Question

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

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{2}-1$
• C. $0$
• D. None of these

Answer 1

The correct option is B $\frac{\pi }{2}-1$

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

Putting $x=\sin \theta$ $dx=\cos \theta d\theta$

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

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

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

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

$={\left(\theta \sin \theta \right)}^{\frac{\pi }{2}}-{\int }_0^{\frac{\pi }{2}}1.\sin \theta d\theta$

$={{\left[\theta \sin \theta +\cos \theta \right]}_0}^{\frac{\pi }{2}}$

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

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