WB Board Class 12 Mathematics – Integration (SN Dey) Solutions
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Integration — Method of Substitution (SN Dey, Class 12)
1. \(\displaystyle \int x^{2}\cos(x^{3})\,dx\)
Put \(t=x^{3}.\)
Differentiate: \(dt=3x^{2}\,dx\Rightarrow x^{2}\,dx=\dfrac{1}{3}dt.\)
\(\therefore\; I=\dfrac{1}{3}\int\cos t\,dt\)
\(=\dfrac{1}{3}\sin t + C = \dfrac{1}{3}\sin(x^{3})+C\)
\(\dfrac{1}{3}\sin(x^{3})+C\)
2. \(\displaystyle \int (2-3x)^{7}\,dx\)
Put \(u=2-3x.\)
Differentiate: \(du=-3\,dx\Rightarrow dx=-\dfrac{1}{3}du.\)
\(\therefore\; I=-\dfrac{1}{3}\int u^{7}\,du\)
\(=-\dfrac{1}{3}\cdot\dfrac{u^{8}}{8}+C = -\dfrac{1}{24}(2-3x)^{8}+C\)
\(-\dfrac{1}{24}(2-3x)^{8}+C\)
3. \(\displaystyle \int \dfrac{dx}{a+bx}\)
Put \(u=a+bx.\)
Differentiate: \(du=b\,dx\Rightarrow dx=\dfrac{du}{b}.\)
\(\therefore\; I=\dfrac{1}{b}\int\dfrac{du}{u}\)
\(=\dfrac{1}{b}\ln|u|+C = \dfrac{1}{b}\ln|a+bx|+C\)
\(\dfrac{1}{b}\ln|a+bx|+C\)
4(i). \(\displaystyle \int 2x\sin(x^{2}+1)\,dx\)
Put \(u=x^{2}+1.\)
Differentiate: \(du=2x\,dx.\)
\(\therefore\; I=\int\sin u\,du = -\cos u + C\)
Back: \(-\cos(x^{2}+1)+C\)
\(-\cos(x^{2}+1)+C\)
4(ii). \(\displaystyle \int \sec^{2}(7-4x)\,dx\)
Put \(u=7-4x.\)
Differentiate: \(du=-4\,dx\Rightarrow dx=-\dfrac{1}{4}du.\)
\(\therefore\; I=-\dfrac{1}{4}\int\sec^{2}u\,du = -\dfrac{1}{4}\tan u + C\)
Back: \(-\dfrac{1}{4}\tan(7-4x)+C\)
\(-\dfrac{1}{4}\tan(7-4x)+C\)
5. \(\displaystyle \int \sin(3-4x)\,dx\)
Put \(u=3-4x.\)
Different: \(du=-4\,dx\Rightarrow dx=-\dfrac{1}{4}du.\)
\(\therefore\; I=-\dfrac{1}{4}\int\sin u\,du = \dfrac{1}{4}\cos u + C\)
Back: \(\dfrac{1}{4}\cos(3-4x)+C\)
\(\dfrac{1}{4}\cos(3-4x)+C\)
6. \(\displaystyle \int \cos^{2}(2x+3)\,dx\)
Put \(u=2x+3\).
Differentiate: \(du=2\,dx\Rightarrow dx=\tfrac{1}{2}du\).
Use identity: \(\cos^{2}u=\tfrac{1+\cos2u}{2}.\)
\(\therefore\; I=\tfrac{1}{2}\int\cos^{2}u\,du=\tfrac{1}{4}\int(1+\cos2u)\,du\)
\(=\tfrac{1}{4}(u+\tfrac{\sin2u}{2})+C = \tfrac{1}{4}(2x+3)+\tfrac{1}{8}\sin(4x+6)+C\)
\(\tfrac{1}{4}(2x+3)+\tfrac{1}{8}\sin(4x+6)+C\)
7. \(\displaystyle \int \dfrac{dx}{x\ln x}\)
Put \(u=\ln x.\)
Differentiate: \(du=\dfrac{dx}{x}.\)
\(\therefore\; I=\int\dfrac{du}{u} = \ln|u|+C\)
Back: \(\ln|\ln x|+C\)
\(\ln|\ln x|+C\)
8. \(\displaystyle \int e^{2-3x}\,dx\)
Put \(u=2-3x.\)
Differentiate: \(du=-3\,dx\Rightarrow dx=-\dfrac{1}{3}du.\)
\(\therefore\; I=-\dfrac{1}{3}\int e^{u}\,du = -\dfrac{1}{3}e^{u}+C\)
Back: \(-\dfrac{1}{3}e^{2-3x}+C\)
\(-\dfrac{1}{3}e^{2-3x}+C\)
9. \(\displaystyle \int 10^{2x-5}\,dx\)
Put \(u=2x-5.\)
Differentiate: \(du=2\,dx\Rightarrow dx=\tfrac{1}{2}du.\)
\(\therefore\; I=\tfrac{1}{2}\int10^{u}\,du=\tfrac{10^{u}}{2\ln10}+C\)
Back: \(\tfrac{10^{2x-5}}{2\ln10}+C\)
\(\tfrac{10^{2x-5}}{2\ln10}+C\)
10. \(\displaystyle \int \dfrac{dx}{\sqrt{ax+b}}\)
Put \(u=ax+b\).
Differentiate: \(du=a\,dx\Rightarrow dx=\dfrac{du}{a}.\)
\(\therefore\; I=\dfrac{1}{a}\int u^{-1/2}\,du = \dfrac{2}{a}u^{1/2}+C\)
Back: \(\dfrac{2}{a}\sqrt{ax+b}+C\)
\(\dfrac{2}{a}\sqrt{ax+b}+C\)
11. \(\displaystyle \int \dfrac{\cos(\ln x)}{x}\,dx\)
Put \(u=\ln x\).
Differentiate: \(du=\dfrac{dx}{x}.\)
\(\therefore\; I=\int\cos u\,du = \sin u + C\)
Back: \(\sin(\ln x)+C\)
\(\sin(\ln x)+C\)
12. \(\displaystyle \int \dfrac{\cos(\sqrt{x})}{\sqrt{x}}\,dx\)
Put \(u=\sqrt{x}\Rightarrow x=u^{2}.\)
Differentiate: \(dx=2u\,du\Rightarrow \dfrac{dx}{\sqrt{x}}=2\,du.\)
\(\therefore\; I=2\int\cos u\,du = 2\sin u + C\)
Back: \(2\sin(\sqrt{x})+C\)
\(2\sin(\sqrt{x})+C\)
13. \(\displaystyle \int \sin x\sin(\cos x)\,dx\)
Put \(u=\cos x\).
Differentiate: \(du=-\sin x\,dx\Rightarrow \sin x\,dx=-du.\)
\(\therefore\; I=-\int\sin u\,du = \cos u + C\)
Back: \(\cos(\cos x)+C\)
\(\cos(\cos x)+C\)
14(i). \(\displaystyle \int \dfrac{e^{\cot^{-1}x}}{1+x^{2}}\,dx\)
Put \(u=\cot^{-1}x\).
Differentiate: \(du=-\dfrac{dx}{1+x^{2}}\Rightarrow \dfrac{dx}{1+x^{2}}=-du.\)
\(\therefore\; I=-\int e^{u}\,du = -e^{u}+C\)
Back: \(-e^{\cot^{-1}x}+C\)
\(-e^{\cot^{-1}x}+C\)
14(ii). \(\displaystyle \int \dfrac{\sin(\tan^{-1}x)}{1+x^{2}}\,dx\)
Put \(u=\tan^{-1}x\).
Differentiate: \(du=\dfrac{dx}{1+x^{2}}.\)
\(\therefore\; I=\int\sin u\,du = -\cos u + C\)
Back: \(-\cos(\tan^{-1}x)+C\)
\(-\cos(\tan^{-1}x)+C\)
15. \(\displaystyle \int \dfrac{e^{a\sin^{-1}x}}{\sqrt{1-x^{2}}}\,dx\)
Put \(u=\sin^{-1}x.\)
Differentiate: \(du=\dfrac{dx}{\sqrt{1-x^{2}}}.\)
\(\therefore\; I=\int e^{au}\,du = \dfrac{1}{a}e^{au}+C\)
Back: \(\dfrac{1}{a}e^{a\sin^{-1}x}+C\)
\(\dfrac{1}{a}e^{a\sin^{-1}x}+C\)
16. \(\displaystyle \int x\sqrt{x^{2}+1}\,dx\)
Put \(u=x^{2}+1\).
Differentiate: \(du=2x\,dx\Rightarrow x\,dx=\tfrac{1}{2}du\).
\(\therefore\; I=\tfrac{1}{2}\int u^{1/2}\,du = \tfrac{1}{3}u^{3/2}+C\)
Back: \(\tfrac{1}{3}(x^{2}+1)^{3/2}+C\)
\(\tfrac{1}{3}(x^{2}+1)^{3/2}+C\)
17. \( \displaystyle \int x^{3}\sqrt[3]{x^{4}-4}\,dx \)
Put \(u=x^{4}-4.\)
Differentiate: \(du=4x^{3}\,dx\Rightarrow x^{3}\,dx=\dfrac{1}{4}du.\)
\(\therefore\; I=\dfrac{1}{4}\int u^{1/3}\,du\)
\(=\dfrac{1}{4}\cdot\dfrac{u^{4/3}}{4/3}+C=\dfrac{3}{16}u^{4/3}+C\)
Back: \(\dfrac{3}{16}(x^{4}-4)^{4/3}+C\)
\(\dfrac{3}{16}(x^{4}-4)^{4/3}+C\)
18. \( \displaystyle \int \dfrac{x\,dx}{\sqrt{3x^{2}+4}} \)
Put \(u=3x^{2}+4.\)
Differentiate: \(du=6x\,dx\Rightarrow x\,dx=\tfrac{1}{6}du.\)
\(\therefore\; I=\tfrac{1}{6}\int\dfrac{du}{\sqrt{u}}\)
\(=\tfrac{1}{3}\sqrt{u}+C = \tfrac{1}{3}\sqrt{3x^{2}+4}+C\)
\(\tfrac{1}{3}\sqrt{3x^{2}+4}+C\)
19(i). \(\displaystyle \int \dfrac{\sin^{2}x}{\cos^{4}x}\,dx\)
Rewrite: \(\dfrac{\sin^{2}x}{\cos^{4}x}=\tan^{2}x\sec^{2}x\).
Put \(u=\tan x\).
Differentiate: \(du=\sec^{2}x\,dx\).
\(\therefore\; I=\int u^{2}\,du = \tfrac{u^{3}}{3}+C = \tfrac{\tan^{3}x}{3}+C\)
\(\tfrac{\tan^{3}x}{3}+C\)
19(ii). \(\displaystyle \int \dfrac{\tan^{4}\sqrt{x}\,\sec^{2}\sqrt{x}}{\sqrt{x}}\,dx\)
Put \(u=\sqrt{x}\Rightarrow x=u^{2},\; dx=2u\,du\).
