How do u integrate (e^2x) *(sin x)dx using integral by parts?

Question

How do u integrate (e^2x) *(sin x)dx using integral by parts?

Answer 1

The idea is to integrate by parts twice.  Start with the original integral,

∫sin(x)e^(2x) dx =

Let u = e^(2x) and dv = sin(x).  Then du = 2e^(2x) and v = -cos(x).  Apply integration by parts to get:

∫u dv = uv - ∫v du

∫e^(2x)sin(x) dx = -e^(2x)cos(x) - ∫-cos(x)2e^(2x) dx

∫e^(2x)sin(x) dx = -cos(x)e^(2x) + 2∫cos(x)e^(2x) dx

Call this last equation *1*.  Now this seems to have made the integral we couldn't compute, ∫e^(2x)sin(x) dx, into yet another integral we can't compute, namely ∫cos(x)e^(2x) dx.  So, apply integration by parts again - this time on just that rightmost integral.

For ∫cos(x)e^(2x) dx, let u = e^(2x) and dv = cos(x).  Then du = 2e^(2x) and v = sin(x).  So, from integration by parts you get:

∫u dv = uv - ∫v du

∫e^(2x)cos(x) dx = sin(x)e^(2x) - ∫sin(x)2e^(2x) dx

∫e^(2x)cos(x) dx = sin(x)e^(2x) - 2∫sin(x)e^(2x) dx

Now plug this substitution of the integral ∫e^(2x)cos(x) dx into equation *1* and see if it becomes cleaner.

∫e^(2x)sin(x) dx = -cos(x)e^(2x) + 2∫cos(x)e^(2x) dx

∫e^(2x)sin(x) dx = -cos(x)e^(2x) + 2[sin(x)e^(2x) - 2∫sin(x)e^(2x) dx]

∫e^(2x)sin(x) dx = -cos(x)e^(2x) + 2sin(x)e^(2x) - 4∫sin(x)e^(2x) dx

∫e^(2x)sin(x) dx + 4∫sin(x)e^(2x) dx = -cos(x)e^(2x) + 2sin(x)e^(2x)

5∫e^(2x)sin(x) dx = -cos(x)e^(2x) + 2sin(x)e^(2x)

5∫e^(2x)sin(x) dx = e^(2x)(2sin(x) - cos(x))

∫e^(2x)sin(x) dx = e^(2x)(2sin(x) - cos(x))/5

Hope this helps you!