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 Exercise 7.3

Find the integrals of the functions in Exercises 1 to 22:

1. sin2(2x+5)

Solun:- Let f(x) = sin2(2x+5)

Integrate f(x):-

{"id":"2-0-0-0-0-0","font":{"color":"#000000","family":"Arial","size":12},"code":"$\\int_{}^{}f\\left(x\\right)dx=\\int_{}^{}\\sin ^{2}\\left(2x+5\\right).dx$","type":"$","ts":1600415133565,"cs":"meUps+pZnDh65G3T+GgeGg==","size":{"width":264,"height":24}}

We know that cos 2x = 1 - 2sin2x

{"type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(\\frac{1-\\cos2\\left(2x+5\\right)}{2}\\right).dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(\\frac{1}{2}\\right).dx-\\int_{}^{}\\left(\\frac{\\cos2\\left(2x+5\\right)}{2}\\right).dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\int_{}^{}1.dx-\\frac{1}{2}\\int_{}^{}\\cos\\left(4x+10\\right).dx}\t\n\\end{align*}","id":"2-0-0-1-0-0","ts":1600415732561,"cs":"OOGkSAx3JwxFOyGoeI+tuA==","size":{"width":346,"height":121}}

Let 4x + 10 = t

Differentiate w.r.t to t:-

⇒ 4.dx = dt

⇒ dx = (1/4).dt

{"id":"2-1-0-0-0-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\int_{}^{}1.dx-\\frac{1}{2}\\int_{}^{}\\cos t.\\frac{dt}{4}}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\int_{}^{}1.dx-\\frac{1}{8}\\int_{}^{}\\cos t.dt}\t\n\\end{align*}","type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1600416025903,"cs":"yg461c8GS/Yvqcuihw0/2g==","size":{"width":268,"height":76}}

We know that:-

{"font":{"color":"#000000","size":12,"family":"Arial"},"code":"$\\int_{}^{}1.dx=x+c$","type":"$","id":"3-0-0-0","ts":1600416087550,"cs":"JqbM4iK6wmB5qrplV4Zs4Q==","size":{"width":128,"height":22}}

{"font":{"size":12,"family":"Arial","color":"#000000"},"id":"3-0-1-0","type":"$","code":"$\\int_{}^{}\\cos x.dx=\\sin x+c$","ts":1600416219068,"cs":"mrzaa7mbwFp4DnM9G5/5eQ==","size":{"width":192,"height":22}}

{"font":{"family":"Arial","color":"#000000","size":12},"id":"2-1-1-0-0-0-0-0","type":"$","code":"$\\int_{}^{}f\\left(x\\right)dx=\\frac{x}{2}-\\frac{\\sin t}{8}+C$","ts":1600416260638,"cs":"TQT+JYucvKl7eReSSOA42w==","size":{"width":222,"height":24}}

Put the value of t:-

{"code":"$\\int_{}^{}f\\left(x\\right)dx=\\frac{x}{2}-\\frac{\\sin \\left(4x+10\\right)}{8}+C$","type":"$","font":{"family":"Arial","size":12,"color":"#000000"},"id":"2-1-1-0-0-1-0-0","ts":1600416317576,"cs":"1fvE7NTakG2af8tvw4oLUQ==","size":{"width":266,"height":28}}

2. sin 3x.cos 4x

Solun:- Let f(x) = sin 3x.cos 4x

Integrate f(x):-

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\sin 3x.\\cos 4x.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\int_{}^{}2\\sin 3x.\\cos 4x.dx}\\\\\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","id":"2-0-0-0-1-0","ts":1600416848731,"cs":"OlFI4iu9L/KT1q9cKY7qkA==","size":{"width":246,"height":76}}

We know that 2sinA.cosB = sin(A+B)+sin(A-B)

{"id":"2-0-0-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\int_{}^{}\\left[\\sin\\left(7x\\right)+\\sin\\left(-x\\right)\\right].dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\int_{}^{}\\left[\\sin\\left(7x\\right)-\\sin\\left(x\\right)\\right].dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\left[\\int_{}^{}\\sin7x.dx-\\int_{}^{}\\sin x.dx\\right]}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","ts":1600418366121,"cs":"wqT8u5Nv0S+q14bAOFS/FQ==","size":{"width":296,"height":118}}

Let 7x = t

Differentiate w.r.t to t:-

⇒ 7.dx = dt

⇒ dx = (1/7).dt

{"font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\left[\\int_{}^{}\\sin t.\\frac{dt}{7}-\\int_{}^{}\\sin x.dx\\right]}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{14}\\int_{}^{}\\sin t.dt-\\frac{1}{2}\\int_{}^{}\\sin x.dx}\t\n\\end{align*}","id":"2-1-0-0-1-0","ts":1600418454123,"cs":"y2Cp5b28uxTc45D9JI43jA==","size":{"width":293,"height":77}}

We know that:-

{"id":"3-0-2-0","font":{"family":"Arial","color":"#000000","size":11},"type":"$","code":"$\\int_{}^{}\\sin x.dx=-\\cos x+c$","ts":1600417976655,"cs":"Ss6/U3H9A7aKsVozL6Ci/Q==","size":{"width":188,"height":20}}

{"code":"$\\int_{}^{}f\\left(x\\right)dx=\\frac{-1}{14}\\cos t+\\frac{1}{2}\\cos x+C$","font":{"color":"#000000","family":"Arial","size":11},"type":"$","id":"2-1-1-0-0-0-1-0-0","ts":1600418490584,"cs":"llL4zTUGOSudFaQn1mXisg==","size":{"width":261,"height":21}}

Put the value of t:-

{"id":"2-1-1-0-0-0-1-1-0","type":"$","code":"$\\int_{}^{}f\\left(x\\right)dx=\\frac{-1}{14}\\cos \\left(7x\\right)+\\frac{1}{2}\\cos x\\,+C$","font":{"color":"#000000","family":"Arial","size":11},"ts":1600418509606,"cs":"xGeyiI2w89VPSCr0GGIMiw==","size":{"width":292,"height":21}}

3. cos 2x.cos 4x.cos 6x

Solun:- Let f(x) = cos 2x.cos 4x.cos 6x

Integrate f(x):-

{"font":{"family":"Arial","color":"#000000","size":10},"id":"2-0-0-0-1-1-0","type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\cos2x.\\cos4x.\\cos6x.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\int_{}^{}2\\cos2x.\\cos4x.\\cos6x.dx}\t\n\\end{align*}","ts":1600418852375,"cs":"Wj6yCS3WXvs1EYwPyVYcKg==","size":{"width":296,"height":76}}

We know that 2cosA.cosB = cos(A+B)+cos(A-B)

{"id":"2-0-0-1-1-1-0-0","type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\int_{}^{}\\left[\\cos\\left(6x\\right)+\\cos\\left(-2x\\right)\\right].\\cos6x.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\int_{}^{}\\left[\\cos\\left(6x\\right)+\\cos\\left(2x\\right)\\right].\\cos6x.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\left[\\int_{}^{}\\cos^{2}6x.dx+\\frac{1}{2}\\int_{}^{}2\\cos 6x.\\cos2x.dx\\right]}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1600424920904,"cs":"DoDL9dZQT7tyn8u0T44cVQ==","size":{"width":390,"height":118}}

We know that:- 2cosA.cosB = cos(A+B)+cos(A-B)

cos 2x = 2cos2x - 1

{"font":{"color":"#000000","family":"Arial","size":9},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\left[\\int_{}^{}\\frac{\\left(1+\\cos12x\\right)}{2}.dx+\\frac{1}{2}\\int_{}^{}\\left(\\cos 8x+\\cos4x\\right).dx\\right]}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\left[\\int_{}^{}\\frac{1}{2}.dx+\\int_{}^{}\\frac{\\cos12x}{2}.dx+\\frac{1}{2}\\int_{}^{}\\cos8x.dx+\\frac{1}{2}\\int_{}^{}\\cos4x.dx\\right]}\t\n\\end{align*}","type":"align*","id":"2-0-0-1-1-1-1-0","ts":1600424967414,"cs":"vSglkLo1JOov1duZDNeLxQ==","size":{"width":540,"height":80}}

We know that:-

{"code":"$\\int_{}^{}\\cos x.dx=\\sin x+c$","id":"3-0-2-1-0","type":"$","font":{"family":"Arial","color":"#000000","size":11},"ts":1600420425191,"cs":"nPu+DlryNZovlhLp6i2YsQ==","size":{"width":170,"height":20}}

Put the value of t:-

{"font":{"color":"#000000","family":"Arial","size":10},"id":"2-1-1-0-0-0-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x}{4}+\\frac{1}{4}\\times\\frac{1}{12}\\sin\\left(12x\\right)+\\frac{1}{4}\\times\\frac{1}{8}\\sin8x+\\frac{1}{4}\\times\\frac{1}{4}\\sin4x+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{x}{4}+\\frac{1}{48}\\sin\\left(12x\\right)+\\frac{1}{32}\\sin8x+\\frac{1}{16}\\sin4x+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{4}\\left(x+\\frac{1}{12}\\sin\\left(12x\\right)+\\frac{1}{8}\\sin8x+\\frac{1}{4}\\sin4x\\right)+C}\t\n\\end{align*}","type":"align*","ts":1600425191316,"cs":"NJcGqDZP7/4YYKcF0rcO/w==","size":{"width":485,"height":118}}

4. sin3(2x + 1)

Solun:- Let f(x) = sin3(2x + 1)

