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. 
 A: When you do it with integration by parts, you have to go in the "same direction" both times. For instance, if you initially differentiate $e^{2 \theta}$, then you need to differentiate $e^{2 \theta}$ again; if you integrate it, you will wind up back where you started. If you do this, you should find something of the form
$$\int e^{2 \theta} \cos(3 \theta) d \theta = f(\theta) + C \int e^{2 \theta} \cos(3 \theta) d \theta$$
where $C$ is not $1$. Therefore you can solve the equation for the desired quantity:
$$\int e^{2 \theta} \cos(3 \theta) d \theta = \frac{f(\theta)}{1-C}.$$
There is also a nice approach with complex numbers: $\cos(3 \theta)=\frac{e^{3 i \theta}+e^{-3 i \theta}}{2}$, so your integral is
$$\frac{1}{2} \int e^{(2+3i) \theta} + e^{(2-3i) \theta} d \theta$$
which are pretty easy integrals. You do some complex number arithmetic and it works out.
A: 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 
A: Here is a plain answer for your reference:
Let $\simeq$ denote the equality sign up to a constant. 
We have
$$
\int e^{2x}\sin 3x dx = \int \sin 3x de^{2x}\frac{1}{2} \simeq \frac{1}{2}[ e^{2x}\sin 3x - 3 \int e^{2x} \cos 3x dx ]\\ = \frac{1}{2}e^{2x}\sin 3x - \frac{3}{2} \int e^{2x}\cos 3x dx;
$$
we have
$$
\int e^{2x}\cos 3x dx = \frac{1}{2}\int \cos 3x de^{2x} \simeq \frac{1}{2}e^{2x}\cos 3x + \frac{3}{2}\int e^{2x}\sin 3x dx;
$$
hence
$$
\frac{13}{4}\int e^{2x}\sin 3x dx \simeq \frac{1}{2}e^{2x}\sin 3x - \frac{3}{4}e^{2x}\cos 3x.
$$
