Question

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

Answer 1

We know ${\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)=2{\tan }^{-1}x$

$\therefore \int {\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)dx=2\int {\tan }^{-1}xdx$.

Let $I=2\left[{\tan }^{-1}x\int dx-\int \left\{\frac{d}{dx}\right({\tan }^{-1}x)\int dx\}dx\right]$

$I=2\left[{\tan }^{-1}a\int dx-\int \{\frac{1}{1+x^2}.x\}dx\right]$

$I=2x{\tan }^{-1}x-\int \frac{2x}{1+x^2}dx$

Let $1+x^2=t\Rightarrow 2x=dt$

$\therefore I=2x{\tan }^{-1}x-\int \frac{dt}{t}$

$=2x{\tan }^{-1}x-\log t+C$ where $c=$ constant of integration

$I=2x{\tan }^{-1}x-\log (1+x{)}^2+C$