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Exercise 7.10

Evaluate the integrals in Exercises 1 to 8 using substitution

{"type":"$","font":{"color":"#000000","family":"Arial","size":10},"code":"$1.\\,\\int_{0}^{1}\\frac{x}{x^{2}+1}.dx$","id":"4-0-0","ts":1602230433486,"cs":"saMJ9qF1UFPc2BKiUM0qUw==","size":{"width":93,"height":22}}

Solun:- Let f(x) = {"type":"$","font":{"color":"#000000","family":"Arial","size":10},"id":"1-0","code":"$\\frac{x}{x^{2}+1}$","ts":1602230473325,"cs":"plgfpFio6lqRMy+9XvYKrA==","size":{"width":32,"height":17}}

Integrate f(x):-

{"font":{"color":"#000000","family":"Arial","size":10},"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-0","code":"$I=\\int_{0}^{1}\\frac{x}{x^{2}+1}.dx$","type":"$","ts":1602230537727,"cs":"SX4H2dOCg7OYitLJLvwLaA==","size":{"width":104,"height":22}}

Let x2+1 = t……...(1)

Differentiate w.r.t. to t:-

⇒ 2x.dx = dt

⇒ x.dx = (1/2)dt

Limits Change:

Put x = 0 in eq. 1:-

⇒ t = 1

Put x = 1 in eq. 1:-

⇒ t = 2

{"font":{"color":"#000000","family":"Arial","size":10},"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-1","code":"\\begin{align*}\n{I}&={\\int_{1}^{2}\\frac{1}{t}.\\frac{dt}{2}}\\\\\n{I}&={\\frac{1}{2}\\int_{1}^{2}\\frac{1}{t}.dt}\t\n\\end{align*}","type":"align*","ts":1602231610441,"cs":"CGTU06P+aUGtgm/5zE+OJg==","size":{"width":108,"height":82}}

We know that:-

{"code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}x+C$","type":"$","font":{"size":10,"color":"#000000","family":"Arial"},"id":"3-0-0-0-0-0-0-0-0-0-0-0","ts":1602231695903,"cs":"ejhJcjEnSRRcd6Hq7pTpog==","size":{"width":136,"height":18}}

{"type":"align*","id":"3-0-0-0-0-0-0-1-0-0-0-0-0","font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{I}&={\\frac{1}{2}\\left[\\log_{}\\left|t\\right|\\right]_{1}^{2}}\\\\\n{I}&={\\frac{1}{2}\\left[\\left(\\log_{}\\left|2\\right|\\right)-\\left(\\log_{}\\left|1\\right|\\right)\\right]}\\\\\n{I}&={\\frac{1}{2}\\log_{}\\left|2\\right|}\t\n\\end{align*}","ts":1602231801731,"cs":"yyvDfU8Hp4mtMnYPTzMmQQ==","size":{"width":177,"height":105}}

{"type":"$","font":{"color":"#000000","size":10,"family":"Arial"},"code":"$2.\\,\\int_{0}^{\\frac{\\Pi}{2}}{\\sqrt[]{\\sin\\phi}}\\cos^{5}\\phi.d\\phi$","id":"4-0-1-0","ts":1602231971048,"cs":"fTwwGtBFDhgVN7BHQZGcmQ==","size":{"width":156,"height":24}}

Solun:- Let f(ɸ) = {"code":"${\\sqrt[]{\\sin\\phi}}\\cos^{5}\\phi$","id":"1-1-0","type":"$","font":{"color":"#000000","family":"Arial","size":10},"ts":1602231996278,"cs":"jiFyn5QPBhM0X/aXKlskAg==","size":{"width":89,"height":20}}

Integrate f(x):-

{"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-2-0","font":{"color":"#000000","size":8.333333333333334,"family":"Arial"},"type":"align*","code":"\\begin{align*}\n{I}&={\\int_{0}^{\\frac{\\Pi}{2}}{\\sqrt[]{\\sin\\phi}}\\cos^{5}\\phi.d\\phi}\\\\\n{I}&={\\int_{0}^{\\frac{\\Pi}{2}}{\\sqrt[]{\\sin\\phi}}\\cos^{4}\\phi.\\cos\\phi.d\\phi}\\\\\n{I}&={\\int_{0}^{\\frac{\\Pi}{2}}{\\sqrt[]{\\sin\\phi}}\\left(\\cos^{2}\\phi\\right)^{2}.\\cos\\phi.d\\phi}\\\\\n{I}&={\\int_{0}^{\\frac{\\Pi}{2}}\\left[{\\sqrt[]{\\sin\\phi}}\\left(1-\\sin^{2}\\phi\\right)^{2}\\right].\\cos\\phi.d\\phi}\t\n\\end{align*}","ts":1602233220768,"cs":"Jxx2CXRURHezoHPS/8gMag==","size":{"width":270,"height":150}}

Let sin ɸ = t……...(1)

Differentiate w.r.t. to t:-

⇒ cos ɸ.dɸ = dt

Limits Change:

