Question

How do you find the integral of `(2-3x) cosx dx`?

Answer 1

`(2-3x)sinx-3cosx+C`

Explanation:

We have:

`int(2-3x)cosxdx`

The integration by parts states that:

`intudv=uv-intvdu`

We let:

`u=2-3x`

`dv=cosx`

`=>du=d/dx(2-3x)`

`=>du=-3`

`=>v=intcosxdx`

`=>v=sinx`

We now have:

`(2-3x)*sinx-intsinx*-3dx`

`=>(2-3x)*sinx+3intsinxdx`

Remember that:

`intsinxdx=-cosx`

`=>(2-3x)* sinx+3* -cosx`

`=>(2-3x)* sinx-3cosx`

Do you `C` why this is incomplete?

`=>(2-3x)sinx-3cosx+C`

That is the answer!Just note that my way of solving is the simplest, but not the only solution.