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}`