Integrate f(x):-

{"font":{"color":"#000000","size":11,"family":"Arial"},"id":"2-0-0-0-0-1-0-0","type":"$","code":"$\\int_{}^{}f\\left(x\\right)dx=\\int_{}^{}\\sin ^{3}\\left(2x+1\\right).dx$","ts":1600425398556,"cs":"fkV3QhF4DOf2CW/eD8skZw==","size":{"width":234,"height":21}}

Method 1:-

We know that:- sin 3x = 3sin x - 4sin3x

{"font":{"family":"Arial","size":10,"color":"#000000"},"id":"2-0-0-1-0-1","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{\\left(3\\sin \\left(2x+1\\right)-\\sin3\\left(2x+1\\right)\\right)}{4}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(\\frac{3}{4}\\sin \\left(2x+1\\right)\\right).dx-\\int_{}^{}\\left(\\frac{\\sin\\left(6x+3\\right)}{4}\\right).dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{3}{4}\\int_{}^{}\\sin \\left(2x+1\\right).dx-\\frac{1}{4}\\int_{}^{}\\sin\\left(6x+3\\right).dx}\t\n\\end{align*}","type":"align*","ts":1600426628820,"cs":"GZHhM1arWbJjmmwLQYUzvQ==","size":{"width":416,"height":120}}

Let 6x + 3 = t

Differentiate w.r.t to t:-

⇒ 6.dx = dt

⇒ dx = (1/6).dt

{"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\left[\\frac{3}{4}\\int_{}^{}\\sin \\left(2x+1\\right).dx-\\frac{1}{4}\\int_{}^{}\\sin t.\\frac{dt}{6}\\right]}\t\n\\end{align*}","id":"2-1-0-0-1-1-0","font":{"family":"Arial","color":"#000000","size":10},"ts":1600426931442,"cs":"HPTUKe1qFjnkyIE3O5OktA==","size":{"width":356,"height":37}}

We know that:-

{"type":"$","font":{"family":"Arial","size":12,"color":"#000000"},"id":"3-0-1-1-0","code":"$\\int_{}^{}\\sin x.dx=-\\cos x+c$","ts":1600426085258,"cs":"dhgTlqDmRR/YkNMDkhztnA==","size":{"width":212,"height":22}}

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{-3}{4}\\times\\frac{\\cos\\left(2x+1\\right)}{2}-\\frac{\\left(-\\cos t\\right)}{24}+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{-3}{8}\\cos\\left(2x+1\\right)+\\frac{\\cos t}{24}+C}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","id":"2-1-1-0-0-0-0-1-0-0","ts":1600427109721,"cs":"JBDJBAQ7B8qYMDyfdW/OOQ==","size":{"width":336,"height":77}}

Put the value of t:-

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{-3}{8}\\cos\\left(2x+1\\right)+\\frac{\\cos \\left(6x+3\\right)}{24}+C}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"id":"2-1-1-0-0-0-0-1-1-0","type":"align*","ts":1600427404651,"cs":"aSqmKGdIpy3kpH0MidFXMw==","size":{"width":333,"height":36}}

Method 2:-

{"font":{"color":"#000000","size":11,"family":"Arial"},"id":"2-0-0-0-0-1-0-1","type":"$","code":"$\\int_{}^{}f\\left(x\\right)dx=\\int_{}^{}\\sin ^{3}\\left(2x+1\\right).dx$","ts":1600425398556,"cs":"s6EF5zUvL1SSlsraDXyVpg==","size":{"width":234,"height":21}}

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\sin ^{2}\\left(2x+1\\right).\\sin \\left(2x+1\\right).dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{} \\left[1-\\cos^{2}\\left(2x+1\\right)\\right].\\sin \\left(2x+1\\right).dx}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"type":"align*","id":"2-0-0-0-0-1-1","ts":1600427848898,"cs":"gBT80zq3cfy2PRSCCXFZkA==","size":{"width":341,"height":76}}

Let cos (2x+1) = t

Differentiate w.r.t to t:-

⇒ -2.sin(2x+1).dx = dt

⇒ sin(2x+1).dx = (-1/2). dt

{"font":{"family":"Arial","color":"#000000","size":10},"id":"2-1-0-0-1-1-1-0","type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={-\\frac{1}{2}\\int_{}^{}\\left(1-t^{2}\\right).dt}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\int_{}^{}\\left(t^{2}-1\\right).dt}\t\n\\end{align*}","ts":1600428292118,"cs":"xUqXzXj6LiGAPs8uhbsc0Q==","size":{"width":209,"height":76}}

We know that:-

{"font":{"size":12,"color":"#000000","family":"Arial"},"type":"$","id":"3-0-1-1-1","code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+c$","ts":1600428088570,"cs":"nnrLasPR3+GS/NUazn/Jlg==","size":{"width":168,"height":28}}

{"type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{t^{3}}{6}-\\frac{t}{2}+C}\t\n\\end{align*}","id":"2-1-1-0-0-0-0-1-0-1-0-0","ts":1600428329878,"cs":"OZ/sM7/t7tdYmIs17ygnEA==","size":{"width":172,"height":37}}

Put the value of t:-

{"type":"align*","font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{\\cos^{3}\\left(2x+1\\right)}{6}-\\frac{1}{2}\\cos\\left(2x+1\\right)+C}\t\n\\end{align*}","id":"2-1-1-0-0-0-0-1-0-1-1-0","ts":1600428346559,"cs":"7mckF7zfpgBv17yYnvMeRA==","size":{"width":328,"height":38}}

5. sin3x.cos3x

Solun:- Let f(x) = sin3x.cos3x

Integrate f(x):-

{"type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"id":"2-0-0-0-0-1-0-2","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\sin ^{3}x.\\cos^{3}x.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\sin ^{3}x.\\cos^{2}x.\\cos x.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\sin ^{3}x.\\left(1-\\sin^{2}x\\right).\\cos x.dx}\t\n\\end{align*}","ts":1600428804891,"cs":"/31tnzZKqYrYSWNdJYQR0Q==","size":{"width":296,"height":116}}

Let sin x = t

Differentiate w.r.t to t:-

⇒ cos x.dx = dt

{"font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","id":"2-1-0-0-1-1-1-1","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}t^{3}\\left(1-t^{2}\\right).dt}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(t^{3}-t^{5}\\right).dt}\t\n\\end{align*}","ts":1600429084390,"cs":"aoAWarox0j7dX1lndy/Oew==","size":{"width":193,"height":76}}

We know that:-

{"font":{"size":12,"color":"#000000","family":"Arial"},"type":"$","id":"3-0-1-1-1","code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+c$","ts":1600428088570,"cs":"nnrLasPR3+GS/NUazn/Jlg==","size":{"width":168,"height":28}}

{"type":"align*","id":"2-1-1-0-0-0-0-1-0-1-0-1-0","font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{t^{4}}{4}-\\frac{t^{6}}{6}+C}\t\n\\end{align*}","ts":1600429122904,"cs":"6K+11BbK/TTDBXQ0aBCFtA==","size":{"width":176,"height":37}}

Put the value of t:-

{"font":{"color":"#000000","size":10,"family":"Arial"},"id":"2-1-1-0-0-0-0-1-0-1-0-1-1","type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{\\sin^{4}x}{4}-\\frac{\\sin^{6}x}{6}+C}\t\n\\end{align*}","ts":1600429237107,"cs":"D4EKMD5D+ygbiJ4W+WMsIg==","size":{"width":228,"height":38}}

6. sin x.sin 2x.sin 3x

Solun:- Let f(x) = sin x.sin 2x.sin 3x

Integrate f(x):-

{"id":"2-0-0-0-1-1-1-0","font":{"color":"#000000","family":"Arial","size":10},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\sin x.\\sin 2x.\\sin3x.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\int_{}^{}2\\sin x.\\sin 2x.\\sin3x.dx}\t\n\\end{align*}","ts":1600430725579,"cs":"UjPEBW2pcQsE5KZFUfGgmw==","size":{"width":282,"height":76}}

We know that 2sinA.sinB = cos(A-B) - cos(A+B)

{"type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"id":"2-0-0-1-1-1-0-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\int_{}^{}\\left[\\cos\\left(-x\\right)-\\cos\\left(3x\\right)\\right].\\sin3x.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\int_{}^{}\\left[\\cos\\left(x\\right)-\\cos\\left(3x\\right)\\right].\\sin3x.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\left[\\int_{}^{}\\cos x.\\sin3x.dx-\\int_{}^{}\\cos 3x.\\sin3x.dx\\right]}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\left[\\frac{1}{2}\\int_{}^{}2\\cos x.\\sin3x.dx-\\frac{1}{2}\\int_{}^{}2\\cos 3x.\\sin3x.dx\\right]}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{4}\\left[\\int_{}^{}2\\cos x.\\sin3x.dx-\\int_{}^{}2\\cos 3x.\\sin3x.dx\\right]}\t\n\\end{align*}","ts":1600434108977,"cs":"8XZd8LnlJ44ffGkghZncOQ==","size":{"width":448,"height":204}}

We know that:- 2cosA.sinB = sin(A+B) - sin(A-B)

{"type":"align*","font":{"color":"#000000","size":9,"family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{4}\\left[\\int_{}^{}\\left[\\sin4x-\\sin\\left(-2x\\right)\\right].dx-\\int_{}^{}\\left(\\sin6x-\\sin\\left(0\\right)\\right).dx\\right]}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{4}\\left[\\int_{}^{}\\left[\\sin4x+\\sin\\left(2x\\right)\\right].dx-\\int_{}^{}\\sin6x.dx\\right]}\t\n\\end{align*}","id":"2-0-0-1-1-1-1-1","ts":1600434123882,"cs":"cNEBrPQc43gViSc6RoF9+w==","size":{"width":464,"height":80}}

