Question

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

Answer 1

We have,

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

Let

$x=\sin \theta$

$dx=\cos \theta d\theta$

Therefore,

${\int }_0^{\frac{\pi }{2}}{\tan }^{-1}\left(\frac{\sin \theta }{\sqrt{1-{\sin }^2\theta }}\right)\cos \theta d\theta$

$\Rightarrow {\int }_0^{\frac{\pi }{2}}{\tan }^{-1}\left(\tan \theta \right)\cos \theta d\theta$

$\Rightarrow {\int }_0^{\frac{\pi }{2}}\theta .\cos \theta d\theta$

Using formula,

$\int I.IIdx=I\int IIdx-\int (\frac{dI}{dx}\int IIdx)dx$

Therefore,

$I={\int }_0^{\frac{\pi }{2}}\theta .\cos \theta d\theta$

Then,

$I=\theta {\int }_0^{\frac{\pi }{2}}\cos \theta d\theta -{\int }_0^{\frac{\pi }{2}}\left(\frac{d\theta }{d\theta }{\int }_0^{\frac{\pi }{2}}\cos \theta d\theta \right)d\theta$

$=\theta {{\left(\sin \theta \right)}_0}^{\frac{\pi }{2}}-{\int }_0^{\frac{\pi }{2}}\sin \theta d\theta$

$=\theta {{\left(\sin \theta \right)}_0}^{\frac{\pi }{2}}-{{(-\cos \theta )}_0}^{\frac{\pi }{2}}$

Taking limit and we get

$=[\frac{\pi }{2}-0][\sin \frac{\pi }{2}-\sin 0]-\left[(-\cos \frac{\pi }{2}+\cos 0)\right]$

$=\frac{\pi }{2}(1-0)-(-0+1)$

$=\frac{\pi }{2}-1$

Hence, this is the answer.