Question

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

Answer 1

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

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

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

Now the integral becoms-

$\int \sin tdt$

$=-\cos t+C$

Substituting the value of $t$ in above equation, we have

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