Question

Evaluate :

$\int x\sin xdx$

Answer 1

$I=\int x\sin xdx$

Taking x as first function and sin x as second function and integrating by parts, we get,

$I=x\int sinxdx-\int [\left(\frac{d}{dx}x\right)\int \sin xdx]dx$

$=x(-\cos x)-\int 1.(-\cos x)dx$

$=-x\cos x+\sin x+C$