt[]{3}}-1}{2}\\right)-\\left(-\\sin^{-1}\\left(\\frac{{\\sqrt[]{3}}-1}{2}\\right)\\right)\\right]}\\\\\n{I}&={\\left[\\sin^{-1}\\left(\\frac{{\\sqrt[]{3}}-1}{2}\\right)+\\sin^{-1}\\left(\\frac{{\\sqrt[]{3}}-1}{2}\\right)\\right]}\\\\\n{I}&={2\\sin^{-1}\\left(\\frac{{\\sqrt[]{3}}-1}{2}\\right)}\t\n\\end{align*}","id":"28-0","type":"align*","ts":1603015972400,"cs":"ZQvEnQVEKTY8qpmlf5zClA==","size":{"width":342,"height":152}}$

${"type":"$","id":"29-0","code":"$29.\\,\\int_{0}^{1}\\frac{dx}{{\\sqrt[]{1+x}}-{\\sqrt[]{x}}}$","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1603016086777,"cs":"lu0GIFykljh//sxXsMzv1Q==","size":{"width":106,"height":25}}$

Solun:- Let f(x) = ${"code":"$\\frac{1}{{\\sqrt[]{1+x}}-{\\sqrt[]{x}}}$","type":"$","font":{"color":"#000000","family":"Arial","size":10},"id":"30","ts":1603188579397,"cs":"VqfZrapIfwZWG8ktGC8YjA==","size":{"width":61,"height":24}}$

Integrate f(x):-

${"font":{"color":"#000000","family":"Arial","size":10},"id":"31","code":"\\begin{align*}\n{I}&={\\int_{0}^{1}\\frac{dx}{{\\sqrt[]{1+x}}-{\\sqrt[]{x}}}}\\\\\n{I}&={\\int_{0}^{1}\\frac{1}{{\\sqrt[]{1+x}}-{\\sqrt[]{x}}}\\times\\frac{{\\sqrt[]{1+x}}+{\\sqrt[]{x}}}{{\\sqrt[]{1+x}}+{\\sqrt[]{x}}}.dx}\\\\\n{I}&={\\int_{0}^{1}\\frac{{\\sqrt[]{1+x}}+{\\sqrt[]{x}}}{1+x-x}.dx}\\\\\n{I}&={\\int_{0}^{1}\\left({\\sqrt[]{1+x}}+{\\sqrt[]{x}}\\right).dx}\\\\\n{I}&={\\int_{0}^{1}\\left[\\left(1+x\\right)^{\\frac{1}{2}}+x^{\\frac{1}{2}}\\right].dx}\t\n\\end{align*}","type":"align*","ts":1603016776107,"cs":"NGBkIFslOrvx+POOoi60vg==","size":{"width":294,"height":221}}$

We know that:-

${"type":"$","font":{"size":10,"family":"Arial","color":"#222222"},"id":"26-1-1","code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+C$","ts":1603016685494,"cs":"3zMfJecmnP8PcpIfqLPIUg==","size":{"width":136,"height":21}}$

${"code":"\\begin{align*}\n{I}&={\\left[\\frac{\\left(1+x\\right)^{\\frac{3}{2}}}{\\frac{3}{2}}+\\frac{x^{\\frac{3}{2}}}{\\frac{3}{2}}\\right]_{0}^{1}}\\\\\n{I}&={\\left[\\frac{2\\left(1+x\\right)^{\\frac{3}{2}}}{3}+\\frac{2x^{\\frac{3}{2}}}{3}\\right]_{0}^{1}}\\\\\n{I}&={\\frac{2}{3}\\left[\\left(1+x\\right)^{\\frac{3}{2}}+x^{\\frac{3}{2}}\\right]_{0}^{1}}\\\\\n{I}&={\\frac{2}{3}\\left[\\left(\\left(2\\right)^{\\frac{3}{2}}+\\left(1\\right)^{\\frac{3}{2}}\\right)-\\left(\\left(1\\right)^{\\frac{3}{2}}+\\left(0\\right)^{\\frac{3}{2}}\\right)\\right]}\\\\\n{I}&={\\frac{2}{3}\\left[\\left(2{\\sqrt[]{2}}+1\\right)-\\left(1+0\\right)\\right]}\\\\\n{I}&={\\frac{2}{3}\\left[2{\\sqrt[]{2}}+1-1+0\\right]}\\\\\n{I}&={\\frac{4{\\sqrt[]{2}}}{3}}\t\n\\end{align*}","type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"id":"27-1-1","ts":1603017163192,"cs":"lzqV/NRwM1jkbNEQ/n1PKw==","size":{"width":274,"height":300}}$

${"type":"$","font":{"color":"#000000","family":"Arial","size":11},"code":"$30.\\,\\int_{0}^{\\frac{\\Pi}{4}}\\frac{\\sin x+\\cos x}{9+16\\sin2x}.dx$","id":"4-0-0-0-1-1-2-1-1-0-1-2-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-1-0","ts":1603017535040,"cs":"APhxg3xB4Ykh+PMXC9x+3Q==","size":{"width":156,"height":28}}$

Solun:- Let f(x) = ${"id":"14-1-1-1-1-1-1-1-1-1-1-1-1-1-0","code":"$\\frac{\\sin x+\\cos x}{9+16\\sin2x}$","type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1603017560470,"cs":"AaZlkhJGIPGL3G9XXloB4g==","size":{"width":61,"height":20}}$

Integrate f(x):-

${"code":"\\begin{align*}\n{I}&={\\int_{0}^{\\frac{\\Pi}{4}}\\frac{\\sin x+\\cos x}{9-16\\left(-\\sin2x\\right)}.dx}\\\\\n{I}&={\\int_{0}^{\\frac{\\Pi}{4}}\\frac{\\sin x+\\cos x}{9-16\\left(1-\\sin2x-1\\right)}.dx}\\\\\n{I}&={\\int_{0}^{\\frac{\\Pi}{4}}\\frac{\\sin x+\\cos x}{9-16\\left[\\left(\\sin^{2}x+\\cos^{2}x-\\sin2x-1\\right)\\right]}.dx}\\\\\n{I}&={\\int_{0}^{\\frac{\\Pi}{4}}\\frac{\\sin x+\\cos x}{9-16\\left[\\left(\\sin x-\\cos x\\right)^{2}-1\\right]}.dx}\\\\\n{I}&={\\int_{0}^{\\frac{\\Pi}{4}}\\frac{\\sin x+\\cos x}{9-16\\left(\\sin x-\\cos x\\right)^{2}+16}.dx}\\\\\n{I}&={\\int_{0}^{\\frac{\\Pi}{4}}\\frac{\\sin x+\\cos x}{25-16\\left(\\sin x-\\cos x\\right)^{2}}.dx}\\\\\n{I}&={\\frac{1}{16}\\int_{0}^{\\frac{\\Pi}{4}}\\frac{\\sin x+\\cos x}{\\frac{25}{16}-\\left(\\sin x-\\cos x\\right)^{2}}.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-1-1-0-1-0-1-1-1-1-1-2-1-1-1-0","font":{"family":"Arial","color":"#000000","size":9.260959500653215},"type":"align*","ts":1603018114829,"cs":"nP9kyGjeCrPFqbnP/boT0w==","size":{"width":342,"height":350}}$

Let sin x - cos x = t…....(1)

Differentiate w.r.t. to t:-

⇒ (cos x + sin x).dx = dt

Limits change:-

Put x = 0 in eq. 1:-

⇒ t = - 1

Put x = π/4 in eq. 1:-

⇒ t = 0

${"type":"align*","font":{"family":"Arial","color":"#000000","size":10},"id":"24-1-1-0","code":"\\begin{align*}\n{I}&={\\frac{1}{16}\\int_{-1}^{0}\\frac{1}{\\left(\\frac{5}{4}\\right)^{2}-t^{2}}.dt}\t\n\\end{align*}","ts":1603018263237,"cs":"CzN+GSTUuqYTqSvGgoOpsg==","size":{"width":173,"height":45}}$

We know that:-

${"code":"$\\int_{}^{}\\frac{1}{a^{2}-x^{2}}.dx=\\frac{1}{2a}\\log_{}\\left|\\frac{a+x}{a-x}\\right|+C$","id":"26-1-0-1-0","type":"$","font":{"size":10,"color":"#222222","family":"Arial"},"ts":1603188899580,"cs":"dPJjSpvXPgkZz638hRJQeQ==","size":{"width":208,"height":20}}$

