Question

If `x^ { 2} \cos x d x = f ( x ) - \int 2x \sin x d x`, what is `f(x)`?

Answer 1

`f(x)=x^2cosx-2xcosx+2sinx+C`

Explanation:

The main thing here is to solve `int2xsinxdx`, which will have to be solved using Integration by Parts.

`u=2x`

`du=2dx`

`dv=sinxdx`

`v=-cosx`

`uv-intvdu=-2xcosx+2intcosxdx=-2xcosx+2sinx+C`

So,

`x^2cosx=f(x)-(-2xcosx+2sinx+C)=f(x)+2xcosx-2sinx+C`

Solving for `f(x)` yields

`f(x)=x^2cosx-2xcosx+2sinx+C`

We're still adding on the arbitrary constant of integration `C.` It's just a constant, we don't know if it's positive or negative, so we'll always add it on regardless of what we do with it.