How do you find the integral of (2-3x) cosx dx(23x)cosxdx?

1 Answer
Apr 17, 2018

(2-3x)sinx-3cosx+C(23x)sinx3cosx+C

Explanation:

We have:

int(2-3x)cosxdx(23x)cosxdx

The integration by parts states that:

intudv=uv-intvduudv=uvvdu

We let:

u=2-3xu=23x

dv=cosxdv=cosx

=>du=d/dx(2-3x)du=ddx(23x)

=>du=-3du=3

=>v=intcosxdxv=cosxdx

=>v=sinxv=sinx

We now have:

(2-3x)*sinx-intsinx*-3dx(23x)sinxsinx3dx

=>(2-3x)*sinx+3intsinxdx(23x)sinx+3sinxdx

Remember that:

intsinxdx=-cosxsinxdx=cosx

=>(2-3x)* sinx+3* -cosx(23x)sinx+3cosx

=>(2-3x)* sinx-3cosx(23x)sinx3cosx

Do you CC why this is incomplete?

=>(2-3x)sinx-3cosx+C(23x)sinx3cosx+C

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