Help evaluating another integral

[afcwestwarrior]

Homework Statement

∫e^2θ sin 3θ dθ

Homework Equations

∫u dv= uv- ∫v du

The Attempt at a Solution

u= e^2θ dv=sin 3θ dθ
du= 2* e^2θ v=(1/3) * - cos 3 θ

e^2θ 1/3) * - cos 3 θ - ∫1/3) * - cos 3θ 2* e^2θ dθ

what now

 

[nicksauce]

Try another IBP (I think that should work)

 

[afcwestwarrior]

∫1/3) * - cos 3θ 2* e^2θ dθ
u= cos 3θ dv= e^2θ
du= 3*sin 3θ v= e^2θ
cos 3θ e^2θ - ∫ e^2θ 3*sin 3θ
e^2θ 1/3) *(2/3) - cos 3 θ +cos 3θ e^2θ - ∫ e^2θ 3*sin 3θ

 

[statdad]

Integrate by parts once more. You will obtain another integral that is a multiple of the original one. Combine common terms (all the integrals on the left) and solve.

 

[afcwestwarrior]

is my v correct for the second igb

 

[afcwestwarrior]

ibp i mean

 

[Defennder]

I'm having some trouble reading your working. I noticed a single "e^2θ 1/3)" term. You should not have any e^(2θ) without an accompanying cos or sin term. Check your working.