${"id":"27-1-0-1-0","font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{I}&={\\frac{1}{16}\\left[\\frac{1}{2\\times\\frac{5}{4}}\\log_{}\\left|\\frac{\\frac{5}{4}+t}{\\frac{5}{4}-t}\\right|\\right]_{-1}^{0}}\\\\\n{I}&={\\frac{1}{16}\\times\\frac{2}{5}\\left[\\log_{}\\left|\\frac{\\frac{5}{4}+t}{\\frac{5}{4}-t}\\right|\\right]_{-1}^{0}}\t\n\\end{align*}","type":"align*","ts":1603018774640,"cs":"SLl+pHEdzgl7neCVyxrmPg==","size":{"width":210,"height":108}}$

${"code":"\\begin{align*}\n{I}&={\\frac{1}{40}\\left[\\log_{}\\left|\\frac{\\frac{5}{4}+0}{\\frac{5}{4}-0}\\right|-\\log_{}\\left|\\frac{\\frac{5}{4}-1}{\\frac{5}{4}+1}\\right|\\right]}\\\\\n{I}&={\\frac{1}{40}\\left[\\log_{}\\left|1\\right|-\\log_{}\\left|\\frac{\\frac{1}{4}}{\\frac{9}{4}}\\right|\\right]}\\\\\n{I}&={\\frac{1}{40}\\left[0-\\log_{}\\left|\\frac{1}{9}\\right|\\right]}\\\\\n{I}&={\\frac{1}{40}\\left[-\\left(\\log_{}\\left|1\\right|-\\log_{}\\left|9\\right|\\right)\\right]}\\\\\n{I}&={\\frac{1}{40}\\left[-\\left(0-\\log_{}\\left|9\\right|\\right)\\right]}\\\\\n{I}&={\\frac{1}{40}\\log_{}\\left|9\\right|}\t\n\\end{align*}","type":"align*","font":{"color":"#222222","family":"Arial","size":10},"id":"28-1-0","ts":1603018858712,"cs":"YUpcWElRoI0U4rrPzGOihw==","size":{"width":252,"height":253}}$

${"code":"$31.\\,\\int_{0}^{\\frac{\\Pi}{2}}\\sin2x\\tan^{-1}\\left(\\sin x\\right).dx$","id":"4-0-0-0-1-1-2-1-1-0-1-2-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-1-1-0-0","type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1603019106220,"cs":"8LeGoY3vJmf8+xcmYPm4xg==","size":{"width":200,"height":24}}$

Solun:- Let f(x) = sin 2x.tan-1(sin x)

Integrate f(x):-

${"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-1-1-0-1-0-1-1-1-1-1-2-1-1-1-1-0-0","type":"align*","font":{"size":9.260959500653215,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{I}&={\\int_{0}^{\\frac{\\Pi}{2}}\\sin2x\\tan^{-1}\\left(\\sin x\\right).dx}\\\\\n{I}&={\\int_{0}^{\\frac{\\Pi}{2}}2\\sin x.\\cos x.\\tan^{-1}\\left(\\sin x\\right).dx}\\\\\n{I}&={2\\int_{0}^{\\frac{\\Pi}{2}}\\sin x.\\cos x.\\tan^{-1}\\left(\\sin x\\right).dx}\t\n\\end{align*}","ts":1603019273899,"cs":"RxOw5hC7j7I90ajjXwgkkg==","size":{"width":256,"height":133}}$

Let sin x = t…....(1)

Differentiate w.r.t. to t:-

⇒ cos x.dx = dt

Limits change:-

Put x = 0 in eq. 1:-

⇒ t = 0

Put x = π/2 in eq. 1:-

⇒ t = 1

${"type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{I}&={2\\int_{0}^{1}t.\\tan^{-1}\\left(t\\right).dt}\t\n\\end{align*}","id":"24-1-1-1","ts":1603019604033,"cs":"tBCN0QWvmaXdw0AgeTWSEw==","size":{"width":156,"height":38}}$

We know that:-

${"code":"$\\int_{}^{}\\left(u.v\\right).dx=u.\\int_{}^{}v.dx-\\int_{}^{}\\left[\\diff{u}{x}.\\int_{}^{}v.dx\\right].dx+C$","type":"$","id":"26-1-0-1-1-0","font":{"size":10,"family":"Arial","color":"#222222"},"ts":1603019721693,"cs":"flebLdpkEXff8o+IHQN/Lw==","size":{"width":326,"height":20}}$

According to ILATE:- u = tan-1t and v = t

${"id":"26-1-0-1-1-1-0","type":"align*","font":{"size":10,"color":"#222222","family":"Arial"},"code":"\\begin{align*}\n{I}&={2\\left[\\tan^{-1}t.\\int_{0}^{1}t.dt-\\int_{}^{}\\left[\\diff{\\left(\\tan^{-1}t\\right)}{t}.\\int_{0}^{1}t.dt\\right].dt\\right]+C}\\\\\n{I}&={2\\left[\\frac{t^{2}}{2}.\\tan^{-1}t-\\frac{1}{2}\\int_{0}^{1}\\frac{t^{2}}{1+t^{2}}.dt\\right]+C}\t\n\\end{align*}","ts":1603021014477,"cs":"2X2tUJL55gPrUoco3No7aA==","size":{"width":402,"height":92}}$

${"id":"33","type":"align*","code":"\\begin{align*}\n{I}&={2\\left[\\frac{t^{2}}{2}.\\tan^{-1}t-\\frac{1}{2}\\int_{0}^{1}\\frac{t^{2}+1-1}{1+t^{2}}.dt\\right]+C}\\\\\n{I}&={2\\left[\\frac{t^{2}}{2}.\\tan^{-1}t-\\frac{1}{2}\\left[\\int_{0}^{1}\\frac{t^{2}+1}{1+t^{2}}.dt-\\int_{0}^{1}\\frac{1}{1+t^{2}}.dt\\right]\\right]+C}\\\\\n{I}&={2\\left[\\frac{t^{2}}{2}.\\tan^{-1}t-\\frac{1}{2}\\left[\\int_{0}^{1}1.dt-\\int_{0}^{1}\\frac{1}{1+t^{2}}.dt\\right]\\right]+C}\\\\\n{I}&={2\\left[\\frac{t^{2}}{2}.\\tan^{-1}t-\\frac{1}{2}\\int_{0}^{1}1.dt+\\frac{1}{2}\\int_{0}^{1}\\frac{1}{1+t^{2}}.dt\\right]+C}\\\\\n{I}&={t^{2}.\\tan^{-1}t-\\int_{0}^{1}1.dt+\\int_{0}^{1}\\frac{1}{1+t^{2}}.dt+C}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#222222"},"ts":1603021105367,"cs":"ylWG/XS2mDB6m5aXLKlNHw==","size":{"width":412,"height":216}}$

We know that:-

${"code":"$\\int_{}^{}\\frac{1}{1+x^{2}}.dx=\\tan^{-1}x+C$","font":{"size":10,"family":"Arial","color":"#222222"},"id":"34","type":"$","ts":1603020425692,"cs":"+zim42qi2RrWG7UKFLmJPQ==","size":{"width":172,"height":20}}$

${"id":"27-1-0-1-1","code":"\\begin{align*}\n{I}&={\\left[t^{2}\\tan^{-1}t-t+\\tan^{-1}t\\right]_{-1}^{0}}\\\\\n{I}&={\\left[\\left(0\\tan^{-1}\\left(0\\right)-0+\\tan^{-1}\\left(0\\right)\\right)-\\left(\\tan^{-1}\\left(-1\\right)+1+\\tan^{-1}\\left(-1\\right)\\right)\\right]}\t\n\\end{align*}","type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1603021143408,"cs":"3r2S0gl15ftUq4yaoqPplw==","size":{"width":462,"height":48}}$

We know that:-

tan-1(-x) = tan-1x

${"font":{"family":"Arial","size":10,"color":"#222222"},"type":"align*","id":"28-1-1","code":"\\begin{align*}\n{I}&={\\left[\\left(0\\right)-\\left(-\\tan^{-1}\\left(1\\right)+1-\\tan^{-1}\\left(1\\right)\\right)\\right]}\\\\\n{I}&={-\\left(-\\frac{\\Pi}{4}+1-\\frac{\\Pi}{4}\\right)}\\\\\n{I}&={\\left(\\frac{\\Pi}{4}-1+\\frac{\\Pi}{4}\\right)}\\\\\n{I}&={\\frac{\\Pi}{2}-1}\t\n\\end{align*}","ts":1603021244376,"cs":"K3Bm95PaM5W8+QYabcQsqw==","size":{"width":280,"height":141}}$

