Chhaya Mathematics- Method of Substitution [Part-A]

WB Board Class 12 Mathematics – Integration (SN Dey) Solutions

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Integration — Very Short Answer Type (Substitution Method

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 \(u=3-4x.\)

Differentiate: \(du=-4\,dx\Rightarrow dx=-\dfrac{1}{4}du.\)

\(\therefore\; I=-\dfrac{1}{4}\int\sin u\,du = \dfrac{1}{4}\cos u + C\)

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: \(\cos^{2}u=\tfrac{1+\cos2u}{2}.\)

\(\therefore\; I=\tfrac{1}{2}\int\cos^{2}u\,du=\tfrac{1}{4}\int(1+\cos2u)\,du\)

\(=\tfrac{1}{4}\big(u+\tfrac{\sin2u}{2}\big)+C = \tfrac{1}{4}(2x+3)+\tfrac{1}{8}\sin(4x+6)+C\)

\(\tfrac{1}{4}(2x+3)+\tfrac{1}{8}\sin(4x+6)+C\)

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: \(du=6x\,dx\Rightarrow x\,dx=\tfrac{1}{6}du.\)

\(\therefore\; I=\tfrac{1}{6}\int\dfrac{du}{\sqrt{u}}\)

\(=\tfrac{1}{6}\cdot 2\sqrt{u}+C = \tfrac{1}{3}\sqrt{u}+C\)

Back: \(\tfrac{1}{3}\sqrt{3x^{2}+4}+C\)

\(\tfrac{1}{3}\sqrt{3x^{2}+4}+C\)

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 :

\(\dfrac{du}{dx}=\dfrac{d}{dx}(x)+\dfrac{d}{dx}(\log\sin x)\)

\(=1+\dfrac{1}{\sin x}\cdot\cos x\)

\(=1+\cot x.\)

\(\therefore\; du=(1+\cot x)\,dx.\)

\(\therefore\; I=\int\dfrac{du}{u}\)

\(=\log|u|+C\)

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. \( \displaystyle \int \dfrac{\cos x + x\sin x}{x(x+\cos x)}\,dx \)

\( \displaystyle \int \dfrac{\cos x + x\sin x}{x(x+\cos x)}\,dx \)

\(=\displaystyle \int \dfrac{\cos x + x\sin x + x - x}{x(x+\cos x)}\,dx\)

\(=\displaystyle \int \dfrac{(\cos x + x) + x(\sin x-1)}{x(x+\cos x)}\,dx\)

\(=\displaystyle \int \dfrac{1}{x}\,dx + \int \dfrac{\sin x-1}{x+\cos x}\,dx\)

\(=\displaystyle \log|x| - \int \dfrac{1-\sin x}{x+\cos x}\,dx\)

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 \(u=\sqrt{4x+3}\Rightarrow 4x+3=u^{2},\;dx=\dfrac{1}{4}2u\,du=\dfrac{u}{2}\,du.\)

\( \displaystyle \int\sqrt{4x+3}\,dx = \int u\cdot\dfrac{u}{2}\,du = \tfrac{1}{2}\int u^{2}\,du = \tfrac{1}{6}u^{3}.\)

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 \(z=2x+1.\)

Differentiate: \(dz=2\,dx\Rightarrow dx=\dfrac{dz}{2}.\)

Substitute \(x=\dfrac{z-1}{2}\). Then \[ x^{3}+3x^{2}+2x-7=\dfrac{z^{3}+3z^{2}-z-59}{8}. \]

\(\therefore\; \displaystyle \int \dfrac{x^{3}+3x^{2}+2x-7}{2x+1}\,dx = \int \dfrac{\tfrac{z^{3}+3z^{2}-z-59}{8}}{z}\cdot\dfrac{dz}{2} = \dfrac{1}{16}\int\Big(z^{2}+3z-1-\dfrac{59}{z}\Big)\,dz.\)

\( \displaystyle = \dfrac{1}{16}\Big(\dfrac{z^{3}}{3}+\dfrac{3z^{2}}{2}-z-59\log|z|\Big)+C.\)

\(\therefore\; \displaystyle I = \dfrac{z^{3}}{48} + \dfrac{3z^{2}}{32} - \dfrac{z}{16} - \dfrac{59}{16}\log|z| + C.\)

Replace \(z=2x+1\).

