Csdrhrt

$$int e^{2x}sin(3x)dx$$

First I set $u = e^{2x}, v' = cos(3x)$
to get:

$$int e^{2x}sin(3x)dx = -frac{e^{3x}}{3}cos(3x) + frac{2}{3}int e^{2x}cos(3x)dx$$

Applying integration by parts again yields:

$$-frac{e^{2x}}{3}cos(3x) + frac{2}{3}left(frac{e^{2x}}{3}sin(3x) -frac{2}{3}int e^{2x}sin(3x)dxright)$$

And it just keeps going on and on and I'm stuck. How can I solve this?

You are almost done. Lets denote your given integral as

$$I=int e^{2x}sin(3x)dx$$

Lets take a closer look at your final formula

$$begin{align}
I&=-frac{e^{2x}}3cos(3x)+frac23left(frac{e^{2x}}3sin(3x)-frac23underbrace{int e^{2x}sin(3x)dx}_{=I}right)\
I&=-frac{e^{2x}}3cos(3x)+frac23left(frac{e^{2x}}3sin(3x)-frac23Iright)\
I&=-frac{e^{2x}}3cos(3x)+frac29e^{2x}sin(3x)-frac49I\
Leftrightarrow I+frac49I&=e^{2x}left(frac29sin(3x)-frac13cos(3x)right)\
I&=frac9{13}e^{2x}left(frac29sin(3x)-frac13cos(3x)right)\
I&=frac{e^{2x}}{13}left(2sin(3x)-3cos(3x)right)+c
end{align}$$

Note that this can be done in a more general way for in order to integrate the function $e^{ax}sin(bx)$ which turns out to equal

$$int e^{ax}sin(bx)dx=frac{e^{ax}}{a^2+b^2}(asin(bx)-bcos(bx))+c$$

You almost have it! Put

$$I:=int e^{2x}sin 3x,dx$$

and now take your last line:

$$I=-frac{e^{2x}}{3}cos3x + frac{2}{3}left(frac{e^{2x}}{3}sin3x -frac{2}{3}int e^{2x}sin3x,dxright)=frac{e^{2x}}{3}cos3x + frac{2}{9}e^{2x}sin3x -color{red}{frac{4}{9}int e^{2x}sin3x}implies$$

$$impliesfrac{13}9I=frac{e^{2x}}{3}cos3x + frac{2}{9}e^{2x}sin3x$$

and now end the exercise.

$displaystyle I=int(e^{2x}sin3x)dx=dfrac{e^{2x}sin 3x}{2}-dfrac{3e^{2x}cos3x}{4}+intdfrac{9e^{2x}(-sin3x)dx}{4}$

$=dfrac{e^{2x}sin 3x}{2}-dfrac{3e^{2x}cos3x}{4}-dfrac{9I}{4}$

$I=dfrac{4}{13}left(dfrac{e^{2x}sin 3x}{2}-dfrac{3e^{2x}cos3x}{4}right)+C$

(Note: The result of indefinite integral is not unique.)

$$I=int e^{ax}sin bx mathrm{d}x$$
$$mathrm{d}v=sin bx mathrm{d}xRightarrow v=-frac1bcos bx\u=e^{ax}Rightarrow mathrm{d}u=ae^{ax}mathrm{d}x$$
$$I=-frac{e^{ax}}bcos bx+frac abint e^{ax}cos bx mathrm{d}x$$
$$mathrm{d}v=cos bx mathrm{d}xRightarrow v=frac1bsin bx\u=e^{ax}Rightarrow mathrm{d}u=ae^{ax}mathrm{d}x$$
$$I=-frac{e^{ax}}bcos bx+frac abbigg(frac{e^{ax}}{b}sin bx-frac abint e^{ax}sin bx mathrm{d}xbigg)$$
$$I=-frac{e^{ax}}bcos bx+frac{ae^{ax}}{b^2}sin bx-frac{a^2}{b^2}I$$
$$bigg(1+frac{a^2}{b^2}bigg)I=frac{e^{ax}}bbigg(frac absin bx-cos bxbigg)$$
$$I=frac{b^2}{a^2+b^2}frac{e^{ax}}bbigg(frac absin bx-cos bxbigg)$$
$$I=frac{e^{ax}}{a^2+b^2}big(asin bx-bcos bxbig)+C$$

how about this:
$$I=int e^{2x}sin(3x)dx=frac{-e^{2x}cos(3x)}{3}+intfrac{2cos(3x)e^{2x}}{3}dx$$
$$=frac{-e^{2x}cos(3x)}{3}+frac{2sin(3x)e^{2x}}{9}-intfrac{4sin(3x)e^{2x}}{9}dx$$
$$I=frac{-e^{2x}cos(3x)}{3}+frac{2sin(3x)e^{2x}}{9}-frac{4}{9}I$$
so:
$$frac{13}{9}I=frac{-e^{2x}cos(3x)}{3}+frac{2sin(3x)e^{2x}}{9}$$