${"code":"$32.\\,\\int_{0}^{\\Pi}\\frac{x\\tan x}{\\sec x+\\tan x}.dx$","id":"4-0-0-0-1-1-2-1-1-0-1-2-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-1-1-1-0","type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1603021393737,"cs":"m9MiUFPFTs3L4Mpcr9SK2A==","size":{"width":134,"height":21}}$

Solun:- Let f(x) = ${"type":"$","id":"35-0","font":{"color":"#000000","family":"Arial","size":10},"code":"$\\frac{x\\tan x}{\\sec x+\\tan x}$","ts":1603021416773,"cs":"JPzsLTtnb+zFZUprPIYSnw==","size":{"width":62,"height":20}}$

Integrate f(x):-

${"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-1-1-0-1-0-1-1-1-1-1-2-1-1-1-1-1-0","code":"\\begin{align*}\n{I}&={\\int_{0}^{\\Pi}\\frac{x\\tan x}{\\sec x+\\tan x}.dx....\\left(1\\right)}\t\n\\end{align*}","font":{"family":"Arial","size":9.260959500653215,"color":"#000000"},"type":"align*","ts":1603021509869,"cs":"khlT8KNfL2Wx85c9I/MjIA==","size":{"width":222,"height":38}}$

By Property:-

${"font":{"family":"Arial","color":"#000000","size":10},"type":"$","id":"42","code":"$\\int_{a}^{b}f\\left(x\\right).dx=\\int_{a}^{b}f\\left(a+b-x\\right).dx$","ts":1602403552089,"cs":"KzhjKI+pmRpH9SU6RjucMg==","size":{"width":220,"height":21}}$

${"id":"36-0","code":"\\begin{align*}\n{I}&={\\int_{0}^{\\Pi}\\frac{\\left(\\Pi -x\\right)\\tan \\left(\\Pi -x\\right)}{\\sec \\left(\\Pi -x\\right)+\\tan \\left(\\Pi -x\\right)}.dx}\\\\\n{I}&={\\int_{0}^{\\Pi}\\frac{-\\left(\\Pi -x\\right)\\tan x}{-\\sec x-\\tan x}.dx}\\\\\n{I}&={\\int_{0}^{\\Pi}\\frac{-\\left(\\Pi -x\\right)\\tan x}{-\\left(\\sec x+\\tan x\\right)}.dx}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","ts":1603022149687,"cs":"IauQqMiC13ElQyASzmJvQw==","size":{"width":260,"height":128}}$

${"font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","code":"\\begin{align*}\n{I}&={\\int_{0}^{\\Pi}\\frac{\\left(\\Pi -x\\right)\\tan x}{\\sec x+\\tan x}.dx}\\\\\n{I}&={\\int_{0}^{\\Pi}\\frac{\\Pi \\tan x}{\\sec x+\\tan x}.dx-\\int_{0}^{\\Pi}\\frac{x\\tan x}{\\sec x+\\tan x}.dx...\\left(2\\right)}\t\n\\end{align*}","id":"37","ts":1603022176551,"cs":"v3R1oICkqAg8NKH9cyRtJQ==","size":{"width":380,"height":82}}$

Add eq. 1 and 2:-

${"code":"\\begin{align*}\n{2I}&={\\int_{0}^{\\Pi}\\frac{\\Pi \\tan x}{\\sec x+\\tan x}.dx-\\int_{0}^{\\Pi}\\frac{x\\tan x}{\\sec x+\\tan x}.dx+\\int_{0}^{\\Pi}\\frac{x\\tan x}{\\sec x+\\tan x}.dx}\\\\\n{2I}&={\\int_{0}^{\\Pi}\\frac{\\Pi \\tan x}{\\sec x+\\tan x}.dx}\\\\\n{2I}&={\\Pi \\int_{0}^{\\Pi}\\frac{\\tan x}{\\sec x+\\tan x}.dx}\\\\\n{I\\,\\,}&={\\frac{\\Pi }{2}\\int_{0}^{\\Pi}\\frac{\\frac{\\sin x}{\\cos x}}{\\frac{1}{\\cos x}+\\frac{\\sin x}{\\cos x}}.dx}\\\\\n{I\\,\\,}&={\\frac{\\Pi }{2}\\int_{0}^{\\Pi}\\frac{\\sin x}{1+\\sin x}.dx}\\\\\n{I\\,\\,}&={\\frac{\\Pi }{2}\\int_{0}^{\\Pi}\\frac{1+\\sin x-1}{1+\\sin x}.dx}\\\\\n{I\\,\\,}&={\\frac{\\Pi }{2}\\left[\\int_{0}^{\\Pi}\\frac{1+\\sin x}{1+\\sin x}.dx-\\int_{0}^{\\Pi}\\frac{1}{1+\\sin x}.dx\\right]}\\\\\n{I\\,\\,}&={\\frac{\\Pi }{2}\\left[\\int_{0}^{\\Pi}1.dx-\\int_{0}^{\\Pi}\\frac{1}{1+\\sin x}\\times\\frac{1-\\sin x}{1-\\sin x}.dx\\right]}\\\\\n{I\\,\\,}&={\\frac{\\Pi }{2}\\left[\\int_{0}^{\\Pi}1.dx-\\int_{0}^{\\Pi}\\frac{1-\\sin x}{1-\\sin ^{2}x}.dx\\right]}\\\\\n{I\\,\\,}&={\\frac{\\Pi }{2}\\left[\\int_{0}^{\\Pi}1.dx-\\int_{0}^{\\Pi}\\frac{1-\\sin x}{ \\cos^{2}x}.dx\\right]}\\\\\n{I\\,\\,}&={\\frac{\\Pi }{2}\\left[\\int_{0}^{\\Pi}1.dx-\\left[\\int_{0}^{\\Pi}\\sec^{2}x.dx-\\int_{0}^{\\Pi}\\sec x.\\tan x.dx\\right]\\right]}\\\\\n{I\\,\\,}&={\\frac{\\Pi }{2}\\left[\\int_{0}^{\\Pi}1.dx-\\int_{0}^{\\Pi}\\sec^{2}x.dx+\\int_{0}^{\\Pi}\\sec x.\\tan x.dx\\right]}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"id":"38","type":"align*","ts":1603024285341,"cs":"LQADV3NdumUg/kzk5Mavkg==","size":{"width":510,"height":532}}$

We know that:-

${"code":"$\\int_{}^{}\\sec^{2}x.dx=\\tan x+C$","font":{"color":"#000000","family":"Arial","size":10},"id":"39","type":"$","ts":1603024322313,"cs":"utuWBt5H8e5RUtdx9Yl9GQ==","size":{"width":164,"height":18}}$

${"code":"$\\int_{}^{}\\sec x.\\tan x.dx=\\sec x+C$","font":{"family":"Arial","color":"#000000","size":10},"type":"$","id":"39","ts":1603024359799,"cs":"NS4Y2LiA4y9Pym3m8IK5xw==","size":{"width":196,"height":17}}$

${"font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","id":"40","code":"\\begin{align*}\n{I\\,\\,}&={\\frac{\\Pi }{2}\\left[x-\\tan x+\\sec x\\right]_{0}^{\\Pi}}\\\\\n{I\\,\\,}&={\\frac{\\Pi }{2}\\left[\\left(\\Pi-\\tan \\Pi+\\sec \\Pi\\right)-\\left(0-\\tan 0+\\sec 0\\right)\\right]}\\\\\n{I\\,\\,}&={\\frac{\\Pi }{2}\\left[\\left(\\Pi-0-1\\right)-\\left(0-0+1\\right)\\right]}\t\n\\end{align*}","ts":1603024710480,"cs":"lpVkUms70/TTSZJmpr0tBQ==","size":{"width":340,"height":106}}$

${"type":"align*","id":"41","code":"\\begin{align*}\n{I\\,\\,}&={\\frac{\\Pi }{2}\\left[\\Pi-1-1\\right]}\\\\\n{I\\,\\,}&={\\frac{\\Pi }{2}\\left(\\Pi-2\\right)}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"ts":1603024815974,"cs":"Jy7ewusHiUWfMfSx7l5TBA==","size":{"width":128,"height":68}}$

${"font":{"family":"Arial","color":"#000000","size":10},"type":"$","id":"4-0-0-0-1-1-2-1-1-0-1-2-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-1-1-1-1-0","code":"$33.\\,\\int_{1}^{4}\\left[\\left|x-1\\right|+\\left|x-2\\right|+\\left|x-3\\right|\\right].dx$","ts":1603025002713,"cs":"liJquftcU9oEE1PupCarVg==","size":{"width":249,"height":20}}$

