Question

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

Answer 1

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

Let ${\tan }^{-1}x=u\Rightarrow \frac{1}{1+x^2}dx=du$

$\therefore \int \frac{\sin \left({\tan }^{-1}x\right)}{1+x^2}dx=\int \sin udu=-\cos u+C$

As $u={\tan }^{-1}x$

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