Integral becomes \(2\int\tan^{4}u\,\sec^{2}u\,du\).
Put \(v=\tan u,\; dv=\sec^{2}u\,du\).
\(\therefore\; I=2\int v^{4}\,dv = \tfrac{2}{5}v^{5}+C = \tfrac{2}{5}\tan^{5}u + C\)
Back: \(\tfrac{2}{5}\tan^{5}(\sqrt{x})+C\)
\(\tfrac{2}{5}\tan^{5}(\sqrt{x})+C\)
20. \(\displaystyle \int \dfrac{(1+\ln x)^{2}}{x}\,dx\)
Put \(u=1+\ln x\).
Differentiate: \(du=\dfrac{dx}{x}\).
\(\therefore\; I=\int u^{2}\,du = \tfrac{u^{3}}{3}+C\)
Back: \(\tfrac{(1+\ln x)^{3}}{3}+C\)
\(\tfrac{(1+\ln x)^{3}}{3}+C\)
21. \( \displaystyle \int \dfrac{1-\sin x}{\sqrt[3]{x+\cos x}}\,dx \)
Put \(t=\sqrt[3]{x+\cos x}\Rightarrow t^{3}=x+\cos x.\)
Differentiate: \(3t^{2}\,dt=dx-\sin x\,dx=(1-\sin x)\,dx.\)
\(\therefore\; \dfrac{1-\sin x}{\sqrt[3]{x+\cos x}}\,dx=\dfrac{3t^{2}\,dt}{t}=3t\,dt\)
\(I=3\int t\,dt = \dfrac{3}{2}t^{2} + C\)
Back: \(\dfrac{3}{2}(x+\cos x)^{2/3} + C\)
\(\dfrac{3}{2}(x+\cos x)^{2/3} + C\)
22. \(\displaystyle \int \dfrac{\cos x}{\sqrt{1+\sin x}}\,dx\)
Put \(u=1+\sin x\).
Differentiate: \(du=\cos x\,dx\).
\(\therefore\; I=\int u^{-1/2}\,du = 2u^{1/2}+C\)
Back: \(2\sqrt{1+\sin x}+C\)
\(2\sqrt{1+\sin x}+C\)
23. \(\displaystyle \int x^{2}\sqrt[3]{x^{3}+8}\,dx\)
Put \(u=x^{3}+8\).
Differentiate: \(du=3x^{2}\,dx\Rightarrow x^{2}\,dx=\tfrac{1}{3}du\).
\(\therefore\; I=\tfrac{1}{3}\int u^{1/3}\,du = \tfrac{1}{4}u^{4/3}+C\)
Back: \(\tfrac{1}{4}(x^{3}+8)^{4/3}+C\)
\(\tfrac{1}{4}(x^{3}+8)^{4/3}+C\)
24. \( \displaystyle \int (x+\cot x)\cot^{2}x\,dx \)
Put \(u=x+\cot x.\)
Differentiate: \(du=(1-\csc^{2}x)\,dx \Rightarrow (\csc^{2}x-1)\,dx=-du.\)
Use identity: \(\cot^{2}x=\csc^{2}x-1.\)
\(\therefore\; (x+\cot x)\cot^{2}x\,dx = u(\csc^{2}x-1)\,dx = -u\,du.\)
Hence \(I=-\int u\,du = -\dfrac{u^{2}}{2}+C.\)
Back-substitute: \(-\dfrac{1}{2}(x+\cot x)^{2}+C.\)
\(-\dfrac{1}{2}(x+\cot x)^{2}+C\)
25. \(\displaystyle \int \dfrac{x\,dx}{\sin^{2}(x^{2})}\)
Put \(u=x^{2}.\)
Differentiate: \(du=2x\,dx\Rightarrow x\,dx=\dfrac{1}{2}du.\)
\(\therefore\; I=\dfrac{1}{2}\int\csc^{2}u\,du\)
\(=\dfrac{1}{2}(-\cot u)+C=-\dfrac{1}{2}\cot u+C\)
Back: \(-\dfrac{1}{2}\cot(x^{2})+C\)
\(-\dfrac{1}{2}\cot(x^{2})+C\)
26. \(\displaystyle \int \dfrac{1+\cot x}{x+\log\sin x}\,dx\)
Put \(u=x+\log\sin x.\)
Differentiate :
\(\dfrac{du}{dx}=1+\dfrac{\cos x}{\sin x}=1+\cot x.\)
\(\therefore\; du=(1+\cot x)\,dx.\)
\(\therefore\; I=\int\dfrac{du}{u}\)
\(=\log|u|+C\)
Back: \(\log|x+\log\sin x|+C\)
\(\log|x+\log\sin x|+C\)
27. \(\displaystyle \int e^{x}\cos^{2}(e^{x})\,dx\)
Put \(u=e^{x}\).
Differentiate: \(du=e^{x}\,dx\Rightarrow e^{x}\,dx=du\).
Use \(\cos^{2}u=\tfrac{1+\cos2u}{2}.\)
\(\therefore\; I=\tfrac{1}{2}\int(1+\cos2u)\,du = \tfrac{1}{2}u + \tfrac{1}{4}\sin2u + C\)
Back: \(\tfrac{1}{2}e^{x} + \tfrac{1}{4}\sin(2e^{x})+C\)
\(\tfrac{1}{2}e^{x} + \tfrac{1}{4}\sin(2e^{x})+C\)
28. \( \displaystyle \int e^{\tfrac{1}{x}}\cdot \dfrac{1}{x^{2}}\,dx \)
Put \(u=\dfrac{1}{x}.\)
Differentiate: \(du=-\dfrac{1}{x^{2}}\,dx \Rightarrow \dfrac{1}{x^{2}}\,dx = -du.\)
\( \therefore\; I=\int e^{u}(-du) = -\int e^{u}\,du\)
\( = -e^{u} + C\)
Back: \(-e^{\tfrac{1}{x}} + C\)
\( -e^{\tfrac{1}{x}} + C \)
29. \(\displaystyle \int x^{2}\cdot 10^{x^{3}}\,dx\)
Put \(u=x^{3}.\)
Differentiate: \(du=3x^{2}\,dx\Rightarrow x^{2}\,dx=\tfrac{1}{3}du.\)
\(\therefore\; I=\tfrac{1}{3}\int10^{u}\,du = \tfrac{10^{u}}{3\log10}+C\)
Back: \(\tfrac{10^{x^{3}}}{3\log10}+C\)
\(\tfrac{10^{x^{3}}}{3\log10}+C\)
30. \( \displaystyle \int \dfrac{\cos x + x\sin x}{x(x+\cos x)}\,dx \)
\( \displaystyle I = \int \dfrac{\cos x + x\sin x}{x(x+\cos x)}\,dx \)
\(=\displaystyle \int \dfrac{(\cos x + x) + x(\sin x-1)}{x(x+\cos x)}\,dx\)
\(=\displaystyle \int \dfrac{1}{x}\,dx + \int \dfrac{\sin x-1}{x+\cos x}\,dx\)
\(=\displaystyle \log|x| - \int \dfrac{1-\sin x}{x+\cos x}\,dx\)
Put \(v=x+\cos x.\)
Differentiate: \(dv=(1-\sin x)\,dx.\)
\( \therefore\; \int \dfrac{1-\sin x}{x+\cos x}\,dx = \int \dfrac{dv}{v} = \log|v| + C\)
\( \therefore\; I = \log|x| - \log|x+\cos x| + C = \log\Big|\dfrac{x}{x+\cos x}\Big| + C\)
\( \log\Big|\dfrac{x}{x+\cos x}\Big| + C \)
31. \( \displaystyle \int \cos\!\Big(2\cot^{-1}\sqrt{\dfrac{1-x}{1+x}}\Big)\,dx \)
Put \(x=\cos\theta.\)
\(\displaystyle \sqrt{\dfrac{1-x}{1+x}}
=\sqrt{\dfrac{1-\cos\theta}{1+\cos\theta}}
=\sqrt{\dfrac{2\sin^{2}(\theta/2)}{2\cos^{2}(\theta/2)}}
=\tan\frac{\theta}{2}.\)
So \(\displaystyle \cos\!\Big(2\cot^{-1}\sqrt{\dfrac{1-x}{1+x}}\Big)
= \cos\!\Big(2\cot^{-1}\big(\tan\tfrac{\theta}{2}\big)\Big).\)
Use \(\displaystyle \cot^{-1}u=\frac{\pi}{2}-\tan^{-1}u.\) Thus
\(\cot^{-1}\big(\tan\tfrac{\theta}{2}\big)=\frac{\pi}{2}-\tan^{-1}\big(\tan\tfrac{\theta}{2}\big)
=\frac{\pi}{2}-\tfrac{\theta}{2}.\)
Hence
\[
2\cot^{-1}\big(\tan\tfrac{\theta}{2}\big)=\pi-\theta,
\qquad
\cos\big(2\cot^{-1}(\tan\tfrac{\theta}{2})\big)=\cos(\pi-\theta)=-\cos\theta.