Solun:- Let f(x) = [|x - 1| + |x - 2| + |x - 3|]

Integrate f(x):-

${"font":{"color":"#000000","size":9.260959500653215,"family":"Arial"},"code":"\\begin{align*}\n{I}&={\\int_{1}^{4}\\left[\\left|x-1\\right|+\\left|x-2\\right|+\\left|x-3\\right|\\right].dx}\\\\\n{I}&={\\int_{1}^{4}\\left|x-1\\right|.dx+\\int_{1}^{4}\\left|x-2\\right|.dx+\\int_{1}^{4}\\left|x-3\\right|.dx}\\\\\n{I}&={I_{1}+I_{2}+I_{3}....\\left(1\\right)}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-1-1-0-1-0-1-1-1-1-1-2-1-1-1-1-1-1-0","type":"align*","ts":1603025232662,"cs":"Uv/UwbGXGE4oH8nOpv9FMg==","size":{"width":344,"height":102}}$

Calculating I1:-

${"font":{"size":10,"family":"Arial","color":"#000000"},"id":"36-1-0","code":"\\begin{align*}\n{I_{1}}&={\\int_{1}^{4}\\left|x-1\\right|.dx}\\\\\n{I_{1}}&={\\int_{1}^{4}\\left(x-1\\right).dx}\\\\\n{I_{1}}&={\\left[\\frac{x^{2}}{2}-x\\right]_{1}^{4}}\\\\\n{I_{1}}&={\\left[\\left(\\frac{16}{2}-4\\right)-\\left(\\frac{1}{2}-1\\right)\\right]}\\\\\n{I_{1}}&={\\left[4+\\frac{1}{2}\\right]}\\\\\n{I_{1}}&={\\frac{9}{2}}\t\n\\end{align*}","type":"align*","ts":1603025540543,"cs":"FvT7jvbPYgQG6S3l96ojaw==","size":{"width":208,"height":252}}$

Calculating I2:-

${"id":"36-1-1","type":"align*","code":"\\begin{align*}\n{I_{2}}&={\\int_{1}^{4}\\left|x-2\\right|.dx}\\\\\n{I_{2}}&={\\int_{1}^{2}-\\left(x-2\\right).dx+\\int_{2}^{4}\\left(x-2\\right).dx}\\\\\n{I_{2}}&={-\\int_{1}^{2}\\left(x-2\\right).dx+\\int_{2}^{4}\\left(x-2\\right).dx}\\\\\n{I_{2}}&={-\\left[\\frac{x^{2}}{2}-2x\\right]_{1}^{2}+\\left[\\frac{x^{2}}{2}-2x\\right]_{2}^{4}}\\\\\n{I_{2}}&={-\\left[\\left(\\frac{4}{2}-4\\right)-\\left(\\frac{1}{2}-2\\right)\\right]+\\left[\\left(\\frac{16}{2}-8\\right)-\\left(\\frac{4}{2}-4\\right)\\right]}\\\\\n{I_{2}}&={-\\left(\\frac{4}{2}-4\\right)+\\left(\\frac{1}{2}-2\\right)+\\left(\\frac{16}{2}-8\\right)-\\left(\\frac{4}{2}-4\\right)}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"ts":1603026030834,"cs":"n/5aThgYOSQLmNAv95wMjA==","size":{"width":408,"height":260}}$

${"type":"align*","code":"\\begin{align*}\n{I_{2}}&={-2\\left(\\frac{4}{2}-4\\right)-\\frac{3}{2}+0}\\\\\n{I_{2}}&={\\frac{5}{2}}\t\n\\end{align*}","id":"44","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1603026053433,"cs":"U+KCw6MC6ptX7q/B7i7D+Q==","size":{"width":181,"height":74}}$

Calculating I3:-

${"font":{"size":10,"family":"Arial","color":"#000000"},"id":"36-1-2","code":"\\begin{align*}\n{I_{3}}&={\\int_{1}^{4}\\left|x-3\\right|.dx}\\\\\n{I_{3}}&={\\int_{1}^{3}-\\left(x-3\\right).dx+\\int_{3}^{4}\\left(x-3\\right).dx}\\\\\n{I_{3}}&={-\\int_{1}^{3}\\left(x-3\\right).dx+\\int_{3}^{4}\\left(x-3\\right).dx}\\\\\n{I_{3}}&={-\\left[\\frac{x^{2}}{2}-3x\\right]_{1}^{3}+\\left[\\frac{x^{2}}{2}-3x\\right]_{3}^{4}}\\\\\n{I_{3}}&={-\\left[\\left(\\frac{9}{2}-9\\right)-\\left(\\frac{1}{2}-3\\right)\\right]+\\left[\\left(\\frac{16}{2}-12\\right)-\\left(\\frac{9}{2}-9\\right)\\right]}\\\\\n{I_{3}}&={-\\left(\\frac{9}{2}-9\\right)+\\left(\\frac{1}{2}-3\\right)+\\left(\\frac{16}{2}-12\\right)-\\left(\\frac{9}{2}-9\\right)}\\\\\n{I_{3}}&={-2\\left(\\frac{9}{2}-9\\right)+\\left(\\frac{-5}{2}\\right)+\\left(-4\\right)}\\\\\n{I_{3}}&={9-\\frac{5}{2}-4}\\\\\n{I_{3}}&={5-\\frac{5}{2}=\\frac{5}{2}}\t\n\\end{align*}","type":"align*","ts":1603026472001,"cs":"uMM0L1zSt1xxs2HCnriXpg==","size":{"width":416,"height":376}}$

Put the value of I1, I2, and I3 in eq. 1:-

${"id":"45","code":"\\begin{align*}\n{I}&={\\frac{9}{2}+\\frac{5}{2}+\\frac{5}{2}}\\\\\n{I}&={\\frac{9}{2}+5}\\\\\n{I}&={\\frac{19}{2}}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","ts":1603026623016,"cs":"tI5fkPtpomMf3zoimQeKVQ==","size":{"width":112,"height":105}}$

Prove the following (Exercises 34 to 39)

${"code":"$34.\\,\\int_{1}^{3}\\frac{dx}{x^{2}\\left(x+1\\right)}=\\frac{2}{3}+\\log_{}\\frac{2}{3}$","type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"id":"4-0-0-0-1-1-2-1-1-0-1-2-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-1-1-1-1-1-0-0","ts":1603096227383,"cs":"f1OXJOOjGqIbI7/10ZMaxQ==","size":{"width":181,"height":24}}$

Solun:- Let f(x) = ${"id":"46-0","type":"$","code":"$\\frac{1}{x^{2}\\left(x+1\\right)}$","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1603191219626,"cs":"PuhZ31Ssbd9F8Zi4Dqt30w==","size":{"width":46,"height":21}}$

Integrate f(x):-

${"font":{"color":"#000000","family":"Arial","size":9.260959500653215},"code":"\\begin{align*}\n{I}&={\\int_{1}^{3}\\frac{dx}{x^{2}\\left(x+1\\right)}}\t\n\\end{align*}","type":"align*","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-1-1-0-1-0-1-1-1-1-1-2-1-1-1-1-1-1-1-0","ts":1603096187756,"cs":"7zsQGePO+3miujvtMSCG9Q==","size":{"width":124,"height":40}}$

By Partial Fraction:-

${"code":"$\\frac{1}{x^{2}\\left(x+1\\right)}=\\frac{A}{x}+\\frac{B}{x^{2}}+\\frac{C}{\\left(x+1\\right)}$","id":"1-0-0-0-0-1","type":"$","font":{"size":12,"color":"#000000","family":"Arial"},"ts":1603096352425,"cs":"ugIJgwmPGu/qQuRN8pX2aw==","size":{"width":226,"height":29}}$

⇒ 1 = A(x)(x + 1) + B(x + 1) + C(x2)......(1)

Put x = 0 in eq. 1:-

⇒ 1 = B(1)

⇒ B = 1

Put x = -1 in eq. 1:-

⇒ 1 = C(1)

⇒ C = 1

Put x = 1 in eq. 1:-

⇒ 1 = A(1)(2) + B(2) + C(1)

⇒ 1 = A(1)(2) + 1(2) + 1(1)

⇒ 1 - 3 = A(1)(2)