Put ɸ = 0 in eq. 1:-

⇒ t = 0

Put ɸ = π/2 in eq. 1:-

⇒ t = 1

{"type":"align*","code":"\\begin{align*}\n{I}&={\\int_{0}^{1}\\left[{\\sqrt[]{t}}\\left(1-t^{2}\\right)^{2}\\right].dt}\\\\\n{I}&={\\int_{0}^{1}\\left[{\\sqrt[]{t}}\\left(1-2t^{2}+t^{4}\\right)\\right].dt}\\\\\n{I}&={\\int_{0}^{1}\\left[\\left({\\sqrt[]{t}}-2t^{2}{\\sqrt[]{t}}+t^{4}{\\sqrt[]{t}}\\right)\\right].dt}\\\\\n{I}&={\\int_{0}^{1}\\left(t^{\\frac{1}{2}}-2t^{\\frac{5}{2}}+t^{\\frac{9}{2}}\\right).dt}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-3-0","ts":1602234215370,"cs":"UgtfFKwPRW44SjPbMvO9iA==","size":{"width":234,"height":170}}

We know that:-

{"type":"$","code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+C$","font":{"size":10,"family":"Arial","color":"#000000"},"id":"3-0-0-0-0-0-0-0-0-0-0-1-0","ts":1602234256655,"cs":"I4wnA3dfs29TmQGeOthGFw==","size":{"width":136,"height":21}}

{"id":"3-0-0-0-0-0-0-1-0-0-0-0-1-0","code":"\\begin{align*}\n{I}&={\\left[\\frac{t^{\\frac{3}{2}}}{\\frac{3}{2}}-\\frac{2t^{\\frac{7}{2}}}{\\frac{7}{2}}+\\frac{t^{\\frac{11}{2}}}{\\frac{11}{2}}\\right]_{0}^{1}}\\\\\n{I}&={\\left[\\frac{2t^{\\frac{3}{2}}}{3}-\\frac{4t^{\\frac{7}{2}}}{7}+\\frac{2t^{\\frac{11}{2}}}{11}\\right]_{0}^{1}}\\\\\n{I}&={\\left[\\left(\\frac{2}{3}-\\frac{4}{7}+\\frac{2}{11}\\right)-0\\right]}\\\\\n{I}&={\\frac{64}{231}}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","ts":1602238601458,"cs":"eUFjjYcqVvKe9E069z2umA==","size":{"width":188,"height":188}}

{"font":{"size":10,"family":"Arial","color":"#000000"},"code":"$3.\\,\\int_{0}^{1}\\sin^{-1}\\left(\\frac{2x}{1+x^{2}}\\right).dx$","id":"4-0-1-1-0","type":"$","ts":1602238895352,"cs":"2vmk4XZTIwBHpUlm1sIl4g==","size":{"width":149,"height":28}}

Solun:- Let f(x) = {"font":{"color":"#000000","family":"Arial","size":10},"code":"$\\sin^{-1}\\left(\\frac{2x}{1+x^{2}}\\right)$","id":"1-1-1-0","type":"$","ts":1602238939332,"cs":"XERyxRzXzAmHu8QjNkHb/A==","size":{"width":88,"height":28}}

Integrate f(x):-

{"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-2-1-0","code":"\\begin{align*}\n{I}&={\\int_{0}^{1}\\sin^{-1}\\left(\\frac{2x}{1+x^{2}}\\right).dx}\t\n\\end{align*}","type":"align*","font":{"size":8.333333333333334,"color":"#000000","family":"Arial"},"ts":1602239071411,"cs":"FJHWE0UNGbytGklOtz/ylA==","size":{"width":161,"height":33}}

Let x = tan t and t = tan-1 x……...(1)

Differentiate w.r.t. to t:-

⇒ dx = sec2t.dt

Limits Change:

Put x = 0 in eq. 1:-

⇒ t = 0

Put x = 1 in eq. 1:-

⇒ t = Ï€/4

{"font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-3-1-0","code":"\\begin{align*}\n{I}&={\\int_{0}^{\\frac{\\Pi}{4}}\\sin^{-1}\\left(\\frac{2\\tan t}{1+\\tan^{2}t}\\right).\\sec ^{2}t.dt}\\\\\n{I}&={\\int_{0}^{\\frac{\\Pi}{4}}\\sin^{-1}\\left(\\sin2t\\right).\\sec ^{2}t.dt}\\\\\n{I}&={2\\int_{0}^{\\frac{\\Pi}{4}}t.\\sec ^{2}t.dt}\t\n\\end{align*}","ts":1602240448656,"cs":"56P8tMDj3qU5amLKDli2+g==","size":{"width":254,"height":134}}

According ILATE:- u = t and v = sec2t

We know that:-

{"code":"$\\int_{}^{}\\left(u.v\\right).dx=u\\int_{}^{}v.dx-\\int_{}^{}\\left[\\diff{u}{x}\\int_{}^{}v.dx\\right].dx+C$","font":{"color":"#000000","size":10,"family":"Arial"},"type":"$","id":"3-0-0-0-0-0-0-0-0-0-0-1-1-0-0","ts":1602240641938,"cs":"z8UUXPsHGRSNPs/MeX9GtA==","size":{"width":318,"height":20}}