\]
\(\therefore\; \cos\!\Big(2\cot^{-1}\sqrt{\dfrac{1-x}{1+x}}\Big)=-\cos\theta.\)
Back substitute \(x=\cos\theta\): \(\;=-x.\)
Thus \(I=\displaystyle\int -x\,dx = -\dfrac{x^{2}}{2} + C.\)
\( -\dfrac{x^{2}}{2} + C \)
1. \( \displaystyle \int \dfrac{dx}{(1+x^{2})\sqrt{\tan^{-1}x+4}} \)
Put \(u=\tan^{-1}x+4.\)
Differentiate: \(du=\dfrac{dx}{1+x^{2}}.\)
\( \therefore\; \displaystyle \int \dfrac{dx}{(1+x^{2})\sqrt{\tan^{-1}x+4}} = \int u^{-1/2}\,du.\)
\( = 2u^{1/2} + C.\)
\( 2\sqrt{\tan^{-1}x+4} + C \)
2. \( \displaystyle \int \sqrt{2+\sin3t}\,\cos3t\,dt \)
Put \(u=2+\sin3t.\)
Differentiate: \(du=3\cos3t\,dt \Rightarrow \cos3t\,dt=\dfrac{1}{3}du.\)
\( \therefore\; \int \sqrt{2+\sin3t}\,\cos3t\,dt = \dfrac{1}{3}\int u^{1/2}\,du.\)
\( = \dfrac{1}{3}\cdot\dfrac{2}{3}u^{3/2} + C.\)
\( \dfrac{2}{9}\big(2+\sin3t\big)^{3/2} + C \)
3. \( \displaystyle \int \tan^{3}2x\,\sec2x\,dx \)
Put \(u=\tan2x.\)
Differentiate: \(du=2\sec^{2}2x\,dx \Rightarrow \sec2x\,dx=\dfrac{du}{2\sec2x}.\)
But \(\sec2x=\sqrt{1+u^{2}}\Rightarrow \sec2x\,dx=\dfrac{du}{2\sqrt{1+u^{2}}}.\)
\( \therefore\; \int \tan^{3}2x\,\sec2x\,dx = \dfrac{1}{2}\int \dfrac{u^{3}}{\sqrt{1+u^{2}}}\,du.\)
Put \(v=1+u^{2}\Rightarrow dv=2u\,du \Rightarrow u^{3}du=\tfrac{1}{2}(v-1)\,dv.\)
\( = \dfrac{1}{4}\int \dfrac{v-1}{\sqrt{v}}\,dv = \dfrac{1}{4}\int(v^{1/2}-v^{-1/2})\,dv.\)
\( = \dfrac{1}{4}\Big(\tfrac{2}{3}v^{3/2}-2v^{1/2}\Big)+C.\)
\( = \dfrac{1}{6}(1+u^{2})^{3/2}-\dfrac{1}{2}\sqrt{1+u^{2}}+C.\)
\( \dfrac{1}{6}(1+\tan^{2}2x)^{3/2}-\dfrac{1}{2}\sqrt{1+\tan^{2}2x}+C \)
4. \( \displaystyle \int \dfrac{x^{5}}{\sqrt{1+x^{3}}}\,dx \)
Put \(u=1+x^{3}.\)
Differentiate: \(du=3x^{2}\,dx \Rightarrow x^{2}\,dx=\dfrac{1}{3}du.\)
Write \(x^{5}=x^{3}\cdot x^{2}=(u-1)\cdot x^{2}.\)
\( \therefore\; \int \dfrac{x^{5}}{\sqrt{1+x^{3}}}\,dx = \dfrac{1}{3}\int \dfrac{u-1}{\sqrt{u}}\,du.\)
\( = \dfrac{1}{3}\int (u^{1/2}-u^{-1/2})\,du = \dfrac{1}{3}\big(\tfrac{2}{3}u^{3/2}-2u^{1/2}\big)+C.\)
\( \dfrac{2}{9}(1+x^{3})^{3/2} - \dfrac{2}{3}(1+x^{3})^{1/2} + C \)
5. \( \displaystyle \int \dfrac{x^{2}}{\sqrt{x+1}}\,dx \)
Put \(u=x+1 \Rightarrow x=u-1,\,dx=du.\)
Then \(x^{2}=(u-1)^{2}=u^{2}-2u+1.\)
\( \therefore\; \int \dfrac{x^{2}}{\sqrt{x+1}}\,dx = \int \dfrac{u^{2}-2u+1}{\sqrt{u}}\,du.\)
\( = \int (u^{3/2}-2u^{1/2}+u^{-1/2})\,du = \tfrac{2}{5}u^{5/2}-\tfrac{4}{3}u^{3/2}+2u^{1/2}+C.\)
\( \tfrac{2}{5}(x+1)^{5/2}-\tfrac{4}{3}(x+1)^{3/2}+2\sqrt{x+1} + C \)
7. \( \displaystyle \int \dfrac{\cot x}{\log\sin x}\,dx \)
Put \(u=\log\sin x.\)
Differentiate: \(du=\dfrac{1}{\sin x}\cos x\,dx=\cot x\,dx.\)
\( \therefore\; \int \dfrac{\cot x}{\log\sin x}\,dx = \int \dfrac{du}{u}.\)
\( = \log|u| + C.\)
\( \log\big|\log\sin x\big| + C \)
9. \( \displaystyle \int \dfrac{\sin6x-\sin2x}{\cos6x-\cos2x}\,dx \)
Use identities:
\( \sin A-\sin B = 2\cos\dfrac{A+B}{2}\sin\dfrac{A-B}{2},\)
\( \cos A-\cos B = -2\sin\dfrac{A+B}{2}\sin\dfrac{A-B}{2}.\)
Apply with \(A=6x,\;B=2x\): numerator \(=2\cos4x\sin2x\),
denominator \(=-2\sin4x\sin2x\).
\( \therefore\; \dfrac{\sin6x-\sin2x}{\cos6x-\cos2x} = -\dfrac{\cos4x}{\sin4x} = -\cot4x.\)
Integrate: \( \int -\cot4x\,dx = -\tfrac{1}{4}\log|\sin4x|+C.\)
\( -\dfrac{1}{4}\log|\sin4x| + C \)
10. \( \displaystyle \int \dfrac{dx}{\csc4x-\cot4x} \)
Use identity \( \csc y - \cot y = \tan\dfrac{y}{2}.\)
For \(y=4x\) denominator \(=\tan2x.\)
\( \therefore\; \dfrac{1}{\csc4x-\cot4x} = \cot2x.\)
Integrate: \( \int \cot2x\,dx = \tfrac{1}{2}\log|\sin2x| + C.\)
\( \tfrac{1}{2}\log|\sin2x| + C \)
11. \( \displaystyle \int \dfrac{\sin2x}{a^{2}\sin^{2}x + b^{2}\cos^{2}x}\,dx \)
Put \(u=a^{2}\sin^{2}x + b^{2}\cos^{2}x.\)
Differentiate:
\(du=2\sin x\cos x\,(a^{2}-b^{2})\,dx = (a^{2}-b^{2})\sin2x\,dx.\)
\( \therefore\; \dfrac{\sin2x\,dx}{a^{2}\sin^{2}x + b^{2}\cos^{2}x}
= \dfrac{du}{(a^{2}-b^{2})u}.\)
\( \displaystyle I = \dfrac{1}{a^{2}-b^{2}}\int \dfrac{du}{u}.\)
\( = \dfrac{1}{a^{2}-b^{2}}\log\big|u\big| + C.\)
\( \dfrac{1}{a^{2}-b^{2}}\log\!\big|a^{2}\sin^{2}x + b^{2}\cos^{2}x\big| + C \)
12. \( \displaystyle \int \dfrac{\tan x\,\sec^{2}x}{\sqrt{a^{2}+b^{2}\tan^{2}x}}\,dx \)
Put \(u=a^{2}+b^{2}\tan^{2}x.\)
Differentiate:
\(du=2b^{2}\tan x\sec^{2}x\,dx \Rightarrow
\tan x\sec^{2}x\,dx=\dfrac{1}{2b^{2}}du.\)
\( \therefore\; \int \dfrac{\tan x\,\sec^{2}x}{\sqrt{a^{2}+b^{2}\tan^{2}x}}\,dx
= \dfrac{1}{2b^{2}}\int u^{-1/2}\,du.\)
\( = \dfrac{1}{2b^{2}}\cdot 2u^{1/2} + C
= \dfrac{1}{b^{2}}\sqrt{a^{2}+b^{2}\tan^{2}x} + C.\)
\( \dfrac{1}{b^{2}}\sqrt{a^{2}+b^{2}\tan^{2}x} + C \)
13(i). \( \displaystyle \int \dfrac{x+1}{\sqrt{1-2x-x^{2}}}\,dx \)
Put \(u=x+1\Rightarrow 1-2x-x^{2}=2-u^{2}.\)
Differentiate: \(du=dx.\)
\( \therefore\; \int \dfrac{x+1}{\sqrt{1-2x-x^{2}}}\,dx
= \int \dfrac{u}{\sqrt{2-u^{2}}}\,du.\)
Put \(v=2-u^{2}\Rightarrow dv=-2u\,du.\)
\( \int \dfrac{u}{\sqrt{2-u^{2}}}\,du = -\tfrac{1}{2}\int v^{-1/2}\,dv
= -\sqrt{v}+C.\)
\( -\sqrt{2-(x+1)^{2}} + C \)
13(ii). \( \displaystyle \int \dfrac{4x-3}{\sqrt[3]{\,4x^{2}-6x+9\,}}\,dx \)
Put \(z^{3}=4x^{2}-6x+9.\)
Differentiate: \(\dfrac{d}{dx}(4x^{2}-6x+9)=8x-6 = 3z^{2}\dfrac{dz}{dx}.\)
Hence \((4x-3)\,dx=\dfrac{3}{2}z^{2}\,dz.\)
\(\therefore\; \displaystyle \int \dfrac{4x-3}{\sqrt[3]{4x^{2}-6x+9}}\,dx
= \dfrac{3}{2}\int \dfrac{z^{2}}{z}\,dz.\)
\(= \dfrac{3}{2}\int z\,dz = \dfrac{3}{4}z^{2} + C.\)
\(= \dfrac{3}{4}\big(4x^{2}-6x+9\big)^{2/3} + C.\)
\( \dfrac{3}{4}\big(4x^{2}-6x+9\big)^{2/3} + C \)
14. \( \displaystyle \int \dfrac{dx}{\sqrt{x+3}-\sqrt{x+2}} \)
\( \displaystyle \int \dfrac{dx}{\sqrt{x+3}-\sqrt{x+2}}
= \int \dfrac{\sqrt{x+3}+\sqrt{x+2}}{(\sqrt{x+3}-\sqrt{x+2})(\sqrt{x+3}+\sqrt{x+2})}\,dx \)
\( = \int \dfrac{\sqrt{x+3}+\sqrt{x+2}}{(x+3)-(x+2)}\,dx
= \int\big(\sqrt{x+3}+\sqrt{x+2}\big)\,dx \)
\( \therefore\; \displaystyle
= \int \sqrt{x+3}\,dx + \int \sqrt{x+2}\,dx \)
\( = \dfrac{2}{3}(x+3)^{3/2} + \dfrac{2}{3}(x+2)^{3/2} + C \)
\( \dfrac{2}{3}\big((x+3)^{3/2}+(x+2)^{3/2}\big)+C \)
15. \( \displaystyle \int \dfrac{dx}{\sqrt{4x+3}-\sqrt{4x-3}} \)
\( \displaystyle \int \dfrac{dx}{\sqrt{4x+3}-\sqrt{4x-3}}
= \int \dfrac{\sqrt{4x+3}+\sqrt{4x-3}}{(4x+3)-(4x-3)}\,dx \)
\( = \dfrac{1}{6}\int\big(\sqrt{4x+3}+\sqrt{4x-3}\big)\,dx \)
\( \therefore\; \displaystyle
= \dfrac{1}{6}\Big(\int\sqrt{4x+3}\,dx + \int\sqrt{4x-3}\,dx\Big)\)
Put \(u=\sqrt{4x+3}\Rightarrow 4x+3=u^{2},\;dx=\dfrac{u}{2}\,du.\)
\( \displaystyle \int\sqrt{4x+3}\,dx = \int u\cdot\dfrac{u}{2}\,du
