Question

If ${\int }_0^1{\tan }^{-1}xdx=\alpha$, then ${\int }_0^{\pi /4}{\tan }^{-1}\left(\frac{2{\cos }^2\theta }{2-\sin 2\theta }\right){\sec }^2\theta d\theta$ is equal to

• A. $\alpha$
• B. $\frac{\alpha }{2}$
• C. $3\alpha$
• D. $2\alpha$

Answer 1

The correct option is D $2\alpha$

Given : ${\int }_0^1{\tan }^{-1}xdx=\alpha$

$I={\int }_0^{\frac{\pi }{4}}{\tan }^{-1}\left[\frac{2{\cos }^2\theta }{2-\left(\sin 2\theta \right)}\right]{\sec }^2\theta d\theta I={\int }_0^{\frac{\pi }{4}}{\tan }^{-1}\left[\frac{2}{2{\sec }^2\theta -2\tan \theta }\right]{\sec }^2\theta d\theta I={\int }_0^{\frac{\pi }{4}}{\tan }^{-1}\left[\frac{1}{{\tan }^2\theta -\tan \theta +1}\right]{\sec }^2\theta d\theta Let:t=\tan \theta \theta \to 0t\to 0dt={\sec }^2\theta d\theta \theta \to \frac{\pi }{4}t\to 1I={\int }_0^1{\tan }^{-1}\left[\frac{1}{(t^2-t+1)}\right]dtI={\int }_0^1{\tan }^{-1}\left[\frac{t-(t-1)}{1+t(t-1)}\right]dtI={\int }_0^1{\tan }^{-1}tdt-{\int }_0^1{\tan }^{-1}(t-1)dtI={\int }_0^1{\tan }^{-1}tdt-{\int }_0^1{\tan }^{-1}(1-t-1)dtI={\int }_0^1{\tan }^{-1}tdt+{\int }_0^1{\tan }^{-1}\left(t\right)dtI=2{\int }_0^1{\tan }^{-1}tdt=2\alpha$

Hence the correct answer is $2\alpha$.