{"id":"3-0-0-0-0-0-0-1-0-0-0-0-1-1-0-0","code":"\\begin{align*}\n{I}&={2\\left[t\\int_{0}^{\\frac{\\Pi}{4}}\\sec^{2}t.dt-\\int_{0}^{\\frac{\\Pi}{4}}\\left[\\diff{t}{t}\\int_{}^{}\\sec^{2}t.dt\\right].dt\\right]}\\\\\n{I}&={2\\left[t.\\tan t-\\int_{0}^{\\frac{\\Pi}{4}}\\tan t.dt\\right]}\\\\\n{I}&={2\\left[t.\\tan t-\\left(-\\log_{}\\left|\\cos t\\right|\\right)\\right]_{0}^{\\frac{\\Pi}{4}}}\\\\\n{I}&={2\\left[t.\\tan t+\\log_{}\\left|\\cos t\\right|\\right]_{0}^{\\frac{\\Pi}{4}}}\\\\\n{I}&={2\\left[\\left(\\frac{\\Pi}{4}.\\tan\\frac{\\Pi}{4}+\\log_{}\\left|\\cos\\frac{\\Pi}{4}\\right|\\right)-\\left(0.\\tan0+\\log_{}\\left|\\cos0\\right|\\right)\\right]}\\\\\n{I}&={2\\left(\\frac{\\Pi}{4}+\\log_{}\\left|\\frac{1}{{\\sqrt[]{2}}}\\right|\\right)}\t\n\\end{align*}","type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1602241505573,"cs":"OIMFOFNAeW1tonJGBHs5lQ==","size":{"width":406,"height":244}}

We know that:-

log (m/n) = log m - log n

log mn = n.log m

{"code":"\\begin{align*}\n{I}&={2\\left(\\frac{\\Pi}{4}+\\log_{}\\left|1\\right|-\\log_{}\\left|{\\sqrt[]{2}}\\right|\\right)}\\\\\n{I}&={2\\left(\\frac{\\Pi}{4}-\\frac{1}{2}\\log_{}\\left|2\\right|\\right)}\\\\\n{I}&={\\frac{\\Pi}{2}-\\log_{}\\left|2\\right|}\t\n\\end{align*}","type":"align*","font":{"color":"#000000","family":"Arial","size":10},"id":"3-0-0-0-0-0-0-1-0-0-0-0-1-1-1","ts":1602241665747,"cs":"T4Jt7LxOra8vAyu5aiyqAA==","size":{"width":213,"height":117}}

{"code":"$4.\\,\\int_{0}^{2}x{\\sqrt[]{x+2}}.dx$","type":"$","id":"4-0-1-1-1-0","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1602241801561,"cs":"eVVXijeFK1Z52FbqBu/QHA==","size":{"width":120,"height":21}}

Solun:- Let f(x) = {"type":"$","id":"1-1-1-1-0","code":"$x{\\sqrt[]{x+2}}$","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1602241822286,"cs":"4/JhppamIDmfKrx0BPIFTw==","size":{"width":58,"height":16}}

Integrate f(x):-

{"font":{"family":"Arial","size":8.333333333333334,"color":"#000000"},"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-2-1-1-0","type":"align*","code":"\\begin{align*}\n{I}&={\\int_{0}^{2}x{\\sqrt[]{x+2}}.dx}\t\n\\end{align*}","ts":1602241854161,"cs":"g7p0jgeZ19dGcSixGEihFw==","size":{"width":117,"height":33}}

Let x + 2 = t2……...(1)

Differentiate w.r.t. to t:-

⇒ dx = 2t.dt

Limits Change:

Put x = 0 in eq. 1:-

⇒ t = √2

Put x = 2 in eq. 1:-

⇒ t = 2

{"font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{I}&={\\int_{{\\sqrt[]{2}}}^{2}\\left(t^{2}-2\\right)t.2tdt}\\\\\n{I}&={2\\int_{{\\sqrt[]{2}}}^{2}\\left(t^{4}-2t^{2}\\right).dt}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-3-1-1-0","type":"align*","ts":1602593375206,"cs":"9MIq+71yyAusNrUH8VFTbg==","size":{"width":150,"height":84}}

{"font":{"family":"Arial","size":10,"color":"#000000"},"type":"$","code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+C$","id":"3-0-0-0-0-0-0-0-0-0-0-1-1-1-0","ts":1602242083097,"cs":"Le4XGBU/QwNqJdv5HFlCHA==","size":{"width":136,"height":21}}

{"type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"id":"3-0-0-0-0-0-0-1-0-0-0-0-1-1-0-1-0","code":"\\begin{align*}\n{I}&={2\\left[\\frac{t^{5}}{5}-\\frac{2t^{3}}{3}\\right]_{{\\sqrt[]{2}}}^{2}}\\\\\n{I}&={2\\left[\\frac{t^{5}}{5}-\\frac{2t^{3}}{3}\\right]_{{\\sqrt[]{2}}}^{2}}\\\\\n{I}&={2\\left[\\left(\\frac{32}{5}-\\frac{16}{3}\\right)-\\left(\\frac{4{\\sqrt[]{2}}}{5}-\\frac{4{\\sqrt[]{2}}}{3}\\right)\\right]}\\\\\n{I}&={4\\left[\\frac{16}{5}-\\frac{8}{3}-\\frac{2{\\sqrt[]{2}}}{5}+\\frac{2{\\sqrt[]{2}}}{3}\\right]}\\\\\n{I}&={4\\left[\\frac{48-40-6{\\sqrt[]{2}}+10{\\sqrt[]{2}}}{15}\\right]}\\\\\n{I}&={4\\left[\\frac{8+4{\\sqrt[]{2}}}{15}\\right]}\\\\\n{I}&={\\frac{4}{15}\\left[4{\\sqrt[]{2}}\\left({\\sqrt[]{2}}+1\\right)\\right]}\\\\\n{I}&={\\frac{16{\\sqrt[]{2}}}{15}\\left({\\sqrt[]{2}}+1\\right)}\t\n\\end{align*}","ts":1602593498529,"cs":"THRQbYhVgiXGaUVBk6x4NA==","size":{"width":280,"height":380}}