= \tfrac{1}{2}\int u^{2}\,du = \tfrac{1}{6}u^{3}.\)
Hence \( \displaystyle \int\sqrt{4x+3}\,dx = \tfrac{1}{6}(4x+3)^{3/2}.\)
Put \(v=\sqrt{4x-3}\Rightarrow 4x-3=v^{2},\;dx=\dfrac{v}{2}\,dv.\)
\( \displaystyle \int\sqrt{4x-3}\,dx = \tfrac{1}{6}(4x-3)^{3/2}.\)
\( \therefore\; \displaystyle \int \dfrac{dx}{\sqrt{4x+3}-\sqrt{4x-3}}
= \dfrac{1}{6}\Big(\tfrac{1}{6}(4x+3)^{3/2} + \tfrac{1}{6}(4x-3)^{3/2}\Big) + C\)
\( \displaystyle \dfrac{1}{36}\big((4x+3)^{3/2}+(4x-3)^{3/2}\big) + C \)
16. \( \displaystyle \int \dfrac{x-a}{x+a}\,dx \)
\( \displaystyle \frac{x-a}{x+a}=\frac{(x+a)-2a}{x+a}=1-\frac{2a}{x+a}.\)
\( \therefore\; \displaystyle \int \dfrac{x-a}{x+a}\,dx
=\int 1\,dx -2a\int \dfrac{dx}{x+a}.\)
\( = x -2a\log|x+a| + C.\)
\( x - 2a\log|x+a| + C \)
17. \( \displaystyle \int \dfrac{2x+3}{3x-4}\,dx \)
\( = 2\int \dfrac{x}{3x-4}\,dx + 3\int \dfrac{dx}{3x-4}.\)
\( = \dfrac{2}{3}\int \dfrac{3x}{3x-4}\,dx + 3\int \dfrac{dx}{3x-4}.\)
\( = \dfrac{2}{3}\int \dfrac{(3x-4)+4}{3x-4}\,dx + 3\int \dfrac{dx}{3x-4}.\)
\( = \dfrac{2}{3}\int dx + \dfrac{8}{3}\int \dfrac{dx}{3x-4} + 3\int \dfrac{dx}{3x-4}.\)
\( = \dfrac{2x}{3} + \Big(\dfrac{8}{3} + 3\Big)\!\int \dfrac{dx}{3x-4}.\)
\( = \dfrac{2x}{3} + \dfrac{17}{3}\!\int \dfrac{dx}{3x-4}.\)
\( = \dfrac{2x}{3} + \dfrac{17}{9}\log|3x-4| + C.\)
\( \dfrac{2x}{3} + \dfrac{17}{9}\log|3x-4| + C \)
18. \( \displaystyle \int x\cdot \sqrt[3]{x+2}\,dx \)
Put \(x+2=z^{3}.\)
Then \(x=z^{3}-2,\quad dx=3z^{2}\,dz.\)
\(\therefore\; \displaystyle \int x\sqrt[3]{x+2}\,dx
=\int (z^{3}-2)\,z\cdot 3z^{2}\,dz = 3\int (z^{6}-2z^{3})\,dz.\)
\( =3\Big(\dfrac{z^{7}}{7}-\dfrac{2z^{4}}{4}\Big)+C
=\dfrac{3}{7}z^{7}-\dfrac{3}{2}z^{4}+C.\)
Substitute \(z=(x+2)^{1/3}:\)
\( \dfrac{3}{7}(x+2)^{7/3}-\dfrac{3}{2}(x+2)^{4/3}+C \)
20. \( \displaystyle \int \dfrac{2x-1}{(x+1)^{2}}\,dx \)
Put \(u=x+1.\)
Then \(2x-1=2(u-1)-1=2u-3,\;dx=du.\)
\( \therefore\; \displaystyle \int \dfrac{2x-1}{(x+1)^{2}}\,dx
=\int \dfrac{2u-3}{u^{2}}\,du.\)
\( =\int\big(2u^{-1}-3u^{-2}\big)\,du = 2\log|u| + \dfrac{3}{u} + C.\)
\( 2\log|x+1| + \dfrac{3}{x+1} + C \)
21. \( \displaystyle \int \dfrac{x}{a+bx}\,dx \)
Put \(u=a+bx \Rightarrow x=\dfrac{u-a}{b},\;dx=\dfrac{du}{b}.\)
\( \therefore\; \displaystyle \int \dfrac{x}{a+bx}\,dx
= \int \dfrac{(u-a)/b}{u}\cdot\dfrac{du}{b}
= \dfrac{1}{b^{2}}\int\Big(1-\dfrac{a}{u}\Big)\,du.\)
\( = \dfrac{1}{b^{2}}\big(u - a\log|u|\big) + C.\)
\( \dfrac{a+bx}{b^{2}} - \dfrac{a}{b^{2}}\log|a+bx| + C \)
22. \( \displaystyle \int \dfrac{p\,x}{\sqrt[3]{q x + r}}\,dx \)
Put \(u= qx + r \Rightarrow x=\dfrac{u-r}{q},\;dx=\dfrac{du}{q}.\)
\( \therefore\; \displaystyle \int \dfrac{p\,x}{\sqrt[3]{qx+r}}\,dx
= \dfrac{p}{q^{2}}\int\big(u^{2/3}-r\,u^{-1/3}\big)\,du.\)
\( = \dfrac{p}{q^{2}}\Big(\dfrac{3}{5}u^{5/3} - \dfrac{3r}{2}u^{2/3}\Big)+C.\)
\( \dfrac{3p}{5q^{2}}(qx+r)^{5/3} - \dfrac{3pr}{2q^{2}}(qx+r)^{2/3} + C \)
23. \( \displaystyle \int \dfrac{x^{6}}{x-1}\,dx \)
Divide:
\( \dfrac{x^{6}}{x-1}=x^{5}+x^{4}+x^{3}+x^{2}+x+1+\dfrac{1}{x-1}.\)
\( \therefore\; \displaystyle \int \dfrac{x^{6}}{x-1}\,dx
=\int\big(x^{5}+x^{4}+x^{3}+x^{2}+x+1\big)\,dx + \int \dfrac{dx}{x-1}.\)
\( = \dfrac{x^{6}}{6} + \dfrac{x^{5}}{5} + \dfrac{x^{4}}{4}
+ \dfrac{x^{3}}{3} + \dfrac{x^{2}}{2} + x + \log|x-1| + C.\)
\( \dfrac{x^{6}}{6}+\dfrac{x^{5}}{5}+\dfrac{x^{4}}{4}
+\dfrac{x^{3}}{3}+\dfrac{x^{2}}{2}+x+\log|x-1|+C \)
24. \( \displaystyle \int \dfrac{dx}{x+\sqrt{x}} \)
Put \(t=\sqrt{x}\Rightarrow x=t^{2},\;dx=2t\,dt.\)
\( \therefore\; \displaystyle \int \dfrac{dx}{x+\sqrt{x}}
= \int \dfrac{2t}{t^{2}+t}\,dt = 2\int \dfrac{dt}{t+1}.\)
\( = 2\log|t+1| + C.\)
\( 2\log\big(\sqrt{x}+1\big) + C \)
25. \( \displaystyle \int \dfrac{dx}{\sqrt{x}-1} \)
Put \(t=\sqrt{x}\Rightarrow x=t^{2},\;dx=2t\,dt.\)
\( \displaystyle \int \dfrac{dx}{\sqrt{x}-1} = \int \dfrac{2t}{t-1}\,dt.\)
\( \dfrac{2t}{t-1}=2\Big(1+\dfrac{1}{t-1}\Big).\)
\( \therefore\; \displaystyle \int \dfrac{2t}{t-1}\,dt
= \int\big(2+ \dfrac{2}{t-1}\big)\,dt.\)
\( = 2t + 2\log|t-1| + C.\)
\( 2\sqrt{x} + 2\log\big|\sqrt{x}-1\big| + C \)
26. \( \displaystyle \int \dfrac{x^{3}+3x^{2}+2x-7}{2x+1}\,dx \)
Put \(z = 2x+1 \Rightarrow x = \dfrac{z-1}{2},\; dx = \dfrac{dz}{2}.\)
Then \(x^{3}+3x^{2}+2x-7 = \dfrac{z^{3}+3z^{2}-z-59}{8}.\)
\(\therefore\; I = \int \dfrac{\tfrac{z^{3}+3z^{2}-z-59}{8}}{z}\cdot \dfrac{dz}{2}
= \dfrac{1}{16}\int\big(z^{2}+3z-1-\tfrac{59}{z}\big)\,dz.\)
\(\displaystyle I = \dfrac{1}{16}\Big(\dfrac{z^{3}}{3}+\dfrac{3z^{2}}{2}-z-59\log|z|\Big)+C.\)
Put \(z=2x+1\): \( \displaystyle
I = \dfrac{(2x+1)^{3}}{48} + \dfrac{3(2x+1)^{2}}{32}
- \dfrac{2x+1}{16} - \dfrac{59}{16}\log|2x+1| + C.\)
27. \( \displaystyle \int \dfrac{e^{2x}}{e^{x}+1}\,dx \)
Put \(u=e^{x}\Rightarrow du=e^{x}dx.\)
Then \(e^{2x}dx = u^{2}\cdot\dfrac{du}{u}=u\,du.\)
\( \therefore\; \displaystyle \int \dfrac{e^{2x}}{e^{x}+1}\,dx
= \int \dfrac{u}{u+1}\,du.\)
\( = \int\Big(1-\dfrac{1}{u+1}\Big)\,du = u - \log|u+1| + C.\)
\( e^{x} - \log\big(e^{x}+1\big) + C \)
28. \( \displaystyle \int \dfrac{x+1}{x}\,(x+\log x)^{2}\,dx \)
Put \(u=x+\log x.\)
Differentiate: \(du=\Big(1+\dfrac{1}{x}\Big)\,dx=\dfrac{x+1}{x}\,dx.\)
\(\therefore\; \int \dfrac{x+1}{x}(x+\log x)^{2}\,dx=\int u^{2}\,du.\)
\(=\dfrac{u^{3}}{3}+C.\)
\( \dfrac{1}{3}(x+\log x)^{3}+C \)
29. \( \displaystyle \int \dfrac{dx}{1+e^{x/2}} \)
Put \(t=e^{x/2}.\)
Differentiate: \(dt=\tfrac{1}{2}e^{x/2}dx=\tfrac{1}{2}t\,dx\Rightarrow dx=\dfrac{2\,dt}{t}.\)
\(\therefore\; \int \dfrac{dx}{1+e^{x/2}}=\int \dfrac{2}{t(1+t)}\,dt.\)
\(=2\int\Big(\dfrac{1}{t}-\dfrac{1}{1+t}\Big)\,dt
= 2\log t - 2\log(1+t)+C.\)
\( x - 2\log\big(1+e^{x/2}\big) + C \)
30. \( \displaystyle \int \dfrac{dx}{1+e^{x}} \)
Put \(u=e^{x}.\)
Differentiate: \(du=e^{x}dx = u\,dx\Rightarrow dx=\dfrac{du}{u}.\)
\(\therefore\; \int \dfrac{dx}{1+e^{x}}=\int \dfrac{du}{u(1+u)}.\)
\(=\int\Big(\dfrac{1}{u}-\dfrac{1}{1+u}\Big)\,du
= \log u - \log(1+u) + C.\)
\( x - \log\big(1+e^{x}\big) + C \)
31. \( \displaystyle \int \dfrac{e^{x}}{\sqrt{e^{x}+8}}\,dx \)
Put \(u=e^{x}+8.\)
Differentiate: \(du=e^{x}dx.\)
\(\therefore\; \int \dfrac{e^{x}}{\sqrt{e^{x}+8}}\,dx=\int u^{-1/2}\,du.\)
\(=2u^{1/2}+C.\)
\( 2\sqrt{e^{x}+8} + C \)
32. \( \displaystyle \int \dfrac{e^{x}}{1+e^{2x}}\,dx \)
Put \(u=e^{x}.\)
Differentiate: \(du=e^{x}dx.\)
\(\therefore\; \int \dfrac{e^{x}}{1+e^{2x}}\,dx=\int \dfrac{du}{1+u^{2}}.\)
\(=\tan^{-1}u + C.\)
\( \tan^{-1}\!\big(e^{x}\big)+C \)