We know that:-

{"type":"$","id":"3-0-2-1-1-0","font":{"color":"#000000","size":11,"family":"Arial"},"code":"$\\int_{}^{}\\sin x.dx=-\\cos x+c$","ts":1600433750984,"cs":"nDG8wlaTOza+5E1ypiaUFw==","size":{"width":188,"height":20}}

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{4}\\left[\\frac{-1}{4}\\cos4x+\\frac{-1}{2}\\cos2x-\\frac{1}{6}\\left(-\\cos6x\\right)\\right]+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{4}\\left[\\frac{1}{6}\\cos6x-\\frac{1}{4}\\cos4x-\\frac{1}{2}\\cos2x\\right]+C}\t\n\\end{align*}","type":"align*","id":"2-1-1-0-0-0-1-1-1-1-0","font":{"color":"#000000","family":"Arial","size":10},"ts":1600434551466,"cs":"dUrmLIHeKZMvwC48SsLElg==","size":{"width":408,"height":80}}

7. sin 4x.sin 8x

Solun:- Let f(x) = sin 4x.sin 8x

Integrate f(x):-

{"type":"align*","id":"2-0-0-0-1-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\sin 4x.\\sin 8x.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\int_{}^{}2\\sin 4x.\\sin 8x.dx}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1600434812573,"cs":"y1hhXgQFwX6HehD7HxTVwQ==","size":{"width":244,"height":76}}

We know that 2sinA.sinB = cos(A-B) - cos(A+B)

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\int_{}^{}\\left[\\cos\\left(-4x\\right)-\\cos\\left(12x\\right)\\right]dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\int_{}^{}\\left[\\cos\\left(4x\\right)-\\cos\\left(12x\\right)\\right]dx}\t\n\\end{align*}","type":"align*","id":"2-0-0-1-1-1-0-1-1-0","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1600434920326,"cs":"cHy/4k07qxmYmNrAzST7qw==","size":{"width":293,"height":76}}

We know that:-

{"font":{"family":"Arial","size":11,"color":"#000000"},"id":"3-0-2-1-1-1-0","type":"$","code":"$\\int_{}^{}\\cos x.dx=\\sin x+c$","ts":1600434968009,"cs":"utac/pO/BGA8TVS/oy4sAQ==","size":{"width":170,"height":20}}

{"id":"2-1-1-0-0-0-1-1-1-1-1-0","font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\left[\\frac{1}{4}\\sin4x-\\frac{1}{12}\\sin12x\\right]+C}\t\n\\end{align*}","type":"align*","ts":1600435264657,"cs":"9CbW9B3o8VcLXkwHnmfMNA==","size":{"width":293,"height":37}}

{"code":"$8.\\,\\frac{1-\\cos x}{1+\\cos x}$","type":"$","font":{"size":12,"family":"Arial","color":"#000000"},"id":"1-0","ts":1600502517035,"cs":"fVWATBYxYAt38EHt07N0UQ==","size":{"width":76,"height":25}}

Solun:- Let f(x) = {"font":{"color":"#000000","family":"Arial","size":10},"id":"3","code":"$\\frac{1-\\cos x}{1+\\cos x}$","type":"$","ts":1600502550866,"cs":"+wgqRRdLGPkGr1btU9/KxA==","size":{"width":44,"height":20}}

Integrate f(x):-

{"type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1-\\cos x}{1+\\cos x}.dx}\t\n\\end{align*}","id":"2-0-0-0-1-1-1-1-1-0","ts":1600502625656,"cs":"8zt8YnYUEjewy1RABDRy6Q==","size":{"width":198,"height":36}}

We know that 1 - cosx = 2sin2(x/2)

⇒ 1 + cos x = 2cos2(x/2)

{"font":{"family":"Arial","size":10,"color":"#000000"},"id":"2-0-0-1-1-1-0-1-1-1-0-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{2\\sin^{2}\\frac{x}{2}}{2\\cos^{2}\\frac{x}{2}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\tan ^{2}\\frac{x}{2}dx}\t\n\\end{align*}","type":"align*","ts":1600502945968,"cs":"rstEuMV2+kkPae7omq+XMA==","size":{"width":192,"height":84}}

We know that:- sec2x - tan2x = 1

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{} \\left(\\sec^{2}\\frac{x}{2}-1\\right).dx}\t\n\\end{align*}","id":"2-0-0-1-1-1-0-1-1-1-1-0","font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","ts":1600503070011,"cs":"siPyx2d27+4NeaFVnQURVg==","size":{"width":220,"height":36}}

We know that:-

{"font":{"color":"#000000","family":"Arial","size":11},"id":"3-0-2-1-1-1-1-0-0","type":"$","code":"$\\int_{}^{}\\sec ^{2}x.dx=\\tan x+c$","ts":1600503112330,"cs":"WllogtCPsd0DRjCDUKD+Tg==","size":{"width":180,"height":20}}

{"id":"3-0-2-1-1-1-1-1-0","code":"$\\int_{}^{}1.dx=x+c$","type":"$","font":{"size":11,"family":"Arial","color":"#000000"},"ts":1600503150229,"cs":"3QA0310SLjGKFi3sTgGHFw==","size":{"width":113,"height":20}}

{"id":"2-1-1-0-0-0-1-1-1-1-1-1-0","font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\left[\\frac{\\tan \\frac{x}{2}}{\\frac{1}{2}}-x\\right]+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={2.\\tan \\frac{x}{2}-x+C}\t\n\\end{align*}","type":"align*","ts":1600503324688,"cs":"F15m/3saIL2w6roeQWc9PQ==","size":{"width":212,"height":88}}

{"font":{"size":12,"family":"Arial","color":"#000000"},"type":"$","code":"$9.\\,\\frac{\\cos x}{1+\\cos x}$","id":"1-1-0","ts":1600503681165,"cs":"1yaXRe7g+Go3sr8P6DaeaA==","size":{"width":76,"height":22}}

Solun:- Let f(x) = {"font":{"size":10,"family":"Arial","color":"#000000"},"code":"$\\frac{\\cos x}{1+\\cos x}$","id":"4-0","type":"$","ts":1600503704887,"cs":"P+HUCDbj5xPXPrTeufj7SQ==","size":{"width":44,"height":17}}

Integrate f(x):-

{"id":"2-0-0-0-1-1-1-1-1-1-0","type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{\\cos x}{1+\\cos x}.dx}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1600503729289,"cs":"XSLA8n0iiLD37J2cCRC5xQ==","size":{"width":198,"height":36}}

We know that cosx = 1 - 2sin2(x/2)

⇒ 1 + cos x = 2cos2(x/2)

{"font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1-2\\sin^{2}\\frac{x}{2}}{2\\cos^{2}\\frac{x}{2}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{} \\frac{1}{2\\cos^{2}\\frac{x}{2}}dx-\\int_{}^{} \\frac{2\\sin^{2}\\frac{x}{2}}{2\\cos^{2}\\frac{x}{2}}dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{} \\frac{1}{2}\\sec^{2}\\frac{x}{2}dx-\\int_{}^{} \\tan^{2}\\frac{x}{2}dx}\t\n\\end{align*}","type":"align*","id":"2-0-0-1-1-1-0-1-1-1-0-1-0-0","ts":1600504115941,"cs":"QqqR3SwfBVV76C5x1PlqYA==","size":{"width":302,"height":132}}

We know that:- sec2x - tan2x = 1

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\int_{}^{} \\sec^{2}\\frac{x}{2}dx-\\int_{}^{} \\left(\\sec^{2}\\frac{x}{2}-1\\right)dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\int_{}^{} \\sec^{2}\\frac{x}{2}dx-\\int_{}^{} \\sec^{2}\\frac{x}{2}dx+\\int_{}^{} 1.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{-1}{2}\\int_{}^{} \\sec^{2}\\frac{x}{2}dx+\\int_{}^{} 1.dx}\t\n\\end{align*}","type":"align*","id":"2-0-0-1-1-1-0-1-1-1-0-1-1-0","font":{"color":"#000000","family":"Arial","size":10},"ts":1600504461808,"cs":"NKyRbHoMBgUEAW39tkapQA==","size":{"width":357,"height":116}}

We know that:-

{"font":{"color":"#000000","family":"Arial","size":11},"id":"3-0-2-1-1-1-1-0-1","type":"$","code":"$\\int_{}^{}\\sec ^{2}x.dx=\\tan x+c$","ts":1600503112330,"cs":"N5OtpgiZbd7YBpwbmeBR5w==","size":{"width":180,"height":20}}

{"id":"3-0-2-1-1-1-1-1-1","code":"$\\int_{}^{}1.dx=x+c$","type":"$","font":{"size":11,"family":"Arial","color":"#000000"},"ts":1600503150229,"cs":"rwAIQ7uxADQsU+AFAWZXdg==","size":{"width":113,"height":20}}

{"type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"id":"2-1-1-0-0-0-1-1-1-1-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{-1}{2}\\times\\frac{\\tan \\frac{x}{2}}{\\frac{1}{2}}+x+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={x-\\tan \\frac{x}{2}+C}\t\n\\end{align*}","ts":1600504766244,"cs":"uAEl3ljnMR5kYMZ6ZHDkYQ==","size":{"width":240,"height":82}}

{"code":"$10.\\,\\sin^{4}x$","font":{"size":12,"color":"#000000","family":"Arial"},"id":"1-1-1-0","type":"$","ts":1600504968402,"cs":"cGerH78LxzekajH7ASAaVg==","size":{"width":80,"height":17}}

