Question

The integral $\int x{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)dx$ is equal to :
(Note : $(x>0)$)

• A. $-x+(1+x^2){\cot }^{-1}x+c$
• B. $-x+(1+x^2){\tan }^{-1}x+c$
• C. $-x-(1+x^2){\tan }^{-1}xc$
• D. $x-(1+x^2){\cot }^{-1}x+c$

Answer 1

The correct option is B $-x+(1+x^2){\tan }^{-1}x+c$

$\int x{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)dx(x>0)$

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

$I=\int \tan \theta {\cos }^{-1}\left(\cos 2\theta \right){\sec }^2\theta d\theta$

$I=\int 2\theta \tan \theta {\sec }^2\theta d\theta$

Using by parts

$\int u.vdx=u\int vdx$ $-\int (\frac{du}{dx}\int vdx)dx$

$I=2\theta \int \tan \theta {\sec }^2\theta d\theta -\int 2\cdot \int \tan \theta {\sec }^2\theta d\theta$

$=2\theta \frac{{\tan }^2\theta }{2}-2\times \frac{1}{2}\int {\tan }^2\theta d\theta$

$=\theta {\tan }^2\theta -[\int (-1+{\sec }^2\theta \left)d\theta \right]$ ................. ${\sec }^{2}x=1+{\tan }^{2}x$

$=\theta {\tan }^2\theta +\theta -\tan \theta +c$

$=\theta (1+{\tan }^2\theta )-\tan \theta +c$

Replacing $\theta$ by $x$,

$I=\left({\tan }^{-1}x\right)(1+x^2)-x+c$

$=-x+(1+x^2){\tan }^{-1}x+c$