33. \( \displaystyle \int \dfrac{2^{x}}{2^{x}+1}\,dx \)
Put \(u=2^{x}.\)
Differentiate: \(du = 2^{x}\log 2\,dx \Rightarrow dx=\dfrac{du}{u\log 2}.\)
\(\therefore\; \int \dfrac{2^{x}}{2^{x}+1}\,dx
=\dfrac{1}{\log 2}\int \dfrac{du}{1+u}.\)
\(=\dfrac{1}{\log 2}\log(1+u)+C.\)
\( \dfrac{1}{\log 2}\log\big(1+2^{x}\big) + C \)
34. \( \displaystyle \int \dfrac{10^{x}-1}{10^{x}+1}\,dx \)
\( \dfrac{10^{x}-1}{10^{x}+1} = 1 - \dfrac{2}{10^{x}+1} \)
\( \displaystyle I = \int 1\,dx - 2\int \dfrac{dx}{10^{x}+1} \)
Multiply by \(10^{-x}\):
\( \displaystyle \int \dfrac{dx}{10^{x}+1}
= \int \dfrac{10^{-x}}{1+10^{-x}}\,dx \)
Put \(u = 1+10^{-x}\Rightarrow du = -10^{-x}\log10\,dx\)
\( \therefore\; \int \dfrac{dx}{10^{x}+1}
= -\dfrac{1}{\log10}\int \dfrac{du}{u}
= -\dfrac{1}{\log10}\log u \)
\( I = x + \dfrac{2}{\log10}\log(1+10^{-x}) + C \)
35. \( \int \dfrac{\sqrt{\tan x}}{\cos^{4}x}\,dx \)
\( = \int \sqrt{\tan x}\,\sec^{4}x\,dx
= \int \sqrt{\tan x}(1+\tan^{2}x)\sec^{2}x\,dx \)
Put \( \tan x = z^{2} \Rightarrow \sec^{2}x\,dx = 2z\,dz.\)
\( \therefore\; I = 2\int(z^{2}+z^{6})\,dz \)
\( = \dfrac{2}{3}z^{3} + \dfrac{2}{7}z^{7} + C \)
\( \displaystyle \dfrac{2}{3}\tan^{3/2}x + \dfrac{2}{7}\tan^{7/2}x + C \)
36. \( \int \sqrt{\dfrac{\sin x}{\cos^{5}x}}\,dx \)
\( = \int \sqrt{\tan x}\sec^{2}x\,dx \)
Put \( \tan x = z^{2} \Rightarrow \sec^{2}x\,dx = 2z\,dz \)
\( \therefore\; I = \int z \cdot 2z\,dz = 2\int z^{2}\,dz \)
\( = \dfrac{2}{3}z^{3} + C \)
\( \dfrac{2}{3}(\tan x)^{3/2} + C \)
37. \( \displaystyle \int \Big(1-\dfrac{1}{x^{2}}\Big)e^{\,x+\tfrac{1}{x}}\,dx \)
Put \(u=x+\dfrac{1}{x}.\)
Differentiate: \(du=\Big(1-\dfrac{1}{x^{2}}\Big)\,dx.\)
\(\therefore\; \int \Big(1-\dfrac{1}{x^{2}}\Big)e^{x+1/x}\,dx
=\int e^{u}\,du = e^{u}+C.\)
\( e^{x+1/x} + C \)
38. \( \displaystyle \int \dfrac{e^{x}(1+x)}{\cos^{2}(x e^{x})}\,dx \) [HS '18]
Put \(u = x e^{x}.\)
Differentiate: \(du = e^{x}(1+x)\,dx.\)
\(\therefore\; \int \dfrac{e^{x}(1+x)}{\cos^{2}(x e^{x})}\,dx
= \int \sec^{2}u\,du = \tan u + C.\)
\( \tan\!\big(xe^{x}\big) + C \)
39. \( \displaystyle \int \dfrac{dx}{x^{2}\sqrt{1-x^{2}}} \)
Put \(x=\sin\theta\Rightarrow dx=\cos\theta\,d\theta.\)
Denominator \(= \sin^{2}\theta\cdot\cos\theta.\)
\(\therefore\; \int \dfrac{dx}{x^{2}\sqrt{1-x^{2}}}
= \int \dfrac{\cos\theta}{\sin^{2}\theta\cos\theta}\,d\theta
= \int \csc^{2}\theta\,d\theta.\)
\( = -\cot\theta + C.\)
\( -\dfrac{\sqrt{1-x^{2}}}{x} + C \)
40. \( \displaystyle \int \dfrac{x^{2}}{(a+bx)^{2}}\,dx \)
Put \(u=a+bx \Rightarrow x=\dfrac{u-a}{b},\;dx=\dfrac{du}{b}.\)
\(\therefore\; \int \dfrac{x^{2}}{(a+bx)^{2}}\,dx
= \dfrac{1}{b^{3}}\int\Big(1-\dfrac{2a}{u}+\dfrac{a^{2}}{u^{2}}\Big)\,du.\)
\(= \dfrac{1}{b^{3}}\Big(u -2a\log|u| - \dfrac{a^{2}}{u}\Big) + C.\)
\( \dfrac{1}{b^{3}}\Big(a+bx -2a\log|a+bx| - \dfrac{a^{2}}{a+bx}\Big) + C \)
41. \( \displaystyle \int \dfrac{dx}{x^{2}\sqrt{a^{2}+x^{2}}} \)
Let \( x = a\tan\theta \)
Then \( dx = a\sec^{2}\theta\,d\theta \)
Also \( \sqrt{a^{2}+x^{2}} = a\sec\theta \)
\( \therefore\; I = \int \dfrac{a\sec^{2}\theta}{a^{2}\tan^{2}\theta \cdot a\sec\theta}\,d\theta \)
\( = \int \dfrac{\sec\theta}{a^{2}\tan^{2}\theta}\,d\theta \)
\( = \dfrac{1}{a^{2}}\int \dfrac{\sec\theta}{\tan^{2}\theta}\,d\theta \)
\( = \dfrac{1}{a^{2}}\int \cot\theta\csc\theta\,d\theta \)
\( = \dfrac{1}{a^{2}}(-\csc\theta)+C \)
Back substitute: \( \theta = \tan^{-1}\dfrac{x}{a} \)
So \( \csc\theta = \sqrt{1+\tan^{2}\theta} \cdot \dfrac{1}{\tan\theta}
= \dfrac{\sqrt{a^{2}+x^{2}}}{x} \)
\( \displaystyle -\dfrac{1}{a^{2}} \cdot \dfrac{\sqrt{a^{2}+x^{2}}}{x} + C \)
42. \( \displaystyle \int \dfrac{dx}{(1+x^{2})\sqrt{1+x^{2}}} \)
Put \(x=\tan\theta\Rightarrow dx=\sec^{2}\theta\,d\theta,\;
\sqrt{1+x^{2}}=\sec\theta.\)
\(\therefore\; \int \dfrac{dx}{(1+x^{2})\sqrt{1+x^{2}}}
= \int \dfrac{\sec^{2}\theta}{\sec^{2}\theta\sec\theta}\,d\theta
= \int \cos\theta\,d\theta.\)
\( = \sin\theta + C.\)
\( \dfrac{x}{\sqrt{1+x^{2}}} + C \)
43. \( \displaystyle \int \dfrac{dx}{(1-x^{2})\sqrt{1-x^{2}}} \)
Put \(x=\sin\theta\Rightarrow dx=\cos\theta\,d\theta.\)
\(\therefore\; \int \dfrac{dx}{(1-x^{2})\sqrt{1-x^{2}}}
= \int \dfrac{\cos\theta}{\cos^{3}\theta}\,d\theta
= \int \sec^{2}\theta\,d\theta.\)
\(=\tan\theta + C.\)
\( \dfrac{x}{\sqrt{1-x^{2}}} + C \)
44. \( \displaystyle \int \dfrac{dx}{x^{3}\sqrt{x^{2}-1}} \)
Let \(x=\sec\theta\Rightarrow dx=\sec\theta\tan\theta\,d\theta\)
\(\sqrt{x^{2}-1}=\tan\theta\)
\(I=\int \dfrac{\sec\theta\tan\theta}{\sec^{3}\theta\tan\theta}\,d\theta
=\int \cos^{2}\theta\,d\theta\)
\(=\int \dfrac{1+\cos2\theta}{2}\,d\theta
=\dfrac{\theta}{2}+\dfrac{\sin2\theta}{4}+C\)
Now: \(\theta=\sec^{-1}x\)
\(\sin2\theta=\dfrac{2\sqrt{x^{2}-1}}{x^{2}}\)
\( \displaystyle \frac{1}{2}\sec^{-1}x + \frac{\sqrt{x^{2}-1}}{2x^{2}} + C \)
45. \( \int \dfrac{dx}{(1-x)\sqrt{1-x^{2}}} \)
Let \(x=\sin\theta \Rightarrow dx=\cos\theta\,d\theta,\;
\sqrt{1-x^{2}}=\cos\theta\)
\( I=\int \dfrac{\cos\theta}{(1-\sin\theta)\cos\theta}\,d\theta
=\int \dfrac{d\theta}{1-\sin\theta} \)
Multiply by \( \dfrac{1+\sin\theta}{1+\sin\theta}
\Rightarrow \int \dfrac{1+\sin\theta}{\cos^{2}\theta}\,d\theta \)
\( =\int \sec^{2}\theta\,d\theta + \int \sec\theta\tan\theta\,d\theta \)
\( =\tan\theta + \sec\theta + C \)
\( =\dfrac{x}{\sqrt{1-x^{2}}} + \dfrac{1}{\sqrt{1-x^{2}}} + C \)
46. \( \displaystyle \int \dfrac{\cos^{5}x}{\sin x}\,dx \)
Put \(u=\sin x\Rightarrow du=\cos x\,dx.\)
Write \(\cos^{4}x=(1-\sin^{2}x)^{2}=(1-u^{2})^{2}.\)
\(\therefore\; \displaystyle \int \dfrac{\cos^{5}x}{\sin x}\,dx
= \int \dfrac{(1-u^{2})^{2}}{u}\,du.\)
\(=\int\Big(\dfrac{1}{u}-2u+u^{3}\Big)\,du
= \log|u| - u^{2} + \dfrac{u^{4}}{4} + C.\)
\( \log|\sin x| - \sin^{2}x + \dfrac{\sin^{4}x}{4} + C \)
47. \( \displaystyle \int \dfrac{x\,dx}{\sqrt{x+1}+\sqrt[3]{x+1}} \)
Put \(t=(x+1)^{1/6}\Rightarrow x+1=t^{6},\;x=t^{6}-1,\;dx=6t^{5}\,dt.\)
Denominator \(=\sqrt{x+1}+\sqrt[3]{x+1}=t^{3}+t^{2}=t^{2}(t+1).\)
\(\therefore\; \displaystyle \int \dfrac{x\,dx}{\sqrt{x+1}+\sqrt[3]{x+1}}
=6\int \dfrac{(t^{6}-1)t^{5}}{t^{2}(t+1)}\,dt
= 6\int \dfrac{t^{9}-t^{5}}{t+1}\,dt.\)
Divide polynomial:
\(\dfrac{t^{9}-t^{5}}{t+1}=t^{8}-t^{7}+t^{6}-t^{5}+t^{4}-t^{3}+t^{2}-t+1
- \dfrac{1}{t+1}.\)
Integrate termwise:
\(6\Big(\dfrac{t^{9}}{9}-\dfrac{t^{8}}{8}+\dfrac{t^{7}}{7}
-\dfrac{t^{6}}{6}+\dfrac{t^{5}}{5}-\dfrac{t^{4}}{4}
+\dfrac{t^{3}}{3}-\dfrac{t^{2}}{2}+t - \log|t+1|\Big)+C.\)
Replace \(t=(x+1)^{1/6}\) to express final answer.