Solun:- Let f(x) = sin4x

Integrate f(x):-

{"id":"2-0-0-0-1-1-1-1-1-1-1-0","type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\sin^{4}x.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\sin^{2}x.\\sin^{2}x.dx}\t\n\\end{align*}","ts":1600505538437,"cs":"Poett476fzQUqsi9wE3c+A==","size":{"width":213,"height":76}}

We know that 2.sin2x = 1 - cos 2x

{"id":"2-0-0-1-1-1-0-1-1-1-0-1-0-1-0-0","font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(\\frac{1-\\cos2x}{2}\\right).\\left(\\frac{1-\\cos2x}{2}\\right).dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{4}\\int_{}^{} \\left(1-\\cos2x-\\cos2x+\\cos^{2}2x\\right)dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{4}\\int_{}^{} \\left(1-2.\\cos2x+\\cos^{2}2x\\right)dx}\t\n\\end{align*}","ts":1600506197202,"cs":"5myQGGZYg34W68Ev41sdpw==","size":{"width":348,"height":118}}

We know that:- 2.cos2x = 1 + cos 2x

{"id":"2-0-0-1-1-1-0-1-1-1-0-1-0-1-1-0","type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{4}\\int_{}^{} \\left(1-2.\\cos2x+\\left(\\frac{1+\\cos4x}{2}\\right)\\right)dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{8}\\int_{}^{} \\left(2-4.\\cos2x+1+\\cos4x\\right)dx}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1600506704763,"cs":"mOgbwvqUCWcEj8POHZ+kXw==","size":{"width":364,"height":77}}

We know that:-

{"code":"$\\int_{}^{}\\cos x.dx=\\sin x+c$","font":{"size":11,"family":"Arial","color":"#000000"},"id":"3-0-2-1-1-1-1-0-2-0","type":"$","ts":1600506743196,"cs":"BEVbHYOdbh8I7EigxWXUuQ==","size":{"width":170,"height":20}}

{"id":"3-0-2-1-1-1-1-1-2","code":"$\\int_{}^{}1.dx=x+c$","type":"$","font":{"size":11,"family":"Arial","color":"#000000"},"ts":1600503150229,"cs":"30Uvpi8xyeU94yPd7MoKVw==","size":{"width":113,"height":20}}

{"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{8}\\left(2x-\\frac{4\\sin2x}{2}+x+\\frac{\\sin4x}{4}\\right)+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{8}\\left(2x-2\\sin2x+x+\\frac{\\sin4x}{4}\\right)+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\left(\\frac{3x}{8}-\\frac{1}{4}\\sin2x+\\frac{\\sin4x}{32}\\right)+C}\t\n\\end{align*}","font":{"size":10,"family":"Arial","color":"#000000"},"id":"2-1-1-0-0-0-1-1-1-1-1-1-1-1-0","ts":1600507231229,"cs":"MVU/7xqRp8AVNBniwuTm4Q==","size":{"width":344,"height":122}}

{"font":{"color":"#000000","size":12,"family":"Arial"},"id":"1-1-1-1-0","code":"$11.\\,\\cos^{4}2x$","type":"$","ts":1600507487747,"cs":"3J3neQ72RLhlPAXx63StkA==","size":{"width":92,"height":16}}

Solun:- Let f(x) = cos42x

Integrate f(x):-

{"font":{"family":"Arial","color":"#000000","size":10},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\cos^{4}2x.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\cos^{2}2x.\\cos^{2}2x.dx}\t\n\\end{align*}","id":"2-0-0-0-1-1-1-1-1-1-1-1-0","ts":1600508020792,"cs":"Ch/XRVM66rJ8Pz7rx+Sv1Q==","size":{"width":232,"height":76}}

We know that 2.cos2x = 1 + cos 2x

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(\\frac{1+\\cos4x}{2}\\right).\\left(\\frac{1+\\cos4x}{2}\\right).dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{4}\\int_{}^{} \\left(1+\\cos4x+\\cos4x+\\cos^{2}4x\\right)dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{4}\\int_{}^{} \\left(1+2.\\cos4x+\\cos^{2}4x\\right)dx}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","id":"2-0-0-1-1-1-0-1-1-1-0-1-0-1-0-1-0","ts":1600508223225,"cs":"OLtdYeosqLSbchZDpAkgkQ==","size":{"width":348,"height":118}}

We know that:- 2.cos2x = 1 + cos 2x

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{4}\\int_{}^{} \\left(1+2.\\cos4x+\\left(\\frac{1+\\cos8x}{2}\\right)\\right)dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{8}\\int_{}^{} \\left(2+4.\\cos4x+1+\\cos8x\\right)dx}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"id":"2-0-0-1-1-1-0-1-1-1-0-1-0-1-1-1-0","type":"align*","ts":1600508610063,"cs":"jRFC2SbORPKPXz6WjeNaSg==","size":{"width":364,"height":77}}

We know that:-

{"code":"$\\int_{}^{}\\cos x.dx=\\sin x+c$","font":{"size":11,"family":"Arial","color":"#000000"},"id":"3-0-2-1-1-1-1-0-2-1","type":"$","ts":1600506743196,"cs":"9rq3ssaEhVGPgQnJ+rPcqw==","size":{"width":170,"height":20}}

{"id":"3-0-2-1-1-1-1-1-3","code":"$\\int_{}^{}1.dx=x+c$","type":"$","font":{"size":11,"family":"Arial","color":"#000000"},"ts":1600503150229,"cs":"Fk46RuxbtVYFEzxiA56FsQ==","size":{"width":113,"height":20}}

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{8}\\left(2x+\\frac{4\\sin4x}{4}+x+\\frac{\\sin8x}{8}\\right)+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{8}\\left(3x+\\sin4x+\\frac{\\sin8x}{8}\\right)+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\left(\\frac{3x}{8}+\\frac{1}{8}\\sin4x+\\frac{\\sin8x}{64}\\right)+C}\t\n\\end{align*}","id":"2-1-1-0-0-0-1-1-1-1-1-1-1-1-1-0","type":"align*","font":{"family":"Arial","color":"#000000","size":10},"ts":1600508576330,"cs":"kj2QKKlCG46ydIONvsvNeg==","size":{"width":344,"height":122}}

{"font":{"size":12,"color":"#000000","family":"Arial"},"code":"$12.\\,\\frac{\\sin^{2}x}{1+\\cos x}$","id":"1-1-1-1-1-0","type":"$","ts":1600508861488,"cs":"4noaSnINEp2Q2xUJSbRDMg==","size":{"width":86,"height":28}}

Solun:- Let f(x) = {"id":"5-0","font":{"family":"Arial","color":"#000000","size":10},"code":"$\\frac{\\sin^{2}x}{1+\\cos x}$","type":"$","ts":1600508883299,"cs":"ETVaASycsLhp4zu2f6eU6w==","size":{"width":44,"height":21}}

Integrate f(x):-

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{\\sin^{2}x}{1+\\cos x}.dx}\t\n\\end{align*}","id":"2-0-0-0-1-1-1-1-1-1-1-1-1-0-0","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","ts":1600508951508,"cs":"PcVsvvq0zm5qqRaFwAgojA==","size":{"width":198,"height":38}}

We know that sin2x = 1 - cos2x

{"type":"align*","font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1-\\cos^{2}x}{1+\\cos x}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{\\left(1-\\cos x\\right)\\left(1+\\cos x\\right)}{1+\\cos x}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(1-\\cos x\\right).dx}\t\n\\end{align*}","id":"2-0-0-0-1-1-1-1-1-1-1-1-1-1-0","ts":1600509139435,"cs":"5zV4GL5tu/YCO7IPBSdjQg==","size":{"width":284,"height":120}}

We know that:-

{"code":"$\\int_{}^{}\\cos x.dx=\\sin x+c$","font":{"size":11,"family":"Arial","color":"#000000"},"id":"3-0-2-1-1-1-1-0-2-2","type":"$","ts":1600506743196,"cs":"hxHcRqNX7C/Z2F2r0fNUyQ==","size":{"width":170,"height":20}}

{"id":"3-0-2-1-1-1-1-1-4","code":"$\\int_{}^{}1.dx=x+c$","type":"$","font":{"size":11,"family":"Arial","color":"#000000"},"ts":1600503150229,"cs":"/0hXm6RgC8BTGjguyS3Flw==","size":{"width":113,"height":20}}

{"font":{"size":10,"color":"#000000","family":"Arial"},"id":"2-1-1-0-0-0-1-1-1-1-1-1-1-1-1-1-0","type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={x-\\sin x+C}\t\n\\end{align*}","ts":1600509240679,"cs":"s0a1pHN0p3dFyNLG+zZX9g==","size":{"width":178,"height":36}}

{"type":"$","font":{"color":"#000000","family":"Arial","size":12},"code":"$13.\\,\\frac{\\cos2x-\\cos2\\alpha}{\\cos x-\\cos\\alpha}$","id":"1-1-1-1-1-1-0","ts":1600509314484,"cs":"zqHF0s36YdwxA//aYWbxeg==","size":{"width":124,"height":25}}

Solun:- Let f(x) = {"code":"$\\frac{\\cos2x-\\cos2\\alpha}{\\cos x-\\cos\\alpha}$","id":"5-1-0","type":"$","font":{"color":"#000000","family":"Arial","size":10},"ts":1600509333858,"cs":"OpaAQZlYQA/8TV2h8PW6Pg==","size":{"width":73,"height":20}}

