Question

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

Answer 1

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

Let $x=\tan y⟹dx=se{c}^{2}ydy$

$=\int {\cos }^{-1}\left(\frac{1-{\tan }^{2}y}{1+{\tan }^{2}y}\right){\sec }^{2}ydy$

$=\int {\cos }^{-1}\cos 2y{\sec }^{2}ydy$

$=\int 2y{\sec }^{2}ydy$

$=2[y\tan y+\ln |cosy|]+c$

$=2[x{\tan }^{-1}x+\ln |\frac{1}{\sqrt{1+x^2}}|]+c$