⇒ Definite Integration

Correct Option is (B)

Let $I=\underset{0}{\overset{\pi }{\int }}\frac{dx}{4+3\cos x}$

Put $\tan \frac{x}{2}=\mathrm{t}$

$\therefore \mathrm{d}x=\frac{2\mathrm{dt}}{1+{\mathrm{t}}^2}$ and $\cos x=\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2}$

$\begin{array}{rl}\therefore \mathrm{I}=\underset{0}{\overset{\pi }{\int }}\frac{\mathrm{d}x}{4+3\cos x}& =\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{2\mathrm{dt}}{7+{\mathrm{t}}^2}\\ \\ & ={[\frac{2}{\sqrt{7}}{\tan }^{-1}\left(\frac{\mathrm{t}}{\sqrt{7}}\right)]}_0^{\mathrm{\infty }}\\ \\ & =\frac{2}{\sqrt{7}}[{\tan }^{-1}\mathrm{\infty }-0]\\ \\ & =\frac{2}{\sqrt{7}}\cdot \frac{\pi }{2}=\frac{\pi }{\sqrt{7}}\end{array}$

Correct Option is (D)

$f(x)=\int \frac{\sqrt{x}}{(1+x{)}^2}\mathrm{d}x,x\ge 0$

$\mathrm{f}(3)-\mathrm{f}(1)=\underset{1}{\overset{3}{\int }}\frac{\sqrt{x}}{(1+x{)}^2}\mathrm{d}x=\mathrm{I}$ (say)

Put $\sqrt{x}=\tan \theta \Rightarrow \mathrm{d}x=2\tan \theta {\sec }^2\theta \mathrm{d}\theta$

When $x=1,\theta =\frac{\pi }{4}$ and when $x=3,\theta =\frac{\pi }{3}$

$\begin{array}{rl}\therefore I& =\underset{\frac{\pi }{4}}{\overset{\frac{\pi }{3}}{\int }}\frac{2{\tan }^2\theta {\sec }^2\theta }{{(1+{\tan }^2\theta )}^2}d\theta \\ & =\underset{\frac{\pi }{4}}{\overset{\frac{\pi }{3}}{\int }}\frac{2{\tan }^2\theta }{1+{\tan }^2\theta }d\theta \\ & =\underset{\frac{\pi }{4}}{\overset{\frac{\pi }{3}}{\int }}\frac{2{\sin }^2\theta }{{\sin }^2\theta +{\cos }^2\theta }d\theta \\ & =\underset{\frac{\pi }{4}}{\overset{\frac{\pi }{3}}{\int }}(1-\cos 2\theta )\mathrm{d}\theta \\ & ={[\theta -\frac{\sin 2\theta }{2}]}_{\frac{\pi }{4}}^{\frac{\pi }{3}}\end{array}$

$\begin{array}{rl}& =(\frac{\pi }{3}-\frac{\pi }{4})-\frac{1}{2}(\frac{\sqrt{3}}{2}-1)\\ & =\frac{\pi }{12}-\frac{\sqrt{3}}{4}+\frac{1}{2}\end{array}$

Correct Option is (D)

$\begin{array}{rl}\text{ Let }\mathrm{I}& =\underset{-1}{\overset{3}{\int }}({\cot }^{-1}\left(\frac{x}{x^2+1}\right)+{\cot }^{-1}\left(\frac{x^2+1}{x}\right))\mathrm{d}x\\ =& \underset{-1}{\overset{3}{\int }}({\tan }^{-1}\left(\frac{x^2+1}{x}\right)+{\cot }^{-1}\left(\frac{x^2+1}{x}\right))\mathrm{d}x\\ & \dots [\because {\cot }^{-1}(x)={\tan }^{-1}\left(\frac{1}{x}\right)]\end{array}$

$\begin{array}{rl}& =\underset{-1}{\overset{3}{\int }}\frac{\pi }{2}\mathrm{d}x\dots [{\tan }^{-1}x+{\cot }^{-1}x=\frac{\pi }{2}]\\ & =\frac{\pi }{2}[x{]}_{-1}^3\\ & =\frac{\pi }{2}(4)\\ \therefore \mathrm{I}& =2\pi \end{array}$

Correct Option is (C)

$\begin{array}{rl}\text{ Let }\mathrm{I}& =\underset{0}{\overset{1}{\int }}({\cos }^{-1}x)(1)\mathrm{d}x\\ & ={[{\cos }^{-1}x\cdot x-\int (\frac{-1}{\sqrt{1-x^2}}\cdot x)\mathrm{d}x]}_0^1\\ & ={[x\cdot {\cos }^{-1}x+\int \frac{x}{\sqrt{1-x^2}}\mathrm{d}x]}_0^1\\ & ={[x{\cos }^{-1}x-\sqrt{1-x^2}]}_0^1\\ & =[1\cdot {\cos }^{-1}(1)-\sqrt{1-(1{)}^2}]-[0\cdot {\cos }^{-1}(0)-\sqrt{1-0^2}]\\ & =0-(-1)\\ & =1\end{array}$

Correct Option is (A)

$\begin{array}{rl}& \text{ Let }\mathrm{I}=\underset{0}{\overset{\frac{1}{2}}{\int }}\frac{x^2}{{(1-x^2)}^{\frac{3}{2}}}\mathrm{d}x\\ & \text{ Put }x=\sin \theta \end{array}$

$\begin{array}{rl}\Rightarrow dx=& \cos \theta d\theta \\ {(1-x^2)}^{\frac{3}{2}}& ={(1-{\sin }^2\theta )}^{\frac{3}{2}}\\ & ={({\cos }^2\theta )}^{\frac{3}{2}}\\ & ={\cos }^3\theta \\ \therefore I& =\underset{0}{\overset{\frac{\pi }{6}}{\int }}\frac{{\sin }^2\theta \cdot \cos \theta d\theta }{{\cos }^3\theta }\\ & =\underset{0}{\overset{\frac{\pi }{6}}{\int }}{\tan }^2\theta d\theta \\ & =\underset{0}{\overset{\frac{\pi }{6}}{\int }}({\sec }^2\theta -1)d\theta \\ & =[\tan \theta {]}_0^{\frac{\pi }{6}}-[\theta {]}_0^{\frac{\pi }{6}}\\ & =(\tan \frac{\pi }{6}-\tan 0)-(\frac{\pi }{6}-0)\\ & =\frac{1}{\sqrt{3}}-\frac{\pi }{6}\end{array}$

$\begin{array}{rl}& =\frac{\sqrt{3}}{3}-\frac{\pi }{6}\\ & =\frac{2\sqrt{3}-\pi }{6}\end{array}$

But, $\underset{0}{\overset{\frac{1}{2}}{\int }}\frac{x^2}{{(1-x^2)}^{\frac{3}{2}}}\mathrm{d}x=\frac{\mathrm{k}}{6}$ $$ ...... [Given]

$\therefore \mathrm{k}=2\sqrt{3}-\pi$