Integrate f(x):-

{"type":"align*","id":"2-0-0-0-1-1-1-1-1-1-1-1-1-0-1-0","font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{\\cos2x-\\cos2\\alpha}{\\cos x-\\cos\\alpha}.dx}\t\n\\end{align*}","ts":1600510003146,"cs":"QeJNEojyP8DLZJMeu6ma+A==","size":{"width":240,"height":36}}

We know that 

{"code":"$\\cos C-\\cos D=-2\\sin\\left(\\frac{C+D}{2}\\right)\\sin\\left(\\frac{C-D}{2}\\right)$","font":{"color":"#000000","size":10,"family":"Arial"},"id":"6","type":"$","ts":1600510095700,"cs":"ZvZxQzrgZH5N934pTh/u7g==","size":{"width":278,"height":20}}

{"font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","id":"2-0-0-0-1-1-1-1-1-1-1-1-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{-2\\sin\\left(\\frac{2x+2\\alpha}{2}\\right)\\sin\\left(\\frac{2x-2\\alpha}{2}\\right)}{-2\\sin\\left(\\frac{x+\\alpha}{2}\\right)\\sin\\left(\\frac{x-\\alpha}{2}\\right)}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{\\sin\\left(x+\\alpha\\right)\\sin\\left(x-\\alpha\\right)}{\\sin\\left(\\frac{x+\\alpha}{2}\\right)\\sin\\left(\\frac{x-\\alpha}{2}\\right)}.dx}\t\n\\end{align*}","ts":1600510317349,"cs":"2DMbsl1EST2VXO4xyWHB5A==","size":{"width":317,"height":90}}

We know that:-

{"id":"7","font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{\\sin2x}&={2\\sin x.\\cos x}\\\\\n{\\sin x\\,\\,}&={2\\sin\\frac{x}{2}.\\cos\\frac{x}{2}}\t\n\\end{align*}","type":"align*","ts":1600510378343,"cs":"molDYTCi99187Bbb2QDCqg==","size":{"width":156,"height":48}}

{"id":"2-0-0-0-1-1-1-1-1-1-1-1-1-1-1-1","type":"align*","font":{"color":"#000000","family":"Arial","size":9},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{\\left[2\\sin\\left(\\frac{x+\\alpha}{2}\\right).\\cos\\left(\\frac{x+\\alpha}{2}\\right)\\right]\\left[2\\sin\\left(\\frac{x-\\alpha}{2}\\right).\\cos\\left(\\frac{x-\\alpha}{2}\\right)\\right]}{\\sin\\left(\\frac{x+\\alpha}{2}\\right)\\sin\\left(\\frac{x-\\alpha}{2}\\right)}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={4\\int_{}^{}\\cos\\left(\\frac{x+\\alpha}{2}\\right).\\cos\\left(\\frac{x-\\alpha}{2}\\right).dx}\t\n\\end{align*}","ts":1600510703273,"cs":"RDCqF6o+tqFivrvLu/wswQ==","size":{"width":466,"height":86}}

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={2\\int_{}^{}2.\\cos\\left(\\frac{x+\\alpha}{2}\\right).\\cos\\left(\\frac{x-\\alpha}{2}\\right).dx}\\\\\n\\end{align*}","type":"align*","id":"8-0-0","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1600510737887,"cs":"ryj4LAoYKLS7PHkobkNLiA==","size":{"width":349,"height":37}}

We know that:-

2cosA.cosB = cos(A+B) + cos(A-B)

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={2\\int_{}^{}\\left[\\cos\\left(x\\right)+\\cos\\left(\\alpha\\right)\\right].dx}\\\\\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"id":"8-1-0","type":"align*","ts":1600511350454,"cs":"1Anwj2A0f3c4V426I6Irww==","size":{"width":258,"height":36}}

We know that:-

{"code":"$\\int_{}^{}\\cos x.dx=\\sin x+c$","font":{"size":11,"family":"Arial","color":"#000000"},"id":"3-0-2-1-1-1-1-0-2-3-0","type":"$","ts":1600506743196,"cs":"sSLTi9aGsJDvDTmisLuiRA==","size":{"width":170,"height":20}}

{"font":{"color":"#000000","family":"Arial","size":10},"id":"8-1-1-0","type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={2\\left(\\sin x+x.\\cos\\left(\\alpha\\right)\\right)}\\\\\n\\end{align*}","ts":1600511817113,"cs":"ahHVxrQaRdhFMJWjGa5uPQ==","size":{"width":220,"height":36}}

{"id":"1-1-1-1-1-1-1-0","font":{"color":"#000000","family":"Arial","size":12},"type":"$","code":"$14.\\,\\frac{\\cos x-\\sin x}{1+\\sin2x}$","ts":1600513494406,"cs":"eplSVQCDidH05CTy9rdugA==","size":{"width":108,"height":25}}

Solun:- Let f(x) = {"font":{"color":"#000000","size":10,"family":"Arial"},"type":"$","code":"$\\frac{\\cos x-\\sin x}{1+\\sin2x}$","id":"5-1-1-0","ts":1600513506110,"cs":"l2IatOQxluE3eqIohyoLCQ==","size":{"width":60,"height":20}}

Integrate f(x):-

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{\\cos x-\\sin x}{1+\\sin2x}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{\\cos x-\\sin x}{\\sin^{2}x+\\cos^{2}x+2\\sin x\\cos x}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{\\cos x-\\sin x}{\\left(\\sin x+\\cos x\\right)^{2}}.dx}\t\n\\end{align*}","id":"2-0-0-0-1-1-1-1-1-1-1-1-1-0-1-1-0","font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","ts":1600513695233,"cs":"2jxuvPhlDsAkBvl4mR/2wA==","size":{"width":332,"height":120}}

Let sin x + cos x = t

Differentiate w.r.t to t:-

⇒ (cos x - sin x).dx = dt

{"id":"2-1-0-0-0-1-0-0","font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{dt}{t^{2}}}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}t^{-2}.dt}\t\n\\end{align*}","type":"align*","ts":1600520274515,"cs":"a2/2T6mToGuLNV0m2RwVqQ==","size":{"width":148,"height":76}}

We know that:-

{"type":"$","font":{"size":11,"color":"#000000","family":"Arial"},"id":"3-0-2-1-1-1-1-0-2-3-1-0","code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+c$","ts":1600514182431,"cs":"q5l1iuLGLeI/B/50euG0Qw==","size":{"width":149,"height":24}}

{"type":"align*","font":{"color":"#000000","family":"Arial","size":10},"id":"2-1-0-0-0-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{t^{-1}}{-1}+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{-1}{t}+C}\t\n\\end{align*}","ts":1600514363377,"cs":"X8gBKbHSUMj6Jx5k0DSyfg==","size":{"width":146,"height":78}}

Put that value of t:-

{"font":{"family":"Arial","color":"#000000","size":10},"id":"8-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{-1}{\\sin x+\\cos x}+C}\\\\\n\\end{align*}","type":"align*","ts":1600514486780,"cs":"MdPsSwvKqWq82ZiPc03ZoA==","size":{"width":209,"height":36}}

{"font":{"color":"#000000","size":12,"family":"Arial"},"code":"$15.\\,\\tan^{3}2x.\\sec2x$","id":"1-1-1-1-1-1-1-1-0","type":"$","ts":1600514745870,"cs":"R0OUK1BF751KEVd7PvNohg==","size":{"width":156,"height":16}}

Solun:- Let f(x) = tan32x.sec2x

Integrate f(x):-

{"font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\tan^{3}2x.\\sec2x.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\tan^{2}2x.\\tan2x.\\sec2x.dx}\t\n\\end{align*}","type":"align*","id":"2-0-0-0-1-1-1-1-1-1-1-1-1-0-1-1-1-0-0","ts":1600518889238,"cs":"W33tw/Bmo4LDErAsNaJVUg==","size":{"width":276,"height":76}}

We know that:-

sec2x - tan2x = 1

{"font":{"size":10,"family":"Arial","color":"#000000"},"id":"2-0-0-0-1-1-1-1-1-1-1-1-1-0-1-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(\\sec^{2}2x-1\\right).\\tan2x.\\sec2x.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\sec^{2}2x.\\tan2x.\\sec2x.dx-\\int_{}^{}\\tan2x.\\sec2x.dx}\t\n\\end{align*}","type":"align*","ts":1600519605144,"cs":"KY8CjqZ9axpiYa3VFZYR2w==","size":{"width":426,"height":76}}

Let sec 2x = t

Differentiate w.r.t to t:-

⇒ 2.sec 2x.tan 2x.dx = dt

⇒ sec 2x.tan 2x.dx = (1/2).dt

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\int_{}^{}t^{2}.dt-\\int_{}^{}\\tan2x.\\sec2x.dt}\t\n\\end{align*}","id":"2-1-0-0-0-1-0-1-0","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","ts":1600521106429,"cs":"+ktoq62c0rQRp5ft2ggppA==","size":{"width":306,"height":36}}

We know that:-

{"type":"$","font":{"size":11,"color":"#000000","family":"Arial"},"id":"3-0-2-1-1-1-1-0-2-3-1-1-0","code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+c$","ts":1600514182431,"cs":"3lCdPA+pTRIFJyac4ANTxw==","size":{"width":149,"height":24}}

{"code":"$\\int_{}^{}\\sec x.\\tan x.dx=\\sec x+c$","id":"9-0","type":"$","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1600521145954,"cs":"gl1By51BbcZBeGQlfOUCpw==","size":{"width":190,"height":17}}

{"id":"2-1-0-0-0-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{2}\\times\\frac{t^{3}}{3}-\\frac{\\sec2x}{2}+C}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"type":"align*","ts":1600521286409,"cs":"T6Gtcu3EAPKnWHRkfdrgaA==","size":{"width":238,"height":37}}