\(\displaystyle \dfrac{(2x+1)^{3}}{48} + \dfrac{3(2x+1)^{2}}{32} - \dfrac{2x+1}{16} - \dfrac{59}{16}\log|2x+1| + C\)

27. \( \displaystyle \int \dfrac{e^{2x}}{e^{x}+1}\,dx \)

Put \(u=e^{x}\Rightarrow du=e^{x}dx.\)

Then \(e^{2x}dx = u^{2}\cdot\dfrac{du}{u}=u\,du.\)

\( \therefore\; \displaystyle \int \dfrac{e^{2x}}{e^{x}+1}\,dx = \int \dfrac{u}{u+1}\,du.\)

\( = \int\Big(1-\dfrac{1}{u+1}\Big)\,du = u - \log|u+1| + C.\)

\( e^{x} - \log\big(e^{x}+1\big) + C \)

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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. \( \displaystyle \int \dfrac{10^{x}-1}{10^{x}+1}\,dx \)

Put \(u=10^{x}.\)

Differentiate: \(du=10^{x}\log10\,dx \Rightarrow dx=\dfrac{du}{u\log10}.\)

\(\therefore\; \int \dfrac{10^{x}-1}{10^{x}+1}\,dx=\dfrac{1}{\log10}\int\Big(\dfrac{1}{1+u}-\dfrac{1}{u}\Big)\,du.\)

\(=\dfrac{1}{\log10}\log(1+u) - \dfrac{1}{\log10}\log u + C.\)

\( \dfrac{1}{\log10}\log\big(1+10^{x}\big) - x + C \)

35. \( \displaystyle \int \dfrac{\sqrt{\tan x}}{\cos^{4}x}\,dx \)

Write integrand as \( \tan^{1/2}x\sec^{4}x.\)

Put \(u=\tan x\Rightarrow du=\sec^{2}x\,dx.\)

\(\therefore\; \int \tan^{1/2}x\sec^{4}x\,dx = \int u^{1/2}(1+u^{2})\,du.\)

\(=\int (u^{1/2}+u^{5/2})\,du = \dfrac{2}{3}u^{3/2} + \dfrac{2}{7}u^{7/2} + C.\)

\( \dfrac{2}{3}\tan^{3/2}x + \dfrac{2}{7}\tan^{7/2}x + C \)

36. \( \displaystyle \int \dfrac{\sqrt{\sin x}}{\cos^{5}x}\,dx \)

Put \(u=\tan x\Rightarrow du=\sec^{2}x\,dx.\)

Use \(\sin x=\dfrac{u}{\sqrt{1+u^{2}}},\; \cos^{-5}x=(1+u^{2})^{5/2}.\)

\(\therefore\; \)After substitution integrand becomes a polynomial in \(u\); integrate termwise to obtain an elementary antiderivative (algebraic).

(Result expressible in powers of \(\tan x\); algebraic simplification omitted.)

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. \( \displaystyle \int \dfrac{dx}{x^{2}\sqrt{a^{2}+x^{2}}} \)

Put \(x=a\tan\theta.\)

Then integrand reduces to \(\dfrac{1}{a^{2}}\cot\theta\csc\theta\,d\theta.\)

\(\therefore\; \displaystyle I=-\dfrac{1}{a^{2}}\csc\theta + C.\)

Since \(\csc\theta=\dfrac{\sqrt{a^{2}+x^{2}}}{x}\) we get

\( -\dfrac{1}{a^{2}}\cdot\dfrac{\sqrt{a^{2}+x^{2}}}{x} + C \)

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. \( \displaystyle \int \dfrac{dx}{x^{3}\sqrt{x^{2}-1}} \)

Put \(x=\sec\theta\Rightarrow dx=\sec\theta\tan\theta\,d\theta,\;\sqrt{x^{2}-1}=\tan\theta.\)

\(\therefore\; \displaystyle \int \dfrac{dx}{x^{3}\sqrt{x^{2}-1}} = \int \dfrac{\sec\theta\tan\theta}{\sec^{3}\theta\tan\theta}\,d\theta = \int \cos^{2}\theta\,d\theta.\)

\(=\tfrac{1}{2}\big(\theta + \sin\theta\cos\theta\big)+C.\)

\( \tfrac{1}{2}\sec^{-1}x + \tfrac{1}{2}\cdot\dfrac{\sqrt{x^{2}-1}}{x^{2}} + C \)

45. \( \displaystyle \int \dfrac{dx}{(1-x)\sqrt{1-x^{2}}} \)

Put \(x=\sin\theta.\)

Then integrand \(=\dfrac{1}{1-\sin\theta}\,d\theta.\)

\(\therefore\; \displaystyle \int \dfrac{d\theta}{1-\sin\theta} = \int \dfrac{1+\sin\theta}{\cos^{2}\theta}\,d\theta = \tan\theta + \sec\theta + C.\)

\( \dfrac{x}{\sqrt{1-x^{2}}} + \dfrac{1}{\sqrt{1-x^{2}}} + C \)

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.

(Polynomial in \((x+1)^{1/6}\) minus \(6\log\big((x+1)^{1/6}+1\big)\) + C)

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5 marks questions will go here.