⇒ A = - 1

${"font":{"size":10,"color":"#000000","family":"Arial"},"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-1-0-0-0-0-1-0","type":"align*","code":"\\begin{align*}\n{I}&={\\int_{}^{}\\left[\\frac{-1}{x}+\\frac{1}{x^{2}}+\\frac{1}{\\left(x+1\\right)}\\right].dx}\\\\\n{I}&={-\\int_{}^{}\\frac{1}{x}.dx+\\int_{}^{}\\frac{1}{x^{2}}.dx+\\int_{}^{}\\frac{1}{\\left(x+1\\right)}.dx}\t\n\\end{align*}","ts":1603096706044,"cs":"1Trs5hEKrD2raFuA1krYEA==","size":{"width":308,"height":80}}$

We know that:-

${"id":"5-0-0-1","type":"$","font":{"color":"#222222","family":"Arial","size":10},"code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}x+C$","ts":1602672753650,"cs":"17IVnzkv6uJxop03I6ksvg==","size":{"width":136,"height":18}}$

${"type":"align*","code":"\\begin{align*}\n{I}&={\\left[-\\log_{}x-\\frac{1}{x}+\\log_{}\\left|x+1\\right|\\right]_{1}^{3}}\\\\\n{I}&={\\left[\\log_{}\\left|\\frac{x+1}{x}\\right|-\\frac{1}{x}\\right]_{1}^{3}}\\\\\n{I}&={\\left[\\left(\\log_{}\\left|\\frac{4}{3}\\right|-\\frac{1}{3}\\right)-\\left(\\log_{}\\left|\\frac{2}{1}\\right|-\\frac{1}{1}\\right)\\right]}\\\\\n{I}&={\\log_{}\\left|\\frac{4}{3}\\right|-\\frac{1}{3}-\\log_{}\\left|2\\right|+1}\\\\\n{I}&={\\log_{}\\left|\\frac{\\frac{4}{3}}{2}\\right|+\\frac{2}{3}}\\\\\n{I}&={\\log_{}\\left|\\frac{2}{3}\\right|+\\frac{2}{3}}\t\n\\end{align*}","id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-0-0-0-1-0","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1603096982726,"cs":"/u58XKdscQhCbNPxcGS8LA==","size":{"width":278,"height":260}}$

${"id":"4-0-0-0-1-1-2-1-1-0-1-2-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-1-1-1-1-1-1","code":"$\\int_{1}^{3}\\frac{dx}{x^{2}\\left(x+1\\right)}=\\frac{2}{3}+\\log_{}\\frac{2}{3}$","font":{"family":"Arial","size":10,"color":"#000000"},"type":"$","ts":1603097037376,"cs":"yqot9znlF4UtDdUcr+5wHw==","size":{"width":156,"height":24}}$

Hence Proved...

${"id":"4-0-0-0-1-1-2-1-1-0-1-2-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-1-1-0-1","font":{"size":10,"color":"#000000","family":"Arial"},"type":"$","code":"$35.\\,\\int_{0}^{1}x.e^{x}.dx=1$","ts":1603191589729,"cs":"nQUoSDAJHASooO2LkPRi0A==","size":{"width":128,"height":21}}$

Solun:- Let f(x) = x.ex

Integrate f(x):-

${"font":{"size":9.260959500653215,"color":"#000000","family":"Arial"},"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-1-1-0-1-0-1-1-1-1-1-2-1-1-1-1-0-1","type":"align*","code":"\\begin{align*}\n{I}&={\\int_{0}^{1}x.e^{x}.dx}\t\n\\end{align*}","ts":1603191632734,"cs":"p9Ksl8h+Q8JL7poma/UTWw==","size":{"width":108,"height":38}}$

We know that:-

According to ILATE:- u = x and v = ex

${"code":"\\begin{align*}\n{I}&={\\left[x.\\int_{0}^{1}e^{x}.dx-\\int_{}^{}\\left[\\diff{x}{x}.\\int_{0}^{1}e^{x}.dx\\right].dx\\right]}\\\\\n{I}&={x.e^{x}-\\int_{0}^{1}e^{x}.dx}\\\\\n{I}&={\\left[x.e^{x}-e^{x}\\right]_{0}^{1}}\\\\\n{I}&={\\left[\\left(\\left(1\\right).e^{1}-e^{1}\\right)-\\left(\\left(0\\right).e^{0}-e^{0}\\right)\\right]}\\\\\n{I}&={0-\\left(0-1\\right)}\\\\\n{I}&={1}\t\n\\end{align*}","id":"54","type":"align*","font":{"family":"Arial","size":10,"color":"#222222"},"ts":1603191984816,"cs":"LoxG2mJvyJsrhXIQMF4enQ==","size":{"width":298,"height":173}}$

${"font":{"color":"#000000","family":"Arial","size":10},"type":"$","code":"$\\int_{0}^{1}x.e^{x}.dx=1$","id":"55","ts":1603192035766,"cs":"pvKKQ0NlQP5tfiXIlSHx7w==","size":{"width":104,"height":21}}$

Hence Proved…....

${"type":"$","id":"4-0-0-0-1-1-2-1-1-0-1-2-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-1-1-1-1-1-0-1-0-0","font":{"color":"#000000","size":10,"family":"Arial"},"code":"$36.\\,\\int_{-1}^{1}x^{17}.\\cos^{4}x.dx=0$","ts":1603098221174,"cs":"mndjmsOw1MoJ1vTyyKPgJQ==","size":{"width":170,"height":21}}$

Solun:- Let f(x) = x17.cos4x

Integrate f(x):-

${"code":"\\begin{align*}\n{I}&={\\int_{-1}^{1}x^{17}.\\cos^{4}x.dx}\t\n\\end{align*}","font":{"family":"Arial","size":9.260959500653215,"color":"#000000"},"type":"align*","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-1-1-0-1-0-1-1-1-1-1-2-1-1-1-1-1-1-1-1-0-0","ts":1603098262592,"cs":"QUxDHJpn4mxp7HfVAHJBAQ==","size":{"width":146,"height":40}}$

By Property:-

${"font":{"color":"#000000","family":"Arial","size":10},"id":"1-2-1-5-1","code":"\\begin{align*}\n{\\int_{-a}^{a}f\\left(x\\right).dx}&={2\\int_{0}^{a}f\\left(x\\right).dx,if\\,f\\left(-x\\right)=f\\left(x\\right)}\\\\\n{\\int_{-a}^{a}f\\left(x\\right).dx}&={0,if\\,f\\left(-x\\right)=-f\\left(x\\right)}\t\n\\end{align*}","type":"align*","ts":1602503490672,"cs":"ju0dUPVndQfGk7GL29ALLg==","size":{"width":309,"height":78}}$

f(-x) = (-x)17.[cos (-x)]4 = - x17.cos4x

f(-x) = - f(x)

${"type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-1-0-1-0-1-1-1-1-0","code":"\\begin{align*}\n{I}&={\\int_{-1}^{1}x^{17}.\\cos^{4}x.dx}\\\\\n{I}&={0}\t\n\\end{align*}","ts":1603098536652,"cs":"Sy1DFL1XwZQEj44Zui9MTA==","size":{"width":146,"height":60}}$

${"font":{"family":"Arial","size":10,"color":"#000000"},"type":"$","id":"4-0-0-0-1-1-2-1-1-0-1-2-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-1-1-1-1-1-0-1-0-1","code":"$\\int_{-1}^{1}x^{17}.\\cos^{4}x.dx=0$","ts":1603098565357,"cs":"wdD3tfD0n5F4E7dLLo2BzQ==","size":{"width":145,"height":21}}$

Hence Proved….

${"code":"$37.\\,\\int_{0}^{\\frac{\\Pi}{2}}\\sin^{3}x.dx=\\frac{2}{3}$","font":{"size":10,"color":"#000000","family":"Arial"},"id":"4-0-0-0-1-1-2-1-1-0-1-2-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-1-1-1-1-1-0-1-0-2-0","type":"$","ts":1603098730110,"cs":"4+R2fwOCimt+QZ0yM0eYig==","size":{"width":145,"height":24}}$

Solun:- Let f(x) = sin3x

Integrate f(x):-

${"code":"\\begin{align*}\n{I}&={\\int_{0}^{\\frac{\\Pi}{2}}\\sin^{3}x.dx}\t\n\\end{align*}","type":"align*","font":{"size":9.260959500653215,"family":"Arial","color":"#000000"},"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-1-1-0-1-0-1-1-1-1-1-2-1-1-1-1-1-1-1-1-0-1-0","ts":1603098769160,"cs":"ylvTyAkIzfKVSmlACmsTpw==","size":{"width":120,"height":40}}$