Put that value of t:-

{"font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","id":"8-1-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{\\sec^{3}2x}{6}-\\frac{\\sec2x}{2}+C}\\\\\n\\end{align*}","ts":1600521271570,"cs":"ZATGxq4gUanHngR3K9IBiQ==","size":{"width":238,"height":37}}

{"code":"$16.\\,\\tan^{4}x$","font":{"family":"Arial","size":12,"color":"#000000"},"type":"$","id":"1-1-1-1-1-1-1-1-1-0","ts":1600521388249,"cs":"V3mjLi3bj0qrcgiVbYEFzA==","size":{"width":84,"height":17}}

Solun:- Let f(x) = tan4x

Integrate f(x):-

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\tan^{4}x.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\tan^{2}x.\\tan ^{2}x.dx}\t\n\\end{align*}","id":"2-0-0-0-1-1-1-1-1-1-1-1-1-0-1-1-1-0-1-0","type":"align*","font":{"color":"#000000","family":"Arial","size":10},"ts":1600521434522,"cs":"X7LW7vh1Hp/qtF/+Co5wtQ==","size":{"width":220,"height":76}}

We know that:-

sec2x - tan2x = 1

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(\\sec^{2}x-1\\right).\\tan ^{2}x.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\sec^{2}x.\\tan ^{2}x.dx-\\int_{}^{}\\tan ^{2}x.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\sec^{2}x.\\tan ^{2}x.dx-\\int_{}^{} \\left(\\sec^{2}x-1\\right).dx}\t\n\\end{align*}","type":"align*","id":"2-0-0-0-1-1-1-1-1-1-1-1-1-0-1-1-1-1-1","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1600521535749,"cs":"TfG01aDkiTgq0RcMJrRZSQ==","size":{"width":358,"height":116}}

Let tan x = t

Differentiate w.r.t to t:-

⇒ sec2x.dx = dt

{"font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}t^{2}.dt-\\int_{}^{}\\left(\\sec^{2}x-1\\right).dx}\t\n\\end{align*}","type":"align*","id":"2-1-0-0-0-1-0-1-1-0","ts":1600607218225,"cs":"yn96KUg6b6Ljw9o4a7O9Sw==","size":{"width":280,"height":36}}

We know that:-

{"type":"$","font":{"size":11,"color":"#000000","family":"Arial"},"id":"3-0-2-1-1-1-1-0-2-3-1-1-1-0","code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+c$","ts":1600514182431,"cs":"HSfWOPQeb3NKItogdQ0Zxg==","size":{"width":149,"height":24}}

{"id":"9-1-0-0","font":{"color":"#000000","size":10,"family":"Arial"},"type":"$","code":"$\\int_{}^{}\\sec ^{2}x.dx=\\tan x+c$","ts":1600521636629,"cs":"PH8C1kzwpFwvMDWuG9e2Vw==","size":{"width":158,"height":18}}

{"type":"align*","id":"2-1-0-0-0-1-1-1-1-0-0","font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{t^{3}}{3}-\\tan x+x+C}\t\n\\end{align*}","ts":1600521682446,"cs":"9Tj/gprKxVezSvJ1cqUm4w==","size":{"width":220,"height":37}}

Put that value of t:-

{"id":"8-1-1-1-1-1-0-0","type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{\\tan^{3}x}{3}-\\tan x+x+C}\\\\\n\\end{align*}","ts":1600521718500,"cs":"p9tCdPPgWdIYa1OcrSQEeg==","size":{"width":249,"height":37}}

{"code":"$17.\\,\\frac{\\sin^{3}x+\\cos^{3}x}{\\sin^{2}x\\cos^{2}x}$","font":{"size":12,"color":"#000000","family":"Arial"},"type":"$","id":"1-1-1-1-1-1-1-1-1-1-0","ts":1600592829192,"cs":"myqv9kz7T/k583eMkiXxtQ==","size":{"width":118,"height":29}}

Solun:- Let f(x) = {"code":"$\\frac{\\sin^{3}x+\\cos^{3}x}{\\sin^{2}x\\cos^{2}x}$","font":{"size":10,"family":"Arial","color":"#000000"},"id":"10-0","type":"$","ts":1600592866423,"cs":"Fv+063B9s94Y+E9au5vrJw==","size":{"width":69,"height":22}}

Integrate f(x):-

{"font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{\\sin^{3}x+\\cos^{3}x}{\\sin^{2}x\\cos^{2}x}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(\\frac{\\sin ^{3}x}{\\sin^{2}x.\\cos^{2}x}+\\frac{ \\cos^{3}x}{\\sin^{2}x.\\cos^{2}x}\\right).dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(\\frac{\\sin x}{\\cos^{2}x}+\\frac{ \\cos x}{\\sin^{2}x}\\right).dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left(\\sec x.\\tan x+\\cot x.\\cos ecx\\right).dx}\t\n\\end{align*}","id":"2-0-0-0-1-1-1-1-1-1-1-1-1-0-1-1-1-0-1-1-0","type":"align*","ts":1600593069477,"cs":"JouSKZap6GqRfhUy7CoHzw==","size":{"width":352,"height":165}}

We know that:-

{"code":"$\\int_{}^{}\\sec x.\\tan x.dx=\\sec x+c$","type":"$","id":"3-0-2-1-1-1-1-0-2-3-1-1-2-0","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1600593368241,"cs":"XE60u9B5eAMUSO6r02qFZA==","size":{"width":190,"height":17}}

{"code":"$\\int_{}^{}\\cos ec\\,x.\\cot x.dx=-\\cos ec\\,x+c$","id":"9-1-1-0","font":{"color":"#000000","family":"Arial","size":10},"type":"$","ts":1600593792286,"cs":"tEeKpGd2c+0g0AnM5lRSEA==","size":{"width":240,"height":17}}

{"font":{"color":"#000000","family":"Arial","size":10},"type":"align*","id":"2-1-0-0-0-1-1-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\sec x-\\cos ec\\,x+C}\t\n\\end{align*}","ts":1600593608266,"cs":"s8Xuzc016AV7x8FPl/tpbg==","size":{"width":220,"height":36}}

{"type":"$","font":{"size":12,"color":"#000000","family":"Arial"},"code":"$18.\\,\\frac{\\cos2x+2\\sin^{2}x}{\\cos^{2}x}$","id":"1-1-1-1-1-1-1-1-1-1-1-0","ts":1600593963796,"cs":"L7a3bItkSkWzyMzJ7maiog==","size":{"width":129,"height":28}}

Solun:- Let f(x) = {"font":{"family":"Arial","color":"#000000","size":10},"id":"10-1-0","type":"$","code":"$\\frac{\\cos2x+2\\sin^{2}x}{\\cos^{2}x}$","ts":1600594050944,"cs":"gv4XRmrp18JYwcBOM2zfyw==","size":{"width":78,"height":21}}

Integrate f(x):-

{"id":"2-0-0-0-1-1-1-1-1-1-1-1-1-0-1-1-1-0-1-1-1-0-0","type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{\\cos2x+2\\sin^{2}x}{\\cos^{2}x}.dx}\t\n\\end{align*}","ts":1600594141621,"cs":"UMtCwmMbKOs5Rh2lVzl/rQ==","size":{"width":246,"height":38}}

We know that:-

cos 2x = 1 - 2sin2x

{"type":"align*","font":{"family":"Arial","color":"#000000","size":10},"id":"2-0-0-0-1-1-1-1-1-1-1-1-1-0-1-1-1-0-1-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1-2\\sin^{2}x+2\\sin^{2}x}{\\cos^{2}x}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{\\cos^{2}x}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\sec^{2}x.dx}\t\n\\end{align*}","ts":1600594347700,"cs":"wGg3z1JWXD2UxzaKiYesVA==","size":{"width":281,"height":120}}

We know that:-

{"font":{"family":"Arial","size":10,"color":"#000000"},"type":"$","code":"$\\int_{}^{}\\sec ^{2}x.dx=\\tan x+c$","id":"3-0-2-1-1-1-1-0-2-3-1-1-2-1","ts":1600594462692,"cs":"jFIke0HPvJOmyw7Z6KOONA==","size":{"width":158,"height":18}}

{"id":"2-1-0-0-0-1-1-1-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\tan x+C}\t\n\\end{align*}","type":"align*","font":{"color":"#000000","family":"Arial","size":10},"ts":1600594525971,"cs":"/QQr1PQj8IIL2qxU+rbg6A==","size":{"width":153,"height":36}}

{"font":{"size":12,"family":"Arial","color":"#000000"},"type":"$","code":"$19.\\,\\frac{1}{\\sin x\\cos^{3}x}$","id":"1-1-1-1-1-1-1-1-1-1-1-1-0","ts":1600596057811,"cs":"ob4eLLJOxI3XyImhlZanvA==","size":{"width":104,"height":24}}

Solun:- Let f(x) = {"code":"$\\frac{1}{\\sin x\\cos^{3}x}$","font":{"color":"#000000","size":10,"family":"Arial"},"type":"$","id":"10-1-1-0","ts":1600596129184,"cs":"NdPH+X5waEgPPf+HrPDkvA==","size":{"width":58,"height":20}}

Integrate f(x):-

{"id":"2-0-0-0-1-1-1-1-1-1-1-1-1-0-1-1-1-0-1-1-1-0-1-0","font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{\\sin x\\cos^{3}x}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{\\frac{\\sin x}{\\cos x}.\\cos^{3}x.\\cos x}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{\\sec^{2}x.\\sec^{2}x}{\\tan x}.dx}\t\n\\end{align*}","ts":1600607399854,"cs":"BM/Lly1JyqouX0I2GUL9GQ==","size":{"width":254,"height":124}}