{"type":"$","id":"4-0-1-1-1-1-0","code":"$5.\\,\\int_{0}^{\\frac{\\Pi}{2}}\\frac{\\sin x}{1+\\cos^{2}x}.dx$","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1602244609511,"cs":"A30YpxK0MLCbl4FtcIurKw==","size":{"width":116,"height":25}}

Solun:- Let f(x) = {"font":{"family":"Arial","size":10,"color":"#000000"},"type":"$","id":"1-1-1-1-1-0","code":"$\\frac{\\sin x}{1+\\cos^{2}x}$","ts":1602244646483,"cs":"Zcq0vSG2pD+iDZ0FU5a/Pg==","size":{"width":48,"height":20}}

Integrate f(x):-

{"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-2-1-1-1-0","code":"\\begin{align*}\n{I}&={\\int_{0}^{\\frac{\\Pi}{2}}\\frac{\\sin x}{1+\\cos^{2}x}.dx}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","ts":1602245400233,"cs":"MSSlh7xN0MjWrPfa5Efm8Q==","size":{"width":156,"height":40}}

Let cos x = t……...(1)

Differentiate w.r.t. to t:-

⇒ - sin x.dx = dt

⇒ sin x.dx = - dt

Limits Change:

Put x = 0 in eq. 1:-

⇒ t = 1

Put x = π/2 in eq. 1:-

⇒ t = 0

{"font":{"color":"#000000","family":"Arial","size":10},"type":"align*","code":"\\begin{align*}\n{I}&={-\\int_{1}^{0}\\frac{dt}{1+t^{2}}}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-3-1-1-1-0","ts":1602244929048,"cs":"uU+C0tvZaQOYDL4UngZrpQ==","size":{"width":116,"height":38}}

{"id":"3-0-0-0-0-0-0-0-0-0-0-1-1-1-1-0","font":{"family":"Arial","color":"#000000","size":10},"type":"$","code":"$\\int_{}^{}\\frac{1}{1+x^{2}}.dx=\\tan^{-1}x+C$","ts":1602244996822,"cs":"6fBFsuG8AA5NUQrBMExOpw==","size":{"width":172,"height":20}}

{"id":"3-0-0-0-0-0-0-1-0-0-0-0-1-1-0-1-1-0","code":"\\begin{align*}\n{I}&={-\\left[\\tan^{-1}t\\right]_{1}^{0}}\\\\\n{I}&={-\\left[\\left(\\tan^{-1}\\left(0\\right)\\right)-\\left(\\tan^{-1}\\left(1\\right)\\right)\\right]}\\\\\n{I}&={-\\left[0-\\frac{\\Pi}{4}\\right]}\\\\\n{I}&={\\frac{\\Pi}{4}}\t\n\\end{align*}","type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1602245115776,"cs":"J+xbHiRE9TvrE6LxWHfb3A==","size":{"width":224,"height":126}}

{"type":"$","font":{"size":10,"family":"Arial","color":"#000000"},"code":"$6.\\,\\int_{0}^{2}\\frac{dx}{x+4-x^{2}}$","id":"4-0-1-1-1-1-1-0","ts":1602245217527,"cs":"Cy8ZfmLOWZc0Lmx3n+tUTA==","size":{"width":84,"height":22}}

Solun:- Let f(x) = {"code":"$\\frac{1}{x+4-x^{2}}$","id":"1-1-1-1-1-1-0","font":{"size":10,"family":"Arial","color":"#000000"},"type":"$","ts":1602245242672,"cs":"uxTV+FXZF5gfjSdUN+RfEg==","size":{"width":46,"height":20}}

Integrate f(x):-

{"code":"\\begin{align*}\n{I}&={\\int_{0}^{2}\\frac{dx}{x+4-x^{2}}}\\\\\n{I}&={\\int_{0}^{2}\\frac{dx}{-\\left(x^{2}-x-4\\right)}}\\\\\n{I}&={\\int_{0}^{2}\\frac{dx}{-\\left(x^{2}-x+\\left(\\frac{1}{2}\\right)^{2}-\\left(\\frac{1}{2}\\right)^{2}-4\\right)}}\\\\\n{I}&={\\int_{0}^{2}\\frac{dx}{-\\left(\\left(x-\\frac{1}{2}\\right)^{2}-\\frac{17}{4}\\right)}}\\\\\n{I}&={\\int_{0}^{2}\\frac{dx}{\\left(\\frac{{\\sqrt[]{17}}}{2}\\right)^{2}-\\left(x-\\frac{1}{2}\\right)^{2}}}\t\n\\end{align*}","type":"align*","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-2-1-1-1-1-0-0","font":{"family":"Arial","color":"#000000","size":10},"ts":1602319103229,"cs":"Y82FIaqGCOLE8KdpIEIDbg==","size":{"width":268,"height":258}}

