Question

How to solve this indefinite integral? ​

Answer 1

Integrate by parts; by the product rule for derivatives, we have

$(u(x)v(x))'=u'(x)v(x)+u(x)v'(x)\implies u(x)v'(x)=(u(x)v(x))'-u'(x)v(x)$

$\implies\displaystyle\int u(x)v'(x)\,\mathrm dx=u(x)v(x)-\int u'(x)v(x)\,\mathrm dx$

or more succinctly,

$\displaystyle\int u\,\mathrm dv=uv-\int v\,\mathrm du$

Let

$u=x\implies\mathrm du=\mathrm dx$

$\mathrm dv=\cos x\,\mathrm dx\implies v=\sin x$

Then

$\displaystyle\int x\cos x\,\mathrm dx=x\sin x-\int\sin x\,\mathrm dx$

so that

$\displaystyle\int x\cos x\,\mathrm dx=\boxed{x\sin x+\cos x+C}$