Question

${\int }_0^1{\tan }^{-1}\left(\frac{2x}{1-x^2}\right)dx=\frac{\pi }{a}-\ln a$. Find $a$.

• A. $2$
• B. $1$
• C. $-1$
• D. None of these

Answer 1

The correct option is A $2$

${\int }_0^1{\tan }^{-1}\left(\frac{2x}{1-x^2}\right)dx$

Putting

$x=\tan \theta$

$dx={\sec }^2\theta d\theta$

$={\int }_0^{\frac{\pi }{4}}{\tan }^{-1}\left(\tan 2\theta \right){\sec }^2\theta d\theta$

$={\int }_0^{\frac{\pi }{4}}2\theta {\sec }^2\theta d\theta$

$=2{\int }_0^{\frac{\pi }{4}}\theta {\sec }^2\theta d\theta$

$=2\left[\theta .\tan \theta -\int 1.\tan \theta d\theta \right]$

$=2{\left[\theta .\tan \theta -\log \left|\sec \theta \right|\right]}^{\frac{\pi }{4}}$

$=2{\left[\theta .\tan \theta -\log cos\theta \right]}^{\frac{\pi }{4}}$

$=2\left[\left(\frac{\pi }{4}+\ln \frac{1}{\sqrt{2}}\right)-\left(0\right)\right]$

$=\frac{2\pi }{4}+2\ln {\left(2\right)}^{\frac{-1}{2}}$

$=\frac{\pi }{2}-\ln 2$

On cmparing with $\frac{\pi }{a}-\ln a$

We get $a=2$