We know that:-

{"font":{"color":"#000000","size":10,"family":"Arial"},"id":"3-0-0-0-0-0-0-0-0-0-0-1-1-0-1-0","code":"$\\int_{}^{}\\frac{1}{a^{2}-x^{2}}.dx=\\frac{1}{2a}\\log_{}\\left|\\frac{a+x}{a-x}\\right|+C$","type":"$","ts":1602322161741,"cs":"ltNQLxHHuF7NHZXgiY56ng==","size":{"width":208,"height":20}}

{"id":"3-0-0-0-0-0-0-1-0-0-0-0-1-1-0-1-1-1-0","code":"\\begin{align*}\n{I}&={\\frac{1}{2.\\frac{{\\sqrt[]{17}}}{2}}\\left[\\log_{}\\left|\\frac{\\frac{{\\sqrt[]{17}}}{2}+x-\\frac{1}{2}}{\\frac{{\\sqrt[]{17}}}{2}-x+\\frac{1}{2}}\\right|\\right]_{0}^{2}}\\\\\n{I}&={\\frac{1}{{\\sqrt[]{17}}}\\left[\\left(\\log_{}\\left|\\frac{\\frac{{\\sqrt[]{17}}}{2}+2-\\frac{1}{2}}{\\frac{{\\sqrt[]{17}}}{2}-2+\\frac{1}{2}}\\right|\\right)-\\left(\\log_{}\\left|\\frac{\\frac{{\\sqrt[]{17}}}{2}+0-\\frac{1}{2}}{\\frac{{\\sqrt[]{17}}}{2}-0+\\frac{1}{2}}\\right|\\right)\\right]}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"type":"align*","ts":1602322688908,"cs":"8JL9cEzqE4bR5Db7eDKuwA==","size":{"width":409,"height":112}}

{"code":"\\begin{align*}\n{I}&={\\frac{1}{{\\sqrt[]{17}}}\\left[\\left(\\log_{}\\left|\\frac{\\frac{{\\sqrt[]{17}}+4-1}{2}}{\\frac{{\\sqrt[]{17}}-4+1}{2}}\\right|\\right)-\\left(\\log_{}\\left|\\frac{\\frac{{\\sqrt[]{17}}-1}{2}}{\\frac{{\\sqrt[]{17}}+1}{2}}\\right|\\right)\\right]}\\\\\n{I}&={\\frac{1}{{\\sqrt[]{17}}}\\left[\\left(\\log_{}\\left|\\frac{{\\sqrt[]{17}}+3}{{\\sqrt[]{17}}-3}\\right|\\right)-\\left(\\log_{}\\left|\\frac{{\\sqrt[]{17}}-1}{{\\sqrt[]{17}}+1}\\right|\\right)\\right]}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","id":"6-0","ts":1602322719274,"cs":"ZMKgXgnyBX5sb91XsUJ37Q==","size":{"width":348,"height":104}}

We know that:-

log m - log n = log (m/n)

{"type":"align*","id":"6-1","font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{I}&={\\frac{1}{{\\sqrt[]{17}}}\\log_{}\\left|\\frac{\\frac{{\\sqrt[]{17}}+3}{{\\sqrt[]{17}}-3}}{\\frac{{\\sqrt[]{17}}-1}{{\\sqrt[]{17}}+1}}\\right|}\\\\\n{I}&={\\frac{1}{{\\sqrt[]{17}}}\\log_{}\\left|\\frac{{\\sqrt[]{17}}+3}{{\\sqrt[]{17}}-3}\\times\\frac{{\\sqrt[]{17}}+1}{{\\sqrt[]{17}}-1}\\right|}\\\\\n{I}&={\\frac{1}{{\\sqrt[]{17}}}\\log_{}\\left|\\frac{17+{\\sqrt[]{17}}+3{\\sqrt[]{17}}+3}{17-{\\sqrt[]{17}}-3{\\sqrt[]{17}}+3}\\right|}\\\\\n{I}&={\\frac{1}{{\\sqrt[]{17}}}\\log_{}\\left|\\frac{17+4{\\sqrt[]{17}}+3}{17-4{\\sqrt[]{17}}+3}\\right|}\\\\\n{I}&={\\frac{1}{{\\sqrt[]{17}}}\\log_{}\\left|\\frac{20+4{\\sqrt[]{17}}}{20-4{\\sqrt[]{17}}}\\right|}\\\\\n{I}&={\\frac{1}{{\\sqrt[]{17}}}\\log_{}\\left|\\frac{4\\left(5+{\\sqrt[]{17}}\\right)}{4\\left(5-{\\sqrt[]{17}}\\right)}\\right|}\\\\\n{I}&={\\frac{1}{{\\sqrt[]{17}}}\\log_{}\\left|\\frac{5+{\\sqrt[]{17}}}{5-{\\sqrt[]{17}}}\\right|}\\\\\n{I}&={\\frac{1}{{\\sqrt[]{17}}}\\log_{}\\left|\\frac{5+{\\sqrt[]{17}}}{5-{\\sqrt[]{17}}}\\times\\frac{5+{\\sqrt[]{17}}}{5+{\\sqrt[]{17}}}\\right|}\\\\\n{I}&={\\frac{1}{{\\sqrt[]{17}}}\\log_{}\\left|\\frac{25+10{\\sqrt[]{17}}+17}{25-17}\\right|}\\\\\n{I}&={\\frac{1}{{\\sqrt[]{17}}}\\log_{}\\left|\\frac{42+10{\\sqrt[]{17}}}{8}\\right|}\\\\\n{I}&={\\frac{1}{{\\sqrt[]{17}}}\\log_{}\\left|\\frac{2\\left(21+5{\\sqrt[]{17}}\\right)}{8}\\right|}\\\\\n{I}&={\\frac{1}{{\\sqrt[]{17}}}\\log_{}\\left|\\frac{21+5{\\sqrt[]{17}}}{4}\\right|}\t\n\\end{align*}","ts":1602323710313,"cs":"Bsyu0/fMLUYCVpHpQV2QJA==","size":{"width":254,"height":632}}

