Question

Evaluate:

$\int x\sin (1+x^2)dx$

Answer 1

$\int x\sin (1+x^2)dx$

$1+x^2=t$

$2xdx=dt$

$=\frac{1}{2}\int \sin tdt$

$\Rightarrow \frac{-\cos t}{2}+c$

$=\frac{-cos(1+x^2)}{2}+c$