We know that:-

sin 3x = 3sinx - 4sin3x

${"type":"align*","font":{"color":"#000000","size":9.260959500653215,"family":"Arial"},"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-1-1-0-1-0-1-1-1-1-1-2-1-1-1-1-1-1-1-1-0-1-1","code":"\\begin{align*}\n{I}&={\\int_{0}^{\\frac{\\Pi}{2}}\\frac{\\left(3\\sin x-\\sin3x\\right)}{4}.dx}\\\\\n{I}&={\\frac{1}{4}\\int_{0}^{\\frac{\\Pi}{2}}\\left(3\\sin x-\\sin3x\\right).dx}\t\n\\end{align*}","ts":1603098888871,"cs":"qmqv/bQIl0VnCqCxgBs9iA==","size":{"width":210,"height":88}}$

We know that:-

${"font":{"color":"#000000","family":"Arial","size":10},"id":"4-0-0-0-1-1-2-1-1-0-1-2-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-1-1-1-1-1-0-1-0-3-0","code":"$\\int_{}^{}\\sin x.dx=-\\cos x+C$","type":"$","ts":1603098939505,"cs":"gEJqgIkzl7vYM0LE+d+dQg==","size":{"width":169,"height":17}}$

${"code":"\\begin{align*}\n{I}&={\\frac{1}{4}\\left[-3\\cos x-\\frac{\\left(-\\cos3x\\right)}{3}\\right]_{0}^{\\frac{\\Pi}{2}}}\\\\\n{I}&={\\frac{1}{4}\\left[-3\\cos x+\\frac{\\cos3x}{3}\\right]_{0}^{\\frac{\\Pi}{2}}}\\\\\n{I}&={\\frac{1}{4}\\left[\\left(-3\\cos \\frac{\\Pi}{2}+\\frac{\\cos\\frac{3\\Pi}{2}}{3}\\right)-\\left(-3\\cos 0+\\frac{\\cos0}{3}\\right)\\right]}\\\\\n{I}&={\\frac{1}{4}\\left[\\left(0+0\\right)-\\left(-3\\times1+\\frac{1}{3}\\right)\\right]}\\\\\n{I}&={\\frac{1}{4}\\left(3-\\frac{1}{3}\\right)}\\\\\n{I}&={\\frac{2}{3}}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-1-0-1-0-1-1-1-1-1","font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","ts":1603099953443,"cs":"W8uVRN3wMUcluGTld6D9SA==","size":{"width":380,"height":268}}$

${"type":"$","code":"$\\int_{0}^{\\frac{\\Pi}{2}}\\sin^{3}x.dx=\\frac{2}{3}$","font":{"color":"#000000","size":10,"family":"Arial"},"id":"4-0-0-0-1-1-2-1-1-0-1-2-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-1-1-1-1-1-0-1-0-2-1","ts":1603100005566,"cs":"F0RKvlP5+QMJcEWte3Lofw==","size":{"width":120,"height":24}}$

Hence Proved…

${"code":"$38.\\,\\int_{0}^{\\frac{\\Pi}{4}}2\\tan^{3}x.dx=1-\\log_{}2$","font":{"size":10,"family":"Arial","color":"#000000"},"id":"4-0-0-0-1-1-2-1-1-0-1-2-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-1-1-1-1-1-0-1-0-2-2-0","type":"$","ts":1603100146879,"cs":"ms5b4RkdAAOK4FCyYjvGLg==","size":{"width":205,"height":24}}$

Solun:- Let f(x) = 2tan3x

Integrate f(x):-

${"type":"align*","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-1-1-0-1-0-1-1-1-1-1-2-1-1-1-1-1-1-1-1-0-1-2-0","font":{"size":9.260959500653215,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{I}&={\\int_{0}^{\\frac{\\Pi}{4}}2\\tan^{3}x.dx}\\\\\n{I}&={\\int_{0}^{\\frac{\\Pi}{4}}2\\tan x.\\tan^{2}x.dx}\\\\\n{I}&={\\int_{0}^{\\frac{\\Pi}{4}}2\\tan x.\\left(\\sec^{2}x-1\\right).dx}\\\\\n{I}&={\\int_{0}^{\\frac{\\Pi}{4}}2\\tan x.\\sec^{2}x.dx-\\int_{0}^{\\frac{\\Pi}{4}}2\\tan x.dx}\\\\\n{I}&={2\\int_{0}^{\\frac{\\Pi}{4}}\\tan x.\\sec^{2}x.dx-2\\int_{0}^{\\frac{\\Pi}{4}}\\tan x.dx}\t\n\\end{align*}","ts":1603100274034,"cs":"qaIZbZFima5dTWn3snVq/w==","size":{"width":293,"height":225}}$

Let tan x = t…....(1)

Differentiate w.r.t. to t:-

⇒ sec2x.dx = dt

Limits Change:-

Put x = 0 in eq. 1:-

⇒ t = 0

Put x = π/4 in eq. 1:-

⇒ t = 1

${"type":"align*","code":"\\begin{align*}\n{I}&={2\\int_{0}^{1}t.dt-2\\left[\\log_{}\\left|\\sec x\\right|\\right]_{0}^{\\frac{\\Pi}{4}}}\\\\\n{I}&={2\\int_{0}^{1}t.dt-2\\left[\\left(\\log_{}\\left|\\sec \\frac{\\Pi}{4}\\right|\\right)-\\left(\\log_{}\\left|\\sec 0\\right|\\right)\\right]}\\\\\n{I}&={2\\int_{0}^{1}t.dt-2\\left[\\left(\\log_{}\\left|{\\sqrt[]{2}}\\right|\\right)-\\left(\\log_{}\\left|1\\right|\\right)\\right]}\\\\\n{I}&={2\\int_{0}^{1}t.dt-2\\left[\\left(\\log_{}\\left|{\\sqrt[]{2}}\\right|\\right)-0\\right]}\\\\\n{I}&={2\\int_{0}^{1}t.dt-2\\left(\\frac{1}{2}\\log_{}\\left|2\\right|\\right)}\\\\\n{I}&={2\\int_{0}^{1}t.dt-\\log_{}\\left|2\\right|}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-1-0-1-0","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1603101506359,"cs":"oO2EE/WhlhgH8qWFSE1Uig==","size":{"width":329,"height":260}}$

We know that:-

${"type":"$","code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+C$","font":{"family":"Arial","size":10,"color":"#222222"},"id":"6-2","ts":1602674707519,"cs":"EFUU5DnUCg9ueRyENZvABA==","size":{"width":136,"height":21}}$

${"code":"\\begin{align*}\n{I}&={2\\left[\\frac{t^{2}}{2}\\right]_{0}^{1}-\\log_{}\\left|2\\right|}\\\\\n{I}&={2\\left[\\frac{1}{2}-0\\right]-\\log_{}\\left|2\\right|}\\\\\n{I}&={1-\\log_{}\\left|2\\right|}\t\n\\end{align*}","type":"align*","id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-1-1-1-0","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1603101569962,"cs":"MBEPbw2Z3P4ZMdMlVDilzQ==","size":{"width":156,"height":105}}$

${"code":"$\\int_{0}^{\\frac{\\Pi}{4}}2\\tan^{3}x.dx=1-\\log_{}2$","id":"4-0-0-0-1-1-2-1-1-0-1-2-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-1-1-1-1-1-0-1-0-2-2-1","font":{"color":"#000000","size":10,"family":"Arial"},"type":"$","ts":1603101601584,"cs":"uHABL1JRIGLOumFDE40LBA==","size":{"width":180,"height":24}}$

Hence Proved…

${"font":{"size":10,"color":"#000000","family":"Arial"},"code":"$39.\\,\\int_{0}^{1}\\sin^{-1}x.dx=\\frac{\\Pi}{2}-1$","type":"$","id":"4-0-0-0-1-1-2-1-1-0-1-2-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-1-1-1-1-1-0-1-0-2-2-2-0","ts":1603101686847,"cs":"DZW2PT29cnU0EatxPJpI4Q==","size":{"width":178,"height":21}}$

Solun:- Let f(x) = sin-1x

Integrate f(x):-

${"type":"align*","code":"\\begin{align*}\n{I}&={\\int_{0}^{1}\\sin^{-1}x.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-1-1-0-1-0-1-1-1-1-1-2-1-1-1-1-1-1-1-1-0-1-2-1-0","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1603101819857,"cs":"XAZQNvjxQqrWZ5re7za2Yg==","size":{"width":124,"height":38}}$

Let sin-1x = t and x = sin t

Differentiate w.r.t. to t:-

⇒ dx = cos t.dt

Limits Change:-

Put x = 0 in eq. 1:-

⇒ t = 0

Put x = 1 in eq. 1:-

⇒ t = π/2

${"type":"align*","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-1-0-1-1-0","font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{I}&={\\int_{0}^{\\frac{\\Pi}{2}}t.\\cos t.dt}\t\n\\end{align*}","ts":1603102044612,"cs":"u08rAEwc3JoPNC9YHNcEOA==","size":{"width":122,"height":40}}$

