Question

The integral $\int x{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)dx$, where $x>0$, is equal to
(where $c$ is constant of integration)

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

Answer 1

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

Let $I=\int x{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)dx$
Putting $x=\tan \theta \Rightarrow dx={\sec }^2\theta d\theta$
$\Rightarrow I=\int \tan \theta {\cos }^{-1}\left(\cos 2\theta \right){\sec }^2\theta d\theta \Rightarrow I=\int 2\theta \tan \theta {\sec }^2\theta d\theta$

Using integration by parts, we get
$\Rightarrow I=2\theta \int \tan \theta {\sec }^2\theta d\theta -\int 2\cdot \left(\int \tan \theta {\sec }^2\theta d\theta \right)d\theta \Rightarrow I=\frac{2\theta \cdot {\tan }^2\theta }{2}-2\times \frac{1}{2}\int {\tan }^2\theta d\theta \Rightarrow I=\theta {\tan }^2\theta -\int ({\sec }^2\theta -1)d\theta \Rightarrow I=\theta {\tan }^2\theta -\tan \theta +\theta +c\Rightarrow I=x^2{\tan }^{-1}x-x+{\tan }^{-1}x+c\therefore I=\left({\tan }^{-1}x\right)(1+x^2)-x+c$
Where $c$ is constant of integration.