1. \( \displaystyle \int \dfrac{x^{3}}{\sqrt{1-x^{2}}}\,dx \)
Put \(t = 1 - x^{2} \Rightarrow dt = -2x\,dx,\; x^{2} = 1 - t.\)
\(x^{3}dx = x^{2}\cdot x\,dx = (1-t)\Big(-\dfrac{1}{2}dt\Big).\)
\( \therefore\; I = -\dfrac{1}{2}\int (1-t)t^{-1/2}\,dt.\)
\( = -\dfrac{1}{2}\Big(\int t^{-1/2}dt - \int t^{1/2}dt\Big)
= -\Big(t^{1/2} - \dfrac{1}{3}t^{3/2}\Big)+C.\)
\( \displaystyle I = -\sqrt{1-x^{2}} + \dfrac{1}{3}(1-x^{2})^{3/2} + C \)
2. \( \displaystyle \int \dfrac{dx}{\sqrt{(a^{2}-x^{2})^{3}}} \)
Let \(x = a\sin\theta \Rightarrow dx = a\cos\theta\,d\theta.\)
Then \(a^{2}-x^{2} = a^{2}\cos^{2}\theta\Rightarrow
\sqrt{(a^{2}-x^{2})^{3}} = \sqrt{a^{6}\cos^{6}\theta}=a^{3}\cos^{3}\theta.\)
\( \therefore\; I = \int \dfrac{a\cos\theta}{a^{3}\cos^{3}\theta}\,d\theta
= \dfrac{1}{a^{2}}\int \sec^{2}\theta\,d\theta.\)
\( = \dfrac{1}{a^{2}}\tan\theta + C.\)
Now \( \tan\theta = \dfrac{x}{\sqrt{a^{2}-x^{2}}}. \)
\( \displaystyle I = \dfrac{x}{a^{2}\sqrt{a^{2}-x^{2}}} + C \)
3. \( \displaystyle \int \dfrac{x^{2}+1}{(x^{2}-1)^{2}}\,dx \)
\( I = \int \dfrac{x^{2}+1}{(x^{2}-1)^{2}}\,dx
= \int \dfrac{1+\dfrac{1}{x^{2}}}{\left(\dfrac{x^{2}-1}{x}\right)^{2}}\,dx \)
[dividing numerator and denominator by \(x^{2}\)]
\( \therefore\; I
= \int \dfrac{1+\dfrac{1}{x^{2}}}{\big(x-\tfrac{1}{x}\big)^{2}}\,dx.
\)
Put \( z = x-\dfrac{1}{x} \Rightarrow dz = \Big(1+\dfrac{1}{x^{2}}\Big)dx. \)
\( \therefore\; I = \int \dfrac{dz}{z^{2}} = \int z^{-2}dz = -\dfrac{1}{z}+C. \)
\( I = -\dfrac{1}{x-\tfrac{1}{x}}+C = -\dfrac{x}{x^{2}-1}+C. \)
\( \displaystyle -\dfrac{x}{x^{2}-1}+C \)
4(i). \( \displaystyle \int \sqrt{\frac{a+x}{\,a-x\,}}\,dx \)
Put \( x = a\cos\theta \).
Then \( dx = -a\sin\theta\,d\theta \).
\( \sqrt{\frac{a+x}{a-x}}
= \sqrt{\frac{a+a\cos\theta}{a-a\cos\theta}}
= \sqrt{\frac{1+\cos\theta}{1-\cos\theta}}. \)
Use
\( 1+\cos\theta = 2\cos^{2}\tfrac{\theta}{2},\;
1-\cos\theta = 2\sin^{2}\tfrac{\theta}{2}. \)
So
\( \sqrt{\frac{1+\cos\theta}{1-\cos\theta}}
= \sqrt{\frac{\cos^{2}\tfrac{\theta}{2}}{\sin^{2}\tfrac{\theta}{2}}}
= \cot\tfrac{\theta}{2}. \)
\( I = \int \cot\tfrac{\theta}{2}\,(-a\sin\theta)\,d\theta. \)
Write \( \sin\theta = 2\sin\tfrac{\theta}{2}\cos\tfrac{\theta}{2} \).
\( I = -a \int \cot\tfrac{\theta}{2}\;2\sin\tfrac{\theta}{2}\cos\tfrac{\theta}{2}\,d\theta. \)
\( I = -a \int 2\cos^{2}\tfrac{\theta}{2}\,d\theta. \)
Use \( \cos^{2}\tfrac{\theta}{2} = \tfrac{1+\cos\theta}{2}. \)
\( I = -a \int (1+\cos\theta)\,d\theta. \)
\( I = -a(\theta+\sin\theta)+C. \)
Back substitute:
\( x = a\cos\theta \Rightarrow \cos\theta = \frac{x}{a}. \)
\( \theta = \cos^{-1}\!\left(\frac{x}{a}\right), \quad
\sin\theta = \sqrt{1-\frac{x^{2}}{a^{2}}}
= \frac{\sqrt{a^{2}-x^{2}}}{a}. \)
\( I = -a\cos^{-1}\!\left(\frac{x}{a}\right) - \sqrt{a^{2}-x^{2}} + C \)
4(ii). \( \displaystyle \int x\sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}}\,dx \)
Put \( x^{2} = a^{2}\cos 2\theta \).
Differentiate:
\( 2x\,dx = -a^{2}\sin 2\theta\, d(2\theta) = -2a^{2}\sin 2\theta\, d\theta \)
\( \Rightarrow x\,dx = -a^{2}\sin 2\theta\, d\theta. \)
\( a^{2}-x^{2}=a^{2}(1-\cos 2\theta)=2a^{2}\sin^{2}\theta. \)
\( a^{2}+x^{2}=a^{2}(1+\cos 2\theta)=2a^{2}\cos^{2}\theta. \)
Hence
\( \sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}}
=\sqrt{\frac{2a^{2}\sin^{2}\theta}{2a^{2}\cos^{2}\theta}}
=\tan\theta. \)
Thus
\( I=\int x\sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}}\,dx
=\int (-a^{2}\sin 2\theta)\tan\theta\,d\theta. \)
\( \sin 2\theta = 2\sin\theta\cos\theta,\quad
\tan\theta=\frac{\sin\theta}{\cos\theta}. \)
So
\( I = -a^{2}\int 2\sin^{2}\theta\, d\theta. \)
Use
\( \sin^{2}\theta = \frac{1-\cos 2\theta}{2}. \)
\( I = -a^{2}\int (1-\cos 2\theta)\,d\theta \)
\( I = -a^{2}\theta + a^{2}\frac{\sin 2\theta}{2} + C. \)
Back substitution:
\( x^{2}=a^{2}\cos 2\theta \Rightarrow \cos 2\theta=\frac{x^{2}}{a^{2}}. \)
\( \theta=\frac{1}{2}\cos^{-1}\!\left(\frac{x^{2}}{a^{2}}\right). \)
\( \sin 2\theta = \sqrt{1-\cos^{2}2\theta}
=\sqrt{1-\frac{x^{4}}{a^{4}}}. \)
\( \displaystyle
I = -\frac{a^{2}}{2}\cos^{-1}\!\left(\frac{x^{2}}{a^{2}}\right)
+ \frac{a^{2}}{2}\sqrt{1-\frac{x^{4}}{a^{4}}}
+ C
\)
5. \( \displaystyle \int \sqrt{\dfrac{x}{a-x}}\,dx \)
Put \(x = a\sin^{2}\theta \Rightarrow dx = 2a\sin\theta\cos\theta\,d\theta.\)
Then \(a-x = a\cos^{2}\theta,\;
\sqrt{\dfrac{x}{a-x}} = \sqrt{\dfrac{\sin^{2}\theta}{\cos^{2}\theta}} = \tan\theta.\)
\( \therefore\; I = \int \tan\theta\cdot 2a\sin\theta\cos\theta\,d\theta
= 2a\int \sin^{2}\theta\,d\theta.\)
\( = 2a\int \dfrac{1-\cos2\theta}{2}\,d\theta
= a\Big(\theta - \dfrac{\sin2\theta}{2}\Big) + C.\)
\( \sin^{2}\theta = \dfrac{x}{a}\Rightarrow \theta = \sin^{-1}\!\sqrt{\dfrac{x}{a}}, \quad
\sin2\theta = \dfrac{2\sqrt{x(a-x)}}{a}. \)
\( \displaystyle I = a\sin^{-1}\!\sqrt{\dfrac{x}{a}} - \sqrt{x(a-x)} + C \)
6. \( \displaystyle \int \dfrac{x^{2}}{\sqrt{1-x^{6}}}\,dx \)
Put \(t = x^{3} \Rightarrow dt = 3x^{2}dx \Rightarrow x^{2}dx = \dfrac{dt}{3}.\)
\( \therefore\; I = \dfrac{1}{3}\int \dfrac{dt}{\sqrt{1-t^{2}}}
= \dfrac{1}{3}\sin^{-1}t + C.\)
\( \displaystyle I = \dfrac{1}{3}\sin^{-1}(x^{3}) + C \)
7. \( \displaystyle \int \tan2x\,\tan3x\,\tan5x\,dx \)
\(\tan5x = \tan(3x + 2x)\)
\( \displaystyle \Rightarrow
\tan5x = \frac{\tan3x + \tan2x}{1 - \tan3x\,\tan2x} \)
\( \Rightarrow \tan5x(1 - \tan3x\,\tan2x)
= \tan3x + \tan2x \)
\( \Rightarrow \tan5x - \tan5x\,\tan3x\,\tan2x
= \tan3x + \tan2x \)
\( \Rightarrow \tan5x - \tan3x - \tan2x
= \tan5x\,\tan3x\,\tan2x \)
\( \therefore\;
\displaystyle \int \tan2x\,\tan3x\,\tan5x\,dx
= \int (\tan5x - \tan3x - \tan2x)\,dx
\)
\( = \int \tan5x\,dx
- \int \tan3x\,dx
- \int \tan2x\,dx \)
\( = \frac{1}{5}\log|\sec5x|
- \frac{1}{3}\log|\sec3x|
- \frac{1}{2}\log|\sec2x|
+ C \)
\( \displaystyle
\frac{1}{5}\log|\sec5x|
- \frac{1}{3}\log|\sec3x|
- \frac{1}{2}\log|\sec2x|
+ C
\)
8(i). \( \displaystyle \int \frac{\cos x}{\cos(x+a)}\,dx \)
\( \displaystyle \int \frac{\cos x}{\cos(x+a)}\,dx
= \int \frac{\cos\{(x+a)-a\}}{\cos(x+a)}\,dx
\)
\( \displaystyle
= \int \frac{\cos(x+a)\cos a + \sin(x+a)\sin a}{\cos(x+a)}\,dx
\)
\( \displaystyle
= \int \cos a\,dx
+ \int \frac{\sin(x+a)\sin a}{\cos(x+a)}\,dx
\)
\( \displaystyle
= \cos a \int dx
+ \sin a \int \tan(x+a)\,dx
\)
\( \displaystyle
x\cos a + \sin a\,\log|\sec(x+a)| + C
\)
8(ii). \( \displaystyle \int \frac{\sin x}{\sin(x+a)}\,dx \)
\( \displaystyle \int \frac{\sin x}{\sin(x+a)}\,dx
= \int \frac{\sin\{(x+a)-a\}}{\sin(x+a)}\,dx
\)
\( \displaystyle
= \int \frac{\sin(x+a)\cos a - \cos(x+a)\sin a}{\sin(x+a)}\,dx
\)
\( \displaystyle
= \int \cos a\,dx
- \int \frac{\cos(x+a)\sin a}{\sin(x+a)}\,dx
\)
\( \displaystyle
= \cos a \int dx
- \sin a \int \cot(x+a)\,dx
\)
\( \displaystyle
x\cos a \;-\; \sin a\,\log|\sin(x+a)| + C
\)
9(i).
