Evaluate the Integral: $\int e^{2\theta}\ \sin 3\theta\ d\theta$ $\int e^{2\theta}\ \sin 3\theta\ d\theta$
After Integrating by parts a second time, It seems that the problem will repeat for ever. Am I doing something wrong. I would love for someone to show me using the method I am using in a clean and clear fashion. Thanks. 

You got $\displaystyle\int e^{2\theta}\sin(3\theta)d\theta=...=\sin(3\theta)\frac12e^{2\theta}-\frac32\Bigl(\frac12e^{2\theta}\cos(3\theta)+\frac32\int e^{2\theta}\sin(3\theta)d\theta\Bigr)$.  
Now you could denote the integral you are looking for by $x$:  
$\displaystyle x=\int e^{2\theta}\sin(3\theta)d\theta$, then copy your equation as  
$\displaystyle x=\sin(3\theta)\frac12e^{2\theta}-\frac32\Bigl(\frac12e^{2\theta}\cos(3\theta)+\frac32x\Bigr)$.  
Then solve for $x$. 
$\displaystyle x=\sin(3\theta)\frac12e^{2\theta}-\frac34e^{2\theta}\cos(3\theta)-\frac94x$.  
$\displaystyle \frac{13}4x=\sin(3\theta)\frac12e^{2\theta}-\frac34e^{2\theta}\cos(3\theta)$.  
$\displaystyle \int e^{2\theta}\sin(3\theta)d\theta=x=\frac4{13}\Bigl(\sin(3\theta)\frac12e^{2\theta}-\frac34e^{2\theta}\cos(3\theta)\Bigr)= 
\frac2{13} \sin(3\theta) e^{2\theta}-\frac3{13}e^{2\theta}\cos(3\theta)+C$.  
This is what you claimed was the answer in another comment, and which is confirmed too by wolframalpha 
