Question

$\underset{0}{\overset{1/2}{\int }}\frac{dx}{(1+x^2)\sqrt{1-x^2}}$ is equal to

• A. $\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\sqrt{\frac{2}{3}}\right)$
• B. $\frac{2}{\sqrt{2}}{\tan }^{-1}\left(\frac{3}{\sqrt{2}}\right)$
• C. $\frac{\sqrt{2}}{2}{\tan }^{-1}\left(\frac{3}{2}\right)$
• D. $\frac{\sqrt{2}}{2}{\tan }^{-1}\left(\frac{\sqrt{3}}{2}\right)$

Answer 1

The correct option is A $\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\sqrt{\frac{2}{3}}\right)$

$\text{Let I}=\underset{0}{\overset{1/2}{\int }}\frac{dx}{(1+x^2)\sqrt{1-x^2}}$
$\text{Put}x=\tan \theta \Rightarrow dx={\sec }^2\theta d\theta$

$\text{Then, I}=\underset{0}{\overset{{\tan }^{-1}1/2}{\int }}\frac{d\theta }{\sqrt{1-{\tan }^2\theta }}$

$=\underset{0}{\overset{{\tan }^{-1}1/2}{\int }}\frac{\cos \theta d\theta }{\sqrt{{\cos }^2\theta -{\sin }^2\theta }}=\underset{0}{\overset{{\tan }^{-1}1/2}{\int }}\frac{\cos \theta d\theta }{\sqrt{1-2{\sin }^2\theta }}$

$\text{Now, put}\sin \theta =t\Rightarrow \cos \theta d\theta =dt$
$\text{When}\theta =0,\text{then}t=0$
$\text{When}\theta ={\tan }^{-1}1/2,\text{then}t=\sin \left({\tan }^{-1}1/2\right)$

$\text{Then, I}=\underset{0}{\overset{\sin \left({\tan }^{-1}1/2\right)}{\int }}\frac{1}{\sqrt{2}}\frac{dt}{\sqrt{\frac{1}{2}-t^2}}$

$=\frac{1}{\sqrt{2}}[{\sin }^{-1}\sqrt{2}t{]}_{t=0}^{t=\sin \left({\tan }^{-1}1/2\right)}$
$=\frac{1}{\sqrt{2}}{\sin }^{-1}\sqrt{2}\sin \left({\tan }^{-1}1/2\right)$
$=\frac{1}{\sqrt{2}}{\sin }^{-1}\sqrt{\frac{2}{5}}$
$=\frac{1}{\sqrt{2}}{\tan }^{-1}\sqrt{\frac{2}{3}}$