{"font":{"family":"Arial","color":"#000000","size":10},"code":"$7.\\,\\int_{-1}^{1}\\frac{dx}{x^{2}+2x+5}$","id":"4-0-1-1-1-1-1-1-0","type":"$","ts":1602321272088,"cs":"6wcSI53wXDMC4kT8VCIgFw==","size":{"width":96,"height":22}}

Solun:- Let f(x) = {"type":"$","code":"$\\frac{1}{x^{2}+2x+5}$","id":"1-1-1-1-1-1-1-0","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1602321300071,"cs":"p/Z05sn8rSSVlzHVhvMF+w==","size":{"width":52,"height":20}}

Integrate f(x):-

{"font":{"color":"#000000","family":"Arial","size":10},"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-2-1-1-1-1-1-0","code":"\\begin{align*}\n{I}&={\\int_{-1}^{1}\\frac{dx}{x^{2}+2x+5}}\\\\\n{I}&={\\int_{-1}^{1}\\frac{dx}{x^{2}+2x+\\left(1\\right)^{2}-\\left(1\\right)^{2}+5}}\\\\\n{I}&={\\int_{-1}^{1}\\frac{dx}{\\left(x+1\\right)^{2}+4}}\\\\\n{I}&={\\int_{-1}^{1}\\frac{dx}{\\left(x+1\\right)^{2}+\\left(2\\right)^{2}}}\t\n\\end{align*}","type":"align*","ts":1602325351228,"cs":"7Q89owPU419xaaWDTTonWQ==","size":{"width":233,"height":182}}

We know that:-

{"code":"$\\int_{}^{}\\frac{1}{x^{2}+a^{2}}.dx=\\frac{1}{a}\\tan^{-1}\\left(\\frac{x}{a}\\right)+C$","font":{"size":10,"family":"Arial","color":"#000000"},"id":"3-0-0-0-0-0-0-0-0-0-0-1-1-0-1-1-0","type":"$","ts":1602325412510,"cs":"49FjSUxHLqravfNqcMa6fQ==","size":{"width":208,"height":20}}

{"code":"\\begin{align*}\n{I}&={\\frac{1}{2}\\left[\\tan^{-1}\\left(\\frac{x+1}{2}\\right)\\right]_{-1}^{1}}\\\\\n{I}&={\\frac{1}{2}\\left[\\left(\\tan^{-1}\\left(\\frac{2}{2}\\right)\\right)-\\left(\\tan^{-1}\\left(\\frac{0}{2}\\right)\\right)\\right]}\\\\\n{I}&={\\frac{1}{2}\\times\\frac{\\Pi}{4}}\\\\\n{I}&={\\frac{\\Pi}{8}}\t\n\\end{align*}","id":"3-0-0-0-0-0-0-1-0-0-0-0-1-1-0-1-1-1-1-0","type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1602325672783,"cs":"nJoGm/SFAmLhmXBq4nkgBg==","size":{"width":284,"height":160}}

{"id":"4-0-1-1-1-1-1-1-1-0","font":{"family":"Arial","color":"#000000","size":10},"type":"$","code":"$8.\\,\\int_{1}^{2}\\left(\\frac{1}{x}-\\frac{1}{2x^{2}}\\right)e^{2x}.dx$","ts":1602325911129,"cs":"AcWoHBhk1l3l5cqFmNLX0Q==","size":{"width":149,"height":21}}

Solun:- Let f(x) = {"type":"$","code":"$\\left(\\frac{1}{x}-\\frac{1}{2x^{2}}\\right)e^{2x}$","font":{"family":"Arial","size":10,"color":"#000000"},"id":"1-1-1-1-1-1-1-1-0","ts":1602325931530,"cs":"n6QAGMkHVwSdnaLPW3lu3Q==","size":{"width":90,"height":20}}

Integrate f(x):-

{"font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{I}&={\\int_{1}^{2}\\left(\\frac{1}{x}-\\frac{1}{2x^{2}}\\right)e^{2x}.dx}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-2-1-1-1-1-1-1-0-0","type":"align*","ts":1602327604633,"cs":"Aep+vvcsInhGkh8U+juWdQ==","size":{"width":184,"height":40}}

Let 2x = t and x = t/2…....(1)