We know that:-

sec2x - tan2x = 1

{"font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","id":"2-0-0-0-1-1-1-1-1-1-1-1-1-0-1-1-1-0-1-1-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{\\sec^{2}x.\\left(\\tan^{2}x+1\\right)}{\\tan x}.dx}\t\n\\end{align*}","ts":1600607449902,"cs":"JYa+G0fjSJE41ZrZ+8Uojw==","size":{"width":266,"height":40}}

Let tan x = t

Differentiate w.r.t to t:-

⇒ sec2x.dx = dt

{"id":"2-1-0-0-0-1-0-1-1-1-0","font":{"family":"Arial","color":"#000000","size":10},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{\\left(t^{2}+1\\right)}{t}.dt}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{t^{2}}{t}.dt+\\int_{}^{}\\frac{1}{t}.dt}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}t.dt+\\int_{}^{}\\frac{1}{t}.dt}\t\n\\end{align*}","ts":1600607472618,"cs":"N4BqsxDH4AEitaNh/mcXSQ==","size":{"width":221,"height":124}}

We know that:-

{"type":"$","font":{"size":11,"color":"#000000","family":"Arial"},"id":"3-0-2-1-1-1-1-0-2-3-1-1-1-1-0","code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+c$","ts":1600514182431,"cs":"wRGP0WT5Edyb4laRLYJnsA==","size":{"width":149,"height":24}}

{"code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}x+c$","id":"11-0","font":{"family":"Arial","size":11,"color":"#000000"},"type":"$","ts":1600597054172,"cs":"btkL8cATH12lQgR7Qu5dMg==","size":{"width":150,"height":21}}

{"type":"align*","font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{t^{2}}{2}+\\log_{}t+C}\t\n\\end{align*}","id":"2-1-0-0-0-1-1-1-1-0-1-0","ts":1600607493059,"cs":"t3pcLpbPgymWZg8ccCf/fg==","size":{"width":186,"height":37}}

Put that value of t:-

{"type":"align*","font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{\\tan^{2}x}{2}+\\log_{}\\left|\\tan x\\right|+C}\\\\\n\\end{align*}","id":"8-1-1-1-1-1-0-1-0","ts":1600607509456,"cs":"wAjlxhcehlZEcmW7boLmxA==","size":{"width":253,"height":37}}

{"type":"$","code":"$20.\\,\\frac{\\cos2x}{\\left(\\cos x+\\sin x\\right)^{2}}$","id":"1-1-1-1-1-1-1-1-1-1-1-1-1-0","font":{"color":"#000000","family":"Arial","size":12},"ts":1600597324284,"cs":"qLUcnv7RpeWeA8RY3zjcXA==","size":{"width":124,"height":32}}

Solun:- Let f(x) = {"code":"$\\frac{\\cos2x}{\\left(\\cos x+\\sin x\\right)^{2}}$","type":"$","font":{"size":10,"family":"Arial","color":"#000000"},"id":"10-1-1-1-0","ts":1600597342617,"cs":"As8XCDb8fpByDaffrmKL5A==","size":{"width":73,"height":24}}

Integrate f(x):-

{"font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{\\cos2x}{\\left(\\cos x+\\sin x\\right)^{2}}.dx}\t\n\\end{align*}","type":"align*","id":"2-0-0-0-1-1-1-1-1-1-1-1-1-0-1-1-1-0-1-1-1-0-1-1-0","ts":1600597499095,"cs":"sBDSUMVAXS++NEKjwAjvUg==","size":{"width":240,"height":40}}

We know that:-

cos 2x = cos2x - sin2x

{"font":{"color":"#000000","family":"Arial","size":10},"type":"align*","id":"2-0-0-0-1-1-1-1-1-1-1-1-1-0-1-1-1-0-1-1-1-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{\\left(\\cos^{2}x-\\sin^{2}x\\right)}{\\left(\\cos x+\\sin x\\right)^{2}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{\\left(\\cos x-\\sin x\\right)\\left(\\cos x+\\sin x\\right)}{\\left(\\cos x+\\sin x\\right)^{2}}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{\\left(\\cos x-\\sin x\\right)}{\\left(\\cos x+\\sin x\\right)}.dx}\t\n\\end{align*}","ts":1600598075016,"cs":"8RccqZwpNq4YLzvTiZuoUg==","size":{"width":329,"height":133}}

Let cos x + sinx = t

Differentiate w.r.t to t:-

⇒ (- sin x+cos x).dx = dt

{"font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{dt}{t}}\t\n\\end{align*}","id":"2-1-0-0-0-1-0-1-1-1-1-0","ts":1600598190430,"cs":"x4oND14/GWJ1QSjsmTOlig==","size":{"width":128,"height":36}}

We know that:-

{"font":{"family":"Arial","color":"#000000","size":11},"id":"3-0-2-1-1-1-1-0-2-3-1-1-1-1-1-0-0","type":"$","code":"$\\int_{}^{}\\frac{1}{x}.dx=\\log_{}x+c$","ts":1600598348778,"cs":"8KZCb3b0tAbOMLzQNGVL7Q==","size":{"width":150,"height":21}}

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\log_{}\\left|t\\right|+C}\t\n\\end{align*}","id":"2-1-0-0-0-1-1-1-1-0-1-1-0-0","type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1600598375432,"cs":"SKzJVYvsQTNCNyGMlpEzZA==","size":{"width":157,"height":36}}

Put value of t in this eq. :-

{"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\log_{}\\left|\\cos x+\\sin x\\right|+C}\t\n\\end{align*}","id":"2-1-0-0-0-1-1-1-1-0-1-1-1-0","font":{"color":"#000000","family":"Arial","size":10},"ts":1600598567393,"cs":"wV3+IzUikhaTTvKbXDnj8Q==","size":{"width":234,"height":36}}

{"font":{"color":"#000000","size":12,"family":"Arial"},"code":"$21.\\,\\sin^{-1}\\left(\\cos x\\right)$","type":"$","id":"1-1-1-1-1-1-1-1-1-1-1-1-1-1-0","ts":1600599125394,"cs":"SAirg8c7LW6XWimiPsQX2w==","size":{"width":138,"height":22}}

Solun:- Let f(x) = {"type":"$","id":"10-1-1-1-1-0","font":{"size":10,"family":"Arial","color":"#000000"},"code":"$\\sin^{-1}\\left(\\cos x\\right)$","ts":1600599161169,"cs":"yJM0HOJBdxn/vLzaKusVQA==","size":{"width":81,"height":17}}

Integrate f(x):-

{"id":"2-0-0-0-1-1-1-1-1-1-1-1-1-0-1-1-1-0-1-1-1-0-1-1-1-0-0","type":"align*","font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\sin^{-1}\\left(\\cos x\\right).dx}\t\n\\end{align*}","ts":1600599222052,"cs":"DtSfZ7AET9Ci0EOOQjM+VA==","size":{"width":213,"height":36}}

We know that:-

{"id":"12-0","code":"$\\sin^{-1}x+\\cos^{-1}x=\\frac{\\Pi}{2}$","font":{"family":"Arial","color":"#000000","size":10},"type":"$","ts":1600599354701,"cs":"5SbBay6xrY7KErB4mJojjQ==","size":{"width":149,"height":18}}

{"font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left[\\frac{\\Pi}{2}-\\cos^{-1}\\left(\\cos x\\right)\\right].dx}\t\n\\end{align*}","type":"align*","id":"2-0-0-0-1-1-1-1-1-1-1-1-1-0-1-1-1-0-1-1-1-0-1-1-1-1-0","ts":1600599472964,"cs":"GAtpB2g4tSfqqtcx6N2Tmw==","size":{"width":268,"height":37}}

{"font":{"family":"Arial","color":"#000000","size":10},"id":"2-0-0-0-1-1-1-1-1-1-1-1-1-0-1-1-1-0-1-1-1-1-1-1-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left[\\frac{\\Pi}{2}-\\cos^{-1}\\left(\\cos x\\right)\\right].dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\left[\\frac{\\Pi}{2}-x\\right].dx}\t\n\\end{align*}","type":"align*","ts":1600599710844,"cs":"lLX/qq/2JJGZjW8S1x7p5A==","size":{"width":268,"height":80}}

We know that:-

{"code":"$\\int_{}^{}x^{n}.dx=\\frac{x^{n+1}}{n+1}+c$","type":"$","font":{"color":"#000000","family":"Arial","size":11},"id":"3-0-2-1-1-1-1-0-2-3-1-1-1-1-1-1-0-0","ts":1600599891629,"cs":"0sLO0hZC3s17TXvYlsev8Q==","size":{"width":149,"height":24}}

{"type":"$","font":{"size":11,"family":"Arial","color":"#000000"},"code":"$\\int_{}^{}1.dx=x+c$","id":"3-0-2-1-1-1-1-0-2-3-1-1-1-1-1-1-1","ts":1600599941595,"cs":"M2WrS4XoG8gOWQfOMzNpXQ==","size":{"width":113,"height":20}}

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{\\Pi.x}{2}-\\frac{x^{2}}{2}+C}\t\n\\end{align*}","type":"align*","id":"2-1-0-0-0-1-1-1-1-0-1-1-2-0","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1600600072961,"cs":"sX27k0lOwmqiTDGl1AlpOA==","size":{"width":196,"height":37}}

