Question

Integrate the function:
$\frac{\sin \left({\tan }^{-1}x\right)}{1+x^2}$

Answer 1

$I=\int \frac{\sin \left({\tan }^{-1}x\right)}{1+x^2}dx$

Let ${\tan }^{-1}x=t$

$\frac{1}{1+x^2}dx=dt$

$\therefore I=\int \sin \left(t\right)dt$

$=-\cos t$

$I=-\cos \left({\tan }^{-1}x\right)$