⇒ 2.dx = dt

⇒ dx = (1/2).dt

Limits Change:-

Put x = 1 in eq. 1:-

⇒ t = 2

Put x = 2 in eq. 1:-

⇒ t = 4

{"type":"align*","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-2-1-1-1-1-1-1-1-0","code":"\\begin{align*}\n{I}&={\\int_{2}^{4}\\left(\\frac{2}{t}-\\frac{4}{2t^{2}}\\right)e^{t}.\\frac{dt}{2}}\\\\\n{I}&={\\frac{2}{2}\\int_{2}^{4}\\left(\\frac{1}{t}-\\frac{1}{t^{2}}\\right)e^{t}.dt}\\\\\n{I}&={\\int_{2}^{4}\\left(\\frac{1}{t}-\\frac{1}{t^{2}}\\right)e^{t}.dt}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"ts":1602328127803,"cs":"PMPTSSYFYGpDDrIwMEffww==","size":{"width":178,"height":128}}

We know that:-

{"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":{"color":"#000000","family":"Arial","size":10},"id":"3-0-0-0-0-0-0-0-0-0-0-1-1-0-1-1-1-0","ts":1602328180523,"cs":"26u4S8lrXj8VfLNh5WSJmA==","size":{"width":246,"height":17}}

Let f(x) = 1/t

Then f’(x) = -1/t2

{"code":"\\begin{align*}\n{I}&={\\left[\\frac{e^{t}}{t}\\right]_{2}^{4}}\\\\\n{I}&={\\left[\\left(\\frac{e^{4}}{4}\\right)-\\left(\\frac{e^{2}}{2}\\right)\\right]}\\\\\n{I}&={\\frac{e^{4}-2e^{2}}{4}}\\\\\n{I}&={\\frac{e^{2}\\left(e^{2}-2\\right)}{4}}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"type":"align*","id":"10-0","ts":1602328475150,"cs":"N8Kh9++SQMLGkop6ukJDVg==","size":{"width":152,"height":168}}

Choose the correct answer to Exercises 9 and 10

9. The value of integrals {"type":"$","code":"$\\int_{\\frac{1}{3}}^{1}\\frac{\\left(x-x^{3}\\right)^{\\frac{1}{3}}}{x^{4}}.dx$","font":{"family":"Arial","color":"#000000","size":10},"id":"4-0-1-1-1-1-1-1-1-1-0","ts":1602328968472,"cs":"Y9UvFWOpJFrRTjiMgFVW6A==","size":{"width":94,"height":33}} is

Solun:- Let f(x) = {"code":"$\\frac{\\left(x-x^{3}\\right)^{\\frac{1}{3}}}{x^{4}}$","font":{"size":10,"family":"Arial","color":"#000000"},"type":"$","id":"1-1-1-1-1-1-1-1-1-0","ts":1602328992363,"cs":"yYyS62ics3YMCxhISyyATA==","size":{"width":50,"height":29}}

Integrate f(x):-

{"font":{"family":"Arial","color":"#000000","size":10},"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-2-1-1-1-1-1-1-0-1-0","code":"\\begin{align*}\n{I}&={\\int_{\\frac{1}{3}}^{1}\\frac{\\left(x-x^{3}\\right)^{\\frac{1}{3}}}{x^{4}}.dx}\\\\\n{I}&={\\int_{\\frac{1}{3}}^{1}\\frac{x^{\\frac{1}{3}}\\left(1-x^{2}\\right)^{\\frac{1}{3}}}{x^{4}}.dx}\t\n\\end{align*}","type":"align*","ts":1602333326582,"cs":"zOV5UHJ4XN70k1FYWBiPNg==","size":{"width":170,"height":104}}

Let x = sin t and t = sin-1(x)…....(1)

Differentiate w.r.t. to t:-

⇒ dx = cos t.dt

Limits Change:-

Put x = 1/3 in eq. 1:-

⇒ t = sin-1(1/3)

Put x = 1 in eq. 1:-

⇒ t = sin-1(1) = Ï€/2

{"type":"align*","code":"\\begin{align*}\n{I}&={\\int_{\\sin^{-1}\\left(\\frac{1}{3}\\right)}^{\\frac{\\Pi}{2}}\\frac{\\sin^{\\frac{1}{3}}t\\left(1-\\sin^{2}t\\right)^{\\frac{1}{3}}}{\\sin^{4}t}.\\cos t.dt}\\\\\n{I}&={\\int_{\\sin^{-1}\\left(\\frac{1}{3}\\right)}^{\\frac{\\Pi}{2}}\\frac{\\sin^{\\frac{1}{3}}t\\cos^{\\frac{2}{3}}t}{\\sin^{2}t.\\sin^{2}t}.\\cos t.dt}\\\\\n{I}&={\\int_{\\sin^{-1}\\left(\\frac{1}{3}\\right)}^{\\frac{\\Pi}{2}}\\frac{\\cos^{\\frac{5}{3}}t}{\\sin^{\\frac{5}{3}}t.\\sin^{2}t}.dt}\\\\\n{I}&={\\int_{\\sin^{-1}\\left(\\frac{1}{3}\\right)}^{\\frac{\\Pi}{2}}\\cot^{\\frac{5}{3}}t.\\cos ec^{2}t.dt}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":9},"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-2-1-1-1-1-1-1-1-1-0","ts":1602334426460,"cs":"vL8JVo/bE1NgmDJMVsEpYw==","size":{"width":272,"height":180}}