We know that:-

${"font":{"size":10,"color":"#222222","family":"Arial"},"id":"5-1-0-1-0-0-1-0-1-1-1-1-1-1-0-1-2","type":"$","code":"$\\int_{}^{}\\left(u.v\\right).dx=u\\int_{}^{}v.dx-\\int_{}^{}\\left[\\diff{u}{x}\\int_{}^{}v.dx\\right].dx+C$","ts":1602925083002,"cs":"nh9ktQ317elAK1DbSQIrTg==","size":{"width":318,"height":20}}$

According to ILATE:- u = t and v = cos t

${"font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{I}&={t\\int_{0}^{\\frac{\\Pi}{2}}\\cos t.dt-\\int_{0}^{\\frac{\\Pi}{2}}\\left[\\diff{t}{t}\\int_{0}^{\\frac{\\Pi}{2}}\\cos t.dt\\right].dt}\t\n\\end{align*}","type":"align*","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-1-0-1-2-0","ts":1603102250448,"cs":"uPk4sXMBl6ochnB404YJMA==","size":{"width":309,"height":46}}$

We know that:-

${"id":"6-3","type":"$","code":"$\\int_{}^{}\\cos x.dx=\\sin x+C$","font":{"family":"Arial","size":10,"color":"#222222"},"ts":1603102277223,"cs":"hYbEp2frHu+LTjwRXoXILQ==","size":{"width":154,"height":17}}$

${"font":{"family":"Arial","size":10,"color":"#222222"},"code":"$\\int_{}^{}\\sin x.dx=-\\cos x+C$","type":"$","id":"6-3","ts":1603102624669,"cs":"tNtY7sPAvDJrET5bmvqUlA==","size":{"width":169,"height":17}}$

${"type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{I}&={t.\\sin t-\\int_{0}^{\\frac{\\Pi}{2}}\\sin t.dt}\\\\\n{I}&={\\left[t.\\sin t+\\cos t\\right]_{0}^{\\frac{\\Pi}{2}}}\\\\\n{I}&={\\left[\\left(\\frac{\\Pi}{2}.\\sin \\frac{\\Pi}{2}+\\cos \\frac{\\Pi}{2}\\right)-\\left(0.\\sin 0+\\cos 0\\right)\\right]}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-1-0-1-2-1","ts":1603102937650,"cs":"zDxEdbATR6Z2vJxoKeBgAA==","size":{"width":325,"height":112}}$

${"type":"align*","id":"3-0-0-0-0-0-0-1-0-0-0-0-0-0-0-1-1-1-1","code":"\\begin{align*}\n{I}&={\\left[\\left(\\frac{\\Pi}{2}\\times1+0\\right)-1\\right]}\\\\\n{I}&={\\frac{\\Pi}{2}-1}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"ts":1603102968784,"cs":"1g8rPsUV0M2PQXXHHoT/mg==","size":{"width":169,"height":74}}$

${"code":"$\\int_{0}^{1}\\sin^{-1}x.dx=\\frac{\\Pi}{2}-1$","type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"id":"4-0-0-0-1-1-2-1-1-0-1-2-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-1-1-1-1-1-0-1-0-2-2-2-1","ts":1603102991265,"cs":"uWbK70gQRfClq3g3JL6Acw==","size":{"width":153,"height":21}}$

Hence Proved…

Choose the correct answer in Exercises 41 to 44.

${"code":"$41.\\,\\int_{}^{}\\frac{dx}{e^{x}+e^{-x}}$","type":"$","font":{"family":"Arial","size":12,"color":"#000000"},"id":"4-0-0-0-1-1-2-1-1-0-1-2-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-1-1-1-1-1-0-1-0-2-2-2-2-0","ts":1603103226802,"cs":"1e1jR6y6ORm7dszfOzDLng==","size":{"width":104,"height":25}}$ is equal to

Solun:- Let f(x) = ${"id":"47-0","code":"$\\frac{1}{e^{x}+e^{-x}}$","font":{"color":"#000000","family":"Arial","size":10},"type":"$","ts":1603103319707,"cs":"UcC76GeXveTQ2aHfhb6m9Q==","size":{"width":41,"height":20}}$

Integrate f(x):-

${"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-1-1-0-1-0-1-1-1-1-1-2-1-1-1-1-1-1-1-1-0-1-2-1-1-0","type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{dx}{e^{x}+e^{-x}}}\\\\\n{I}&={\\int_{}^{}\\frac{1}{e^{x}+\\frac{1}{e^{x}}}.dx}\\\\\n{I}&={\\int_{}^{}\\frac{1}{\\frac{e^{2x}+1}{e^{x}}}.dx}\\\\\n{I}&={\\int_{}^{}\\frac{e^{x}}{\\left(e^{x}\\right)^{2}+1}.dx}\t\n\\end{align*}","ts":1603103656491,"cs":"6b/wn6MCN6Qg0BxU+dredQ==","size":{"width":140,"height":172}}$

Let ex = t

Differentiate w.r.t. to t:-

⇒ ex.dx = dt

${"font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{dt}{1+t^{2}}}\t\n\\end{align*}","type":"align*","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-1-0-1-1-1-0-0","ts":1603104295835,"cs":"xVvRV0DKdUFrZtycKrj9Kg==","size":{"width":94,"height":36}}$

We know that:-

${"id":"5-1-0-1-0-0-1-0-1-1-1-1-1-1-0-1-3-0","font":{"color":"#222222","family":"Arial","size":10},"code":"$\\int_{}^{}\\frac{1}{1+x^{2}}.dx=\\tan^{-1}x+C$","type":"$","ts":1603104352031,"cs":"ca8n2STBk4gpLKcACd1NCA==","size":{"width":172,"height":20}}$

${"code":"\\begin{align*}\n{I}&={\\tan^{-1}t+C}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-1-0-1-1-1-1-0","font":{"family":"Arial","color":"#000000","size":10},"type":"align*","ts":1603104403452,"cs":"pGZIrwYGHRqwbFh+13K9Fg==","size":{"width":106,"height":17}}$

Put the value of t:-

${"code":"\\begin{align*}\n{I}&={\\tan^{-1}\\left(e^{x}\\right)+C}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-1-0-1-1-1-1-1","font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","ts":1603104449316,"cs":"HlawghVOvUZeFWWWKbHIxQ==","size":{"width":128,"height":17}}$

The correct answer is A.

${"type":"$","id":"4-0-0-0-1-1-2-1-1-0-1-2-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-1-1-1-1-1-0-1-0-2-2-2-2-1","font":{"color":"#000000","family":"Arial","size":12},"code":"$42.\\,\\int_{}^{}\\frac{\\cos2x}{\\left(\\sin x+\\cos x\\right)^{2}}.dx$","ts":1603104740255,"cs":"G86Ww+v+W2qekJo7yMnSdQ==","size":{"width":176,"height":32}}$ is equal to

Solun:- Let f(x) = ${"type":"$","code":"$\\frac{\\cos2x}{\\left(\\sin x+\\cos x\\right)^{2}}$","id":"47-1","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1603104791144,"cs":"qj9+0UPLohGvz/OENkWKXw==","size":{"width":73,"height":24}}$

Integrate f(x):-

${"font":{"color":"#000000","size":10,"family":"Arial"},"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-1-1-0-1-0-1-1-1-1-1-2-1-1-1-1-1-1-1-1-0-1-2-1-1-1-0-0","type":"align*","code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{\\cos2x}{\\left(\\sin x+\\cos x\\right)^{2}}.dx}\t\n\\end{align*}","ts":1603104865657,"cs":"ny3EvaOh8eL2/9/IyYSGvQ==","size":{"width":181,"height":40}}$

We know that:-

cos 2x = cos2x - sin2x

${"type":"align*","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-1-1-0-1-0-1-1-1-1-1-2-1-1-1-1-1-1-1-1-0-1-2-1-1-1-1","font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{\\cos^{2}x-\\sin^{2}x}{\\left(\\sin x+\\cos x\\right)^{2}}.dx}\\\\\n{I}&={\\int_{}^{}\\frac{\\left(\\cos x-\\sin x\\right)\\left(\\cos x+\\sin x\\right)}{\\left(\\sin x+\\cos x\\right)^{2}}.dx}\\\\\n{I}&={\\int_{}^{}\\frac{\\left(\\cos x-\\sin x\\right)}{\\left(\\sin x+\\cos x\\right)}.dx}\t\n\\end{align*}","ts":1603105093339,"cs":"fWNeBFkDoI6Ay1vDfv2Kmw==","size":{"width":270,"height":132}}$