\( \displaystyle \int \frac{\sin(x-a)}{\sin(x+a)}\,dx \)
\( = \int \frac{\sin((x+a)-2a)}{\sin(x+a)}\,dx \)
\( = \int
\frac{\sin(x+a)\cos2a - \cos(x+a)\sin2a}
{\sin(x+a)}\,dx \)
\( = \int \cos2a\,dx
- \int \frac{\cos(x+a)\sin2a}{\sin(x+a)}\,dx \)
\( = \cos2a\int dx
- \sin2a\int \cot(x+a)\,dx \)
\( = x\cos2a - \sin2a\,\log|\sin(x+a)| + C \)
9(ii).
\( \displaystyle \int \frac{\tan x - \tan a}{\tan x + \tan a}\,dx \)
\( = \int \frac{\frac{\sin x}{\cos x}-\frac{\sin a}{\cos a}}
{\frac{\sin x}{\cos x}+\frac{\sin a}{\cos a}}\,dx \)
\( = \int
\frac{\sin x\cos a - \sin a\cos x}
{\sin x\cos a + \cos x\sin a}\,dx \)
\( = \int
\frac{\sin(x-a)}{\sin(x+a)}\,dx
\)
Now apply the same result as in 9(i).
10. \( \displaystyle \int e^{x}\Big(\frac{1}{x}-\frac{1}{x^{2}}\Big)\,dx \)
Put \(t = \dfrac{e^{x}}{x}\).
Differentiate:
\( \displaystyle dt = \Big(e^{x}\frac{1}{x} - e^{x}\frac{1}{x^{2}}\Big)\,dx = e^{x}\Big(\frac{1}{x}-\frac{1}{x^{2}}\Big)\,dx. \)
\( \displaystyle dt = \Big(e^{x}\frac{1}{x} - e^{x}\frac{1}{x^{2}}\Big)\,dx = e^{x}\Big(\frac{1}{x}-\frac{1}{x^{2}}\Big)\,dx. \)
\( \displaystyle
\therefore\; \int e^{x}\Big(\frac{1}{x}-\frac{1}{x^{2}}\Big)\,dx
= \int dt = t + C.
\)
\( \displaystyle \frac{e^{x}}{x} + C \)
11. \( \displaystyle \int \frac{e^{x}}{x}(x\log x + 1)\,dx \)
Put \(t = e^{x}\log x.\)
Differentiate:
\( \displaystyle dt = e^{x}\log x\,dx + e^{x}\cdot\frac{1}{x}\,dx = \frac{e^{x}}{x}(x\log x + 1)\,dx. \)
\( \displaystyle dt = e^{x}\log x\,dx + e^{x}\cdot\frac{1}{x}\,dx = \frac{e^{x}}{x}(x\log x + 1)\,dx. \)
\( \therefore\; \displaystyle
\int \frac{e^{x}}{x}(x\log x + 1)\,dx
= \int dt = t + C.
\)
\( \displaystyle e^{x}\log x + C \)
12(i). \( \displaystyle \int \frac{dx}{\sqrt{\sin^{3}x\,\sin(x+\alpha)}} \)
\( I = \int \frac{dx}{\sqrt{\sin^{3}x(\sin x\cos\alpha+\cos x\sin\alpha)}} \)
\( = \int \frac{dx}{\sin^{2}x\sqrt{\cos\alpha+\sin\alpha\cot x}} \)
\( = \int \frac{\csc^{2}x\,dx}{\sqrt{\cos\alpha+\sin\alpha\cot x}} \)
Put \( z^{2}=\cos\alpha+\sin\alpha\cot x \)
\( 2z\,dz=\sin\alpha(-\csc^{2}x)\,dx \)
\( \csc^{2}x\,dx = -\dfrac{2z}{\sin\alpha}\,dz \)
\( I = \int \frac{-2z/\sin\alpha}{z}\,dz \)
\( = -\dfrac{2}{\sin\alpha}\int dz \)
\( = -\dfrac{2}{\sin\alpha}z + C \)
\( = -2\csc\alpha\sqrt{\cos\alpha+\sin\alpha\cot x} + C \)
12(ii). \( \displaystyle \int \dfrac{\sin x + \cos x}{\sin(x-a)}\,dx \)
Put \(t=x-a,\quad dx=dt\).
\(\sin x=\sin(t+a)=\sin t\cos a+\cos t\sin a.\)
\(\cos x=\cos(t+a)=\cos t\cos a-\sin t\sin a.\)
\(\sin x+\cos x=(\sin t+\cos t)\cos a+(\cos t-\sin t)\sin a.\)
\(\dfrac{\sin x+\cos x}{\sin(x-a)}=\dfrac{(\sin t+\cos t)\cos a+(\cos t-\sin t)\sin a}{\sin t}.\)
\(\;=\cos a(1+\cot t)+\sin a(\cot t-1).\)
\(\displaystyle \int \dfrac{\sin x+\cos x}{\sin(x-a)}dx
=\cos a\int(1+\cot t)dt+\sin a\int(\cot t-1)dt.\)
\(\displaystyle =(\cos a-\sin a)\,t + (\cos a+\sin a)\,\log|\sin t|+C.\)
\(\displaystyle (x-a)(\cos a-\sin a) + (\cos a+\sin a)\log|\sin(x-a)| + C.\)
13. \( \displaystyle \int e^{x}(\cos x - \sin x)\,dx \)
Put \( z = e^{x}\cos x \).
\( \dfrac{dz}{dx} = e^{x}(-\sin x) + e^{x}(\cos x) \)
\( \Rightarrow dz = e^{x}(\cos x - \sin x)\,dx \)
\( \int e^{x}(\cos x - \sin x)\,dx = \int dz \)
\( = z + C \)
\( e^{x}\cos x + C \)
14. \( \displaystyle \int e^{x}\big[\log(\sec x + \tan x)+\sec x\big]\,dx \)
Put \( z = e^{x}\log(\sec x + \tan x) \).
\( \dfrac{dz}{dx} = e^{x}\log(\sec x+\tan x) + e^{x}\dfrac{d}{dx}(\log(\sec x+\tan x)) \)
\( \dfrac{d}{dx}(\log(\sec x+\tan x)) = \dfrac{\sec x\tan x + \sec^{2}x}{\sec x+\tan x} = \sec x \)
\( \Rightarrow \dfrac{dz}{dx} = e^{x}\log(\sec x+\tan x) + e^{x}\sec x \)
\( = e^{x}[\log(\sec x+\tan x)+\sec x] \)
\( \Rightarrow dz = e^{x}[\log(\sec x+\tan x)+\sec x]\,dx \)
\( \int e^{x}[\log(\sec x+\tan x)+\sec x]\,dx = \int dz = z + C \)
\( e^{x}\log(\sec x+\tan x) + C \)
15. \( \displaystyle \int \dfrac{e^{x}(1+x^{2})}{(1+x)^{2}}\,dx \)
\( = \int e^{x}\dfrac{(x^{2}-1)+2}{(1+x)^{2}}\,dx \)
\( = \int e^{x}\left[\dfrac{x-1}{x+1} + \dfrac{2}{(x+1)^{2}}\right]dx \)
Let \( z = e^{x}\dfrac{x-1}{x+1}. \)
\(\dfrac{dz}{dx} = e^{x}\dfrac{2}{(x+1)^{2}} + e^{x}\dfrac{x-1}{x+1}\)
\( \Rightarrow dz = e^{x}\left[\dfrac{x-1}{x+1} + \dfrac{2}{(x+1)^{2}}\right]dx \)
\( \int e^{x}\left[\dfrac{x-1}{x+1} + \dfrac{2}{(x+1)^{2}}\right]dx = \int dz = z + C \)
\( e^{x}\dfrac{x-1}{x+1} + C \)
16. \( \displaystyle \int \sqrt{\sin x}\,\cos^{3}x\,dx \)
Let \( z=\sin x \).