{"id":"1-1-1-1-1-1-1-1-1-1-1-1-1-1-1","font":{"size":12,"family":"Arial","color":"#000000"},"code":"$22.\\,\\frac{1}{\\cos\\left(x-a\\right).\\cos\\left(x-b\\right)}$","type":"$","ts":1600600272109,"cs":"JsS/2/OxJsCIOQHcqUwwRg==","size":{"width":156,"height":28}}

Solun:- Let f(x) = {"font":{"family":"Arial","size":10,"color":"#000000"},"id":"10-1-1-1-1-1-0","type":"$","code":"$\\frac{1}{\\cos\\left(x-a\\right).\\cos\\left(x-b\\right)}$","ts":1600600291342,"cs":"gEXiKu4wn0CcMxutHavPYw==","size":{"width":97,"height":21}}

Integrate f(x):-

{"type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{\\cos\\left(x-a\\right).\\cos\\left(x-b\\right)}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\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{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{\\sin\\left(a-b\\right)}\\int_{}^{}\\frac{\\sin \\left[\\left(x-b\\right)-\\left(x-a\\right)\\right]}{\\cos\\left(x-a\\right).\\cos\\left(x-b\\right)}.dx}\t\n\\end{align*}","id":"2-0-0-0-1-1-1-1-1-1-1-1-1-0-1-1-1-0-1-1-1-0-1-1-1-0-1-0","ts":1600600531593,"cs":"HPXhK92G8N4Mzv7CPR/Pbg==","size":{"width":368,"height":122}}

We know that:-

sin (A - B) = sinA.cosB - cosA.sinB

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{\\sin\\left(a-b\\right)}\\int_{}^{}\\frac{\\sin \\left(x-b\\right).\\cos\\left(x-a\\right)-\\cos\\left(x-b\\right).\\sin\\left(x-a\\right)}{\\cos\\left(x-a\\right).\\cos\\left(x-b\\right)}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{\\sin\\left(a-b\\right)}\\int_{}^{}\\left[\\tan\\left(x-b\\right)-\\tan\\left(x-a\\right)\\right].dx}\t\n\\end{align*}","id":"2-0-0-0-1-1-1-1-1-1-1-1-1-0-1-1-1-0-1-1-1-0-1-1-1-0-1-1","type":"align*","font":{"size":8,"color":"#000000","family":"Arial"},"ts":1600600794105,"cs":"X1/Aw2LjNRnd6fy0STXS2A==","size":{"width":458,"height":68}}

We know that:-

{"type":"$","font":{"family":"Arial","size":10,"color":"#000000"},"code":"$\\int_{}^{}\\tan x.dx=-\\log_{}\\left|\\cos x\\right|+c$","id":"3-0-2-1-1-1-1-0-2-3-1-1-1-1-1-1-0-1-0","ts":1600600843319,"cs":"h2p1bsc+5uhHe07c4S/W8Q==","size":{"width":200,"height":17}}

{"font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","id":"2-1-0-0-0-1-1-1-1-0-1-1-2-1-0","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{\\sin\\left(a-b\\right)}\\left[-\\log_{}\\left|\\cos\\left(x-b\\right)\\right|+\\log_{}\\left|\\cos\\left(x-a\\right)\\right|\\right]+C}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\frac{1}{\\sin\\left(a-b\\right)}\\left[\\log_{}\\left|\\frac{\\cos\\left(x-a\\right)}{\\cos\\left(x-b\\right)}\\right|\\right]+C}\t\n\\end{align*}","ts":1600600966578,"cs":"voWr469WHKeqXPmsB/uXAA==","size":{"width":448,"height":80}}

Choose the correct answer in Exercises 23 and 24.

{"font":{"size":12,"family":"Arial","color":"#000000"},"code":"$23.\\,\\int_{}^{}\\frac{\\sin^{2}x-\\cos^{2}x}{\\sin ^{2}x.\\cos^{2}x}dx$","id":"13-0","type":"$","ts":1600601129751,"cs":"XuuB30IuDT6SGJ3GUST6LA==","size":{"width":162,"height":29}} is equal to

(A) tan x + cot x + C

(B) tan x + cosec x + C

(C) - tan x + cot x + C

(D) tan x + sec x + C

Solun:- Let f(x) = {"font":{"size":10,"family":"Arial","color":"#000000"},"id":"10-1-1-1-1-1-1-0","code":"$\\frac{\\sin ^{2}x-\\cos ^{2}x}{\\sin ^{2}x.\\cos^{2}x}$","type":"$","ts":1600601389230,"cs":"dfSJFw5SqaBdtJ2IOKOJ3Q==","size":{"width":69,"height":22}}

Integrate f(x):-

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{\\sin ^{2}x-\\cos ^{2}x}{\\sin ^{2}x.\\cos^{2}x}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{\\sin^{2}x}{\\sin ^{2}x.\\cos^{2}x}.dx-\\int_{}^{}\\frac{\\cos^{2}x}{\\sin ^{2}x.\\cos^{2}x}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{1}{\\cos^{2}x}.dx-\\int_{}^{}\\frac{1}{\\sin^{2}x}.dx}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\sec^{2}x.dx-\\int_{}^{}\\cos ec^{2}x.dx}\t\n\\end{align*}","type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"id":"2-0-0-0-1-1-1-1-1-1-1-1-1-0-1-1-1-0-1-1-1-0-1-1-1-0-1-2-0","ts":1600601521350,"cs":"qTMGY+wAMDB/wvGxCrnXMA==","size":{"width":376,"height":162}}

We know that:-

{"code":"$\\int_{}^{}\\sec ^{2}x.dx=\\tan x+c$","id":"3-0-2-1-1-1-1-0-2-3-1-1-1-1-1-1-0-1-1-0","type":"$","font":{"family":"Arial","color":"#000000","size":10},"ts":1600601795713,"cs":"oP/0BG4fQ5mdgX3AkozGgg==","size":{"width":158,"height":18}}

{"id":"3-0-2-1-1-1-1-0-2-3-1-1-1-1-1-1-0-1-1-1","font":{"size":10,"color":"#000000","family":"Arial"},"type":"$","code":"$\\int_{}^{}\\cos ec^{2}x.dx=-\\cot x+c$","ts":1600601695771,"cs":"q7VU/Z9jSeGLJUXLq3jUgw==","size":{"width":186,"height":18}}

{"font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\tan x+\\cot x+C}\t\n\\end{align*}","id":"2-1-0-0-0-1-1-1-1-0-1-1-2-1-1","type":"align*","ts":1600601841237,"cs":"af5l4NtBKPSW4Nr05OeVjg==","size":{"width":205,"height":36}}

The correct answer is A.

{"font":{"color":"#000000","family":"Arial","size":12},"code":"$24.\\,\\int_{}^{}\\frac{e^{x}\\left(1+x\\right)}{\\cos^{2}\\left(e^{x}.x\\right)}dx$","id":"13-1","type":"$","ts":1600602024825,"cs":"N9u1oA+T4Zdh01U3vn3wbg==","size":{"width":144,"height":32}} is equal to

(A) - cot (e.xx) + C

(B) tan (x.ex) + C

(C) tan (ex) + C

(D) cot (ex) + C

Solun:- Let f(x) = {"type":"$","code":"$\\frac{e^{x}\\left(1+x\\right)}{\\cos^{2}\\left(e^{x}.x\\right)}$","id":"10-1-1-1-1-1-1-1","font":{"family":"Arial","color":"#000000","size":10},"ts":1600602053407,"cs":"pU8tTc3EfxwMweztGTnwXA==","size":{"width":54,"height":24}}

Integrate f(x):-

{"code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{e^{x}\\left(1+x\\right)}{\\cos^{2}\\left(e^{x}.x\\right)}.dx}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"type":"align*","id":"2-0-0-0-1-1-1-1-1-1-1-1-1-0-1-1-1-0-1-1-1-0-1-1-1-0-1-2-1","ts":1600602084595,"cs":"jQRA7oK6cii512C3fYplvA==","size":{"width":212,"height":37}}

Let ex.x = t

Differentiate w.r.t to t:-

⇒ (ex.1 + x.ex).dx = dt

⇒ ex.(1 + x).dx = dt

{"font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","id":"2-1-0-0-0-1-0-1-1-1-1-1","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{}\\frac{dt}{\\cos ^{2}t}}\\\\\n{\\int_{}^{}f\\left(x\\right)dx}&={\\int_{}^{} \\sec^{2}t.dt}\t\n\\end{align*}","ts":1600602450078,"cs":"O8u1qEkc1062sTajCXX1lA==","size":{"width":164,"height":76}}

We know that:-

{"code":"$\\int_{}^{}\\sec^{2}x.dx=\\tan x+c$","type":"$","id":"3-0-2-1-1-1-1-0-2-3-1-1-1-1-1-0-1","font":{"color":"#000000","size":11,"family":"Arial"},"ts":1600602480546,"cs":"YkSg0aaGKs3J/Aw1PJSyaA==","size":{"width":180,"height":20}}

{"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\tan t+C}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#000000"},"id":"2-1-0-0-0-1-1-1-1-0-1-1-0-1","ts":1600602505120,"cs":"S/ieggaY4idnTuZ11QcO8w==","size":{"width":150,"height":36}}

Put the value of t in this eq. :-

{"type":"align*","code":"\\begin{align*}\n{\\int_{}^{}f\\left(x\\right)dx}&={\\tan\\left(e^{x}.x\\right)+C}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"id":"2-1-0-0-0-1-1-1-1-0-1-1-1-1","ts":1600602550594,"cs":"vuDwdtpkfU+R4obxlQLdhw==","size":{"width":188,"height":36}}

The correct answer is B.


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