Let sin x + cos x = t

Differentiate w.r.t. to t:-

⇒ (cos x - sin x).dx = dt

${"code":"\\begin{align*}\n{I}&={\\int_{}^{}\\frac{dt}{t}}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-1-0-1-1-1-0-1","type":"align*","ts":1603105178589,"cs":"7F5mwQxIYAJaym5ooOglLg==","size":{"width":69,"height":36}}$

We know that:-

${"code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}x+C$","font":{"family":"Arial","size":10,"color":"#222222"},"id":"5-1-0-1-0-0-1-0-1-1-1-1-1-1-0-1-3-1","type":"$","ts":1603105204431,"cs":"ta6r/Q88B9FGkrmIge933w==","size":{"width":136,"height":18}}$

${"code":"\\begin{align*}\n{I}&={\\log_{}\\left|t\\right|+C}\t\n\\end{align*}","type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-1-0-1-1-1-1-2","ts":1603105260454,"cs":"1N/czbC60KCVNp6fSqwCQA==","size":{"width":98,"height":16}}$

Put the value of t:-

${"code":"\\begin{align*}\n{I}&={\\log_{}\\left|\\sin x+\\cos x\\right|+C}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-1-0-1-1-1-1-3","type":"align*","ts":1603105296261,"cs":"lv7cY9SAxXI05kRnCMBzIA==","size":{"width":176,"height":16}}$

The correct answer is B.

43. If f(a+b-x) = f(x), then ${"id":"4-0-0-0-1-1-2-1-1-0-1-2-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-1-1-1-1-1-0-1-0-2-2-2-2-2-0","code":"$\\int_{a}^{b}xf\\left(x\\right).dx$","font":{"color":"#000000","size":10,"family":"Arial"},"type":"$","ts":1603105560617,"cs":"vUudzvW3U/tp7HLE1I7nVg==","size":{"width":81,"height":21}}$ is equal to

Solun:- Given

${"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-1-1-0-1-0-1-1-1-1-1-2-1-1-1-1-1-1-1-1-0-1-2-1-1-1-0-1-0","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","code":"\\begin{align*}\n{I}&={\\int_{a}^{b}xf\\left(x\\right).dx....\\left(1\\right)}\t\n\\end{align*}","ts":1603105940862,"cs":"KhLWPSkkptqjV9m7F1zMWA==","size":{"width":164,"height":38}}$

By Property:-

${"font":{"family":"Arial","color":"#000000","size":10},"type":"$","id":"51","code":"$\\int_{a}^{b}f\\left(x\\right).dx=\\int_{a}^{b}f\\left(a+b-x\\right).dx$","ts":1602403552089,"cs":"YmBBP+xtCbzJBNVpNLG+Qg==","size":{"width":220,"height":21}}$

${"code":"\\begin{align*}\n{I}&={\\int_{a}^{b}\\left(a+b-x\\right).f\\left(a+b-x\\right).dx}\t\n\\end{align*}","type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"id":"49-0","ts":1603105800139,"cs":"C1WHR5BRKVcbH8hqmW6yog==","size":{"width":240,"height":38}}$

Given f(a+b-x) = f(x)

${"id":"49-1","code":"\\begin{align*}\n{I}&={\\int_{a}^{b}\\left(a+b-x\\right).f\\left(x\\right).dx}\\\\\n{I}&={\\int_{a}^{b}\\left(a+b\\right).f\\left(x\\right).dx-\\int_{a}^{b}x.f\\left(x\\right).dx....\\left(2\\right)}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"type":"align*","ts":1603105886260,"cs":"g9oIPn2h5j4iqy4cwssSdA==","size":{"width":321,"height":82}}$

Add eq. 1 and 2:-

${"font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","code":"\\begin{align*}\n{2I}&={\\left(a+b\\right)\\int_{a}^{b} f\\left(x\\right).dx}\\\\\n{I\\,\\,}&={\\frac{\\left(a+b\\right)}{2}\\int_{a}^{b} f\\left(x\\right).dx}\t\n\\end{align*}","id":"50-0","ts":1603106016682,"cs":"Qygxr4vMglZAQC7kp6eX1g==","size":{"width":172,"height":82}}$

The correct answer is D.

44. The value of ${"id":"4-0-0-0-1-1-2-1-1-0-1-2-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-1-1-1-1-1-0-1-0-2-2-2-2-2-1","font":{"size":10,"color":"#000000","family":"Arial"},"code":"$\\int_{0}^{1}\\tan^{-1}\\left(\\frac{2x-1}{1+x-x^{2}}\\right).dx$","type":"$","ts":1603106470490,"cs":"hQej3OkNK7FycjynuHc9jA==","size":{"width":150,"height":28}}$ is

Solun:- Given

${"font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{I}&={\\int_{0}^{1}\\tan^{-1}\\left(\\frac{2x-1}{1+x-x^{2}}\\right).dx}\\\\\n{I}&={\\int_{0}^{1}\\tan^{-1}\\left(\\frac{x-\\left(1-x\\right)}{1+x\\left(1-x\\right)}\\right).dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-1-1-0-0-1-0-1-1-1-1-0-1-1-1-1-1-0-1-0-1-1-1-1-1-2-1-1-1-1-1-1-1-1-0-1-2-1-1-1-0-1-1","type":"align*","ts":1603106586947,"cs":"0Ymxafe3SF20Qr/abX/RaQ==","size":{"width":233,"height":84}}$

We know that:-

${"id":"52","code":"$\\tan^{-1}\\left(\\frac{A-B}{1+AB}\\right).dx=\\tan^{-1}\\left(A\\right)-\\tan^{-1}\\left(B\\right)$","type":"$","font":{"family":"Arial","color":"#000000","size":10},"ts":1603106700073,"cs":"6rZ72XZlC9pa8dtNfEIZWg==","size":{"width":288,"height":20}}$

${"id":"49-2-0","font":{"color":"#000000","family":"Arial","size":10},"type":"align*","code":"\\begin{align*}\n{I}&={\\int_{0}^{1}\\left[\\tan^{-1}x-\\tan^{-1}\\left(1-x\\right)\\right].dx.....\\left(1\\right)}\t\n\\end{align*}","ts":1603106776124,"cs":"ezHozOl7iJyovNiN5snncg==","size":{"width":302,"height":38}}$

By Property:-

${"code":"\\begin{align*}\n{I}&={\\int_{0}^{1}\\left[\\tan^{-1}\\left(1-x\\right)-\\tan^{-1}\\left(1-\\left(1-x\\right)\\right)\\right].dx}\\\\\n{I}&={\\int_{0}^{1}\\left[\\tan^{-1}\\left(1-x\\right)-\\tan^{-1}\\left(x\\right)\\right].dx.....\\left(2\\right)}\t\n\\end{align*}","id":"49-2-1","type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1603106866975,"cs":"eFVZI2rjGV5vzjeyZjcOcg==","size":{"width":326,"height":82}}$

Add eq. 1 and 2:-

${"type":"align*","id":"50-1","font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{2I}&={\\int_{0}^{1}\\left[\\tan^{-1}\\left(x\\right)-\\tan^{-1}\\left(1-x\\right)\\right].dx+\\int_{0}^{1}\\left[\\tan^{-1}\\left(1-x\\right)-\\tan^{-1}\\left(x\\right)\\right].dx}\\\\\n{I\\,\\,}&={\\int_{0}^{1}\\left[\\tan^{-1}\\left(x\\right)-\\tan^{-1}\\left(1-x\\right)+\\tan^{-1}\\left(1-x\\right)-\\tan^{-1}\\left(x\\right)\\right].dx}\\\\\n{I\\,\\,}&={0}\t\n\\end{align*}","ts":1603106993094,"cs":"BdDSqNdubs23++auziOp8Q==","size":{"width":517,"height":104}}$

The correct answer is B.

Download PDF of Miscellaneous Exercise

Important Note

Integrals(Miscellaneous Ex.)

Miscellaneous Exercise

Integrate the functions in Exercises 1 to 24.

Evaluate the definite integrals in Exercises 25 to 33.

Prove the following (Exercises 34 to 39)

Hence Proved...

Hence Proved….

Hence Proved…

Hence Proved…

Hence Proved…

Choose the correct answer in Exercises 41 to 44.

is equal to

is equal to

43. If f(a+b-x) = f(x), then is equal to

44. The value of is

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