\(\dfrac{dz}{dx}=\cos x \;\Rightarrow\; \cos x\,dx = dz\).
\( \displaystyle \int \sqrt{\sin x}\cos^{3}x\,dx
= \int \sqrt{z}\,\cos^{2}x\,dz \)
\( = \int \sqrt{z}(1-z^{2})\,dz \)
\( = \int z^{1/2}dz - \int z^{5/2}dz \)
\( = \dfrac{2}{3}z^{3/2} - \dfrac{2}{7}z^{7/2} + C \)
\( \dfrac{2}{3}\sin^{3/2}x - \dfrac{2}{7}\sin^{7/2}x + C \)
17. \( \displaystyle \int \frac{dx}{\sin x + \cos x} \)
\( = \frac{1}{\sqrt{2}} \int \frac{dx}{\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x} \)
\( = \frac{1}{\sqrt{2}} \int \frac{dx}{\sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4}} \)
\( = \frac{1}{\sqrt{2}} \int \frac{dx}{\sin\left(x + \frac{\pi}{4}\right)} \)
\( = \frac{1}{\sqrt{2}} \int \csc\left(x + \frac{\pi}{4}\right)\, dx \)
\( = \frac{1}{\sqrt{2}} \log \left| \tan\left(\frac{x + \frac{\pi}{4}}{2}\right) \right| + C \)
\( = \frac{1}{\sqrt{2}} \log \left| \tan\left(\frac{x}{2} + \frac{\pi}{8}\right) \right| + C \)
18. \( \displaystyle \int \frac{dx}{3\sin x - 4\cos x} \)
Let \( 3 = r\cos\theta \), \( 4 = r\sin\theta \).
\( (r\cos\theta)^{2} + (r\sin\theta)^{2} = 3^{2} + 4^{2} \Rightarrow r^{2} = 25 \Rightarrow r=5. \)
Again, \( \tan\theta = \dfrac{4}{3} \Rightarrow \theta = \tan^{-1}\dfrac{4}{3}. \)
Thus \( 3\sin x - 4\cos x = 5(\sin x \cos\theta - \cos x \sin\theta) = 5\sin(x - \theta). \)
\( \int \frac{dx}{3\sin x - 4\cos x}
= \frac{1}{5} \int \csc(x - \theta)\,dx \)
\( = \frac{1}{5} \log\left|\tan\left(\frac{x - \theta}{2}\right)\right| + C \)
\( = \frac{1}{5} \log\left|\tan\left(\frac{x}{2} -
\frac{1}{2}\tan^{-1}\frac{4}{3}\right)\right| + C \)
19. \( \displaystyle \int \frac{dx}{\cos2x+\sqrt{3}\sin2x} \)
\( = \frac{1}{2}\int \frac{dx}{\frac{1}{2}\cos2x+\frac{\sqrt{3}}{2}\sin2x} \)
\( = \frac{1}{2}\int \frac{dx}{\sin2x\frac{\sqrt{3}}{2}+\cos2x\frac{1}{2}} \)
\( = \frac{1}{2}\int \frac{dx}{\sin2x\cos\frac{\pi}{6}+\cos2x\sin\frac{\pi}{6}} \)
\( = \frac{1}{2}\int \frac{dx}{\sin\left(2x+\frac{\pi}{6}\right)} \)
\( = \frac{1}{2}\int \csc\left(2x+\frac{\pi}{6}\right)\,dx \)
\( = \frac{1}{4}\log\left|\tan\left(\frac{2x+\frac{\pi}{6}}{2}\right)\right| + C \)
\( = \frac{1}{4}\log\left|\tan\left(x+\frac{\pi}{12}\right)\right| + C \)
20. \( \displaystyle \int \frac{dx}{\sqrt{1+\sin x}} \)
\( = \int \frac{dx}{\sqrt{\sin^{2}\frac{x}{2}+\cos^{2}\frac{x}{2}+2\sin\frac{x}{2}\cos\frac{x}{2}}} \)
\( = \int \frac{dx}{\sqrt{\left(\sin\frac{x}{2}+\cos\frac{x}{2}\right)^{2}}} \)
\( = \int \frac{dx}{\sin\frac{x}{2}+\cos\frac{x}{2}} \)
\( = \frac{1}{\sqrt{2}}\int \frac{dx}{\sin\frac{x}{2}\frac{1}{\sqrt{2}}+\cos\frac{x}{2}\frac{1}{\sqrt{2}}} \)
\( = \frac{1}{\sqrt{2}}\int \frac{dx}{\sin\frac{x}{2}\cos\frac{\pi}{4}+\cos\frac{x}{2}\sin\frac{\pi}{4}} \)
\( = \frac{1}{\sqrt{2}}\int \frac{dx}{\sin\left(\frac{x}{2}+\frac{\pi}{4}\right)} \)
\( = \frac{1}{\sqrt{2}}\int \csc\left(\frac{x}{2}+\frac{\pi}{4}\right)\,dx \)
\( = \frac{2}{\sqrt{2}} \log\left|\tan\left(\frac{x}{4}+\frac{\pi}{8}\right)\right| + C \)
\( = \sqrt{2}\,\log\left|\tan\left(\frac{x}{4}+\frac{\pi}{8}\right)\right| + C \)
21. \( \displaystyle \int \frac{dx}{x\log x \,[\log(\log x)]} \)
Let \( z = \log(\log x). \)
\( \displaystyle \frac{d}{dx}(\log(\log x)) = \frac{1}{x\log x} = \frac{dz}{dx} \)
\( \Rightarrow \frac{dx}{x\log x} = dz \)
\( \displaystyle \int \frac{dx}{x\log x[\log(\log x)]} = \int \frac{dz}{z} \)
\( = \log|z| + C = \log|\log(\log x)| + C \)
22. \( \displaystyle \int \frac{\sin x + x\cos x}{x\sin x}\,dx \)
Let \( z = x\sin x. \)
\( \displaystyle \frac{d}{dx}(x\sin x) = x\cos x + \sin x = \frac{dz}{dx} \)
\( \Rightarrow (x\cos x + \sin x)\,dx = dz \)
\( \displaystyle \int \frac{\sin x + x\cos x}{x\sin x}\,dx = \int \frac{dz}{z} \)
\( = \log|z| + C = \log|x\sin x| + C \)
23. \( \displaystyle \int \frac{\tan^{2}x}{\sec x - 1}\,dx \)
\( = \int \frac{\sec^{2}x - 1}{\sec x - 1}\,dx \)
\( = \int \frac{(\sec x + 1)(\sec x - 1)}{(\sec x - 1)}\,dx \)
\( = \int (\sec x + 1)\,dx \)
\( = \int \sec x\,dx \;+\; \int dx \)
\( = \log\left|\tan\!\left(\frac{x}{2} + \frac{\pi}{4}\right)\right| + x + C \)
24. \( \displaystyle \int \frac{1+\sin x}{\sin x(1+\cos x)}\,dx \)
\( = \int \frac{\sin^{2}\frac{x}{2}+\cos^{2}\frac{x}{2}+2\sin\frac{x}{2}\cos\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}\cdot 2\cos^{2}\frac{x}{2}}\,dx \)
\( = \int \frac{\sin^{2}\frac{x}{2}}{4\sin\frac{x}{2}\cos^{3}\frac{x}{2}}\,dx
\;+\; \int \frac{\cos^{2}\frac{x}{2}}{4\sin\frac{x}{2}\cos^{3}\frac{x}{2}}\,dx
\;+\; \int \frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{4\sin\frac{x}{2}\cos^{3}\frac{x}{2}}\,dx \)
\( = \frac{1}{4}\int \frac{\tan\frac{x}{2}}{\cos^{2}\frac{x}{2}}\,dx
\;+\; \frac{1}{4}\int \frac{1}{\sin\frac{x}{2}}\,dx
\;+\; \frac{1}{2}\int \sec^{2}\frac{x}{2}\,dx \)
\( = \frac{1}{4}\cdot 2\int \tan\frac{x}{2}\,d(\tan\frac{x}{2})
\;+\; \frac{1}{2}\int \csc\frac{x}{2}\,dx
\;+\; \frac{1}{2}\int \sec^{2}\frac{x}{2}\,dx \)
\( = \frac{1}{4}\tan^{2}\frac{x}{2}
\;+\; \frac{1}{2}\log\left|\tan\frac{x}{2}\right|
\;+\; \frac{1}{2}\tan\frac{x}{2} + C \)
\( = \frac{1}{4}\tan^{2}\frac{x}{2}
+ \frac{1}{2}\log\left|\tan\frac{x}{2}\right|
+ \frac{1}{2}\tan\frac{x}{2} + C \)
25. \( \displaystyle \int \frac{1+\tan x}{1-\tan x}\,dx \)
\( = \int \frac{\dfrac{\sin x}{\cos x} + 1}{\,1 - \dfrac{\sin x}{\cos x}\,}\,dx \)
\( = \int \frac{\sin x + \cos x}{\cos x - \sin x}\,dx \)
Let \( z = \cos x - \sin x \Rightarrow \dfrac{dz}{dx} = -\sin x - \cos x \).
\( \Rightarrow (\cos x + \sin x)\,dx = -dz \)
\( \int \frac{\cos x + \sin x}{\cos x - \sin x}\,dx = \int \frac{-dz}{z} = -\log|z|+C \)
\( = -\log|\cos x - \sin x| + C \)
26. \( \displaystyle \int \dfrac{x^{4}-1}{x^{2}\sqrt{x^{4}+x^{2}+1}}\,dx \)
Write \( x^{4}+x^{2}+1 = x^{2}\!\left(x^{2}+1+\dfrac{1}{x^{2}}\right) \).
Put \( z^{2} = x^{2}+1+\dfrac{1}{x^{2}} \Rightarrow z = \sqrt{x^{2}+1+\dfrac{1}{x^{2}}}. \)
Differentiate:
\( z^{2} = x^{2}+1+\dfrac{1}{x^{2}} \Rightarrow 2z\,dz = \left( 2x - \dfrac{2}{x^{3}} \right) dx \)
\( \Rightarrow (x - \dfrac{1}{x^{3}})dx = z\,dz \)
Convert integrand:
\( \displaystyle \frac{x^{4}-1}{x^{2}\sqrt{x^{4}+x^{2}+1}}
= \frac{x^{4}-1}{x^{3}z}
= \frac{x-\dfrac{1}{x^{3}}}{z}. \)
\( I = \int \frac{x-\dfrac{1}{x^{3}}}{z}\,dx
= \int \frac{z\,dz}{z} = \int dz = z + C \)
\( \displaystyle = \sqrt{x^{2}+1+\dfrac{1}{x^{2}}} + C \)
