Evaluate the integral of (e^2x)*sin^3 x dx

Question

I let u = e^2x, du = (1/2)e^2x dx
v= (-1/3)cos^3 x , dv =sin^3 x dx

When I used integration by parts and solved it all out I got:

(37/36)intgral of (e^2x)*sin^3 x dx =
(-1/3)(e^2x)*cos^3 x + (1/18)(e^2x)*sin^3 x

However, the answer provided in the book is (1/13)e^2x (2sin^3 x - 3cos^3 x)

So, I don't what know what I did wrong and got a weird answer?

Answer 1

Let
u = sin^3(x)
du = 3 sin^2(x) cos(x) dx
dv = e^(2x)
v = 1/2 e^(2x)

∫u dv = uv - ∫v du
= 1/2 e^(2x) sin^3(x) - 3/2 ∫e^(2x) sin^2(x) cosx dx
= 1/2 e^(2x) sin^3(x) - 3/2 ∫e^(2x)cosx dx - 3/2 ∫e^(2x) cos^3(x) dx

Pull the same trick again, and you will have 3/4 of the original integral on the right hand side, and you can eliminate it.

check all my fractions, and I assume you can integrate e^(2x) cosx dx using integration by parts.