Question

How do you use integration by parts to evaluate the integral `2xsin(x)dx`?

Answer 1

I tried this:

Explanation:

Have a look:

Answer 2

`sin(x)-xcos(x)+C`

Explanation:

We have `int2xsin(x)dx`

Take the constant out:

`2intxsin(x)dx`

According to integration by parts, where `u` and `v` are functions,

`intuvdx=uintvdx-intu'(intvdx)dx`

Here, `u=x` and `v=sin(x)`

So we have:

`x intsin(x)dx-int(x)'(intsin(x)dx)dx`

`x(-cos(x))-int(-cos(x))dx`

`-xcos(x)+intcos(x)dx`

`-xcos(x)+sin(x)`

Add the constant of integration:

`sin(x)-xcos(x)+C`

The answer.

Answer 3

`I=2(sinx-xcosx)+C`

Explanation:

Method of Integration by Parts:
`intu*vdx=uintvdx-int((du)/(dx)intvdx)dx`
Let,`u=2xandv=sinx`
So,`(du)/(dx)=2andintvdx=intsinxdx=-cosx`
`I=int2xsinxdx=2x(-cosx)-int(2)(-cosx)dx`
`I=-2xcosx+2intcosxdx`
`I=-2xcosx+2sinx+C`
`I=2(sinx-xcosx)+C`