Let cot t = y…....(2)

Differentiate w.r.t. to y:-

⇒ -cosec2t.dt = dy

⇒ cosec2t.dt = -dy

Again limits change:-

Put t = sin-1 (1/3) and sin t = ⅓ in eq. 2:-

Then cot t = 2√2

So y = 2√2

Put t = π/2 in eq. 1:-

⇒ y = 0

{"code":"\\begin{align*}\n{I}&={-\\int_{2{\\sqrt[]{2}}}^{0}y^{\\frac{5}{3}}.dy}\t\n\\end{align*}","id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-2-1-1-1-1-1-1-1-1-1","type":"align*","font":{"color":"#000000","family":"Arial","size":10},"ts":1602335492296,"cs":"y50sHkBYSFbhxRqJOsrX2A==","size":{"width":117,"height":40}}

We know that:-

{"code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+C$","id":"3-0-0-0-0-0-0-0-0-0-0-1-1-0-1-1-1-1-0","font":{"size":10,"family":"Arial","color":"#000000"},"type":"$","ts":1602335568640,"cs":"GSmU3D+M/YE+yGkqCeUpLQ==","size":{"width":136,"height":21}}

{"id":"10-1-0","type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{I}&={-\\left[\\frac{y^{\\frac{8}{3}}}{\\frac{8}{3}}\\right]_{2{\\sqrt[]{2}}}^{0}}\\\\\n{I}&={-\\frac{3}{8}\\left[y^{\\frac{8}{3}}\\right]_{2{\\sqrt[]{2}}}^{0}}\\\\\n{I}&={-\\frac{3}{8}\\left[0-\\left(2{\\sqrt[]{2}}\\right)^{\\frac{8}{3}}\\right]}\\\\\n{I}&={\\frac{3}{8}\\left[\\left(2\\right)^{\\frac{3}{2}\\times\\frac{8}{3}}\\right]}\\\\\n{I}&={\\frac{3\\times16}{8}}\\\\\n{I}&={6}\t\n\\end{align*}","ts":1602335824459,"cs":"F1fX/wUlmnzDFwvRewSeCw==","size":{"width":157,"height":228}}

The correct answer is A.

10. If {"font":{"family":"Arial","size":10,"color":"#000000"},"code":"$f\\left(x\\right)=\\int_{0}^{x}t\\sin t.dt$","type":"$","id":"4-0-1-1-1-1-1-1-1-1-1","ts":1602335974317,"cs":"R6laF+wHs9HrXgaf/ihlVQ==","size":{"width":128,"height":18}} then f’(x) is

Solun:- Let f(t) = t.sin t

Integrate f(t):-

{"type":"align*","code":"\\begin{align*}\n{f\\left(x\\right)}&={\\int_{0}^{x}t.\\sin t.dt}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"id":"2-0-0-0-0-0-0-0-0-0-0-0-0-0-2-1-1-1-1-1-1-0-1-1","ts":1602336380773,"cs":"kwaxJb1O9oQUwLghqemmiw==","size":{"width":138,"height":36}}

According to ILATE:-

Let u = t and v = sin t

We know that:-

{"font":{"size":10,"color":"#000000","family":"Arial"},"id":"3-0-0-0-0-0-0-0-0-0-0-1-1-0-1-1-1-1-1","code":"$\\int_{}^{}\\left(u.v\\right).dx=u\\int_{}^{}v.dx-\\int_{}^{}\\left[\\diff{u}{x}.\\int_{}^{}v.dx\\right].dx+C$","type":"$","ts":1602336142596,"cs":"ZVCScgMzGe0emrBZAFtbaw==","size":{"width":322,"height":20}}

{"id":"10-1-1","code":"\\begin{align*}\n{f\\left(x\\right)}&={t\\int_{0}^{x}\\sin t.dt-\\int_{0}^{x}\\left[\\diff{t}{t}.\\int_{0}^{x}\\sin t.dt\\right].dt}\\\\\n{f\\left(x\\right)}&={-t\\cos t-\\int_{0}^{x}\\left[-\\cos t\\right].dt}\\\\\n{f\\left(x\\right)}&={\\left[-t\\cos t+\\sin t\\right]_{0}^{x}}\\\\\n{f\\left(x\\right)}&={\\left[\\left(-x\\cos x+\\sin x\\right)-\\left(0\\right)\\right]}\\\\\n{f\\left(x\\right)}&={-x\\cos x+\\sin x}\t\n\\end{align*}","type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1602336541147,"cs":"+xLoaqPJfSi5vil7Z7ie5Q==","size":{"width":317,"height":142}}

f(x) = -(x.cos x - sin x)

Then f’(x) is equal to:-

f’(x) = -[(-x.sin x+cos x) - cos x]

f’(x) = x.sin x - cos x + cos x

f’(x) = x.sin x

The correct answer is B.


Download PDF of Exercise 7.10

See Also:-

Notes of Integrals

Exercise 7.1

Exercise 7.2

Exercise 7.3

Exercise 7.4

Exercise 7.5

Exercise 7.6

Exercise 7.7

Exercise 7.9


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