Integration By Parts

Question

Integrate e^4xcos2x dx

Answer 1

You have to integrate by parts twice.

u = cos(x)

du = -sin(x)dx

v = (1/4)e^4x

dv = e^4x dx

∫ u dv = uv - ∫v du

You will get some terms and then another ∫e^4x sin(x) dx type of term.

u = sin(x)

du = cos(x)

v = 1/4 e^4x

dv = e^4x dx

Do the same thing, and you will end up with another ∫e^4x cos(x) dx term. Pretend that integral is its own term and isolate it on one side.