I am suppose to use integration by parts but I have no idea what to do for this problem
$$\int e^{2x} \sin3x dx$$
$u = \sin3x dx$ $du = 3\cos3x$
$dv = e^{2x} $ $ v = \frac{ e^{2x}}{2}$
From this I get something really weird that makes it just as complicated
$\frac{e^{2x}\sin3x}{2} - \int \frac{e^{2x}}{2}3\cos2x$
This looks like it will again require integration by parts which from what I saw will require the same again, and it does not help solve the problem.
Another problem I am having is that I do not know what the dx in $u = \sin3x dx$ means. I know it is suppose to be the shorthand representation for the derivative with repsect to x I think but I am not sure when and why it goes away, basically I have just memorized that it dissapears and it not important in the answer so I can ignore it for the most part. It turns into a 1 pretty much.
Answer
You're correct. The integral does indeed require integration by parts. But, it's a little trick. You have to use the method twice, each time using what you consider the differentiated term the trig one or exp it doesn't matter as long as you're consistent. Here's the sketch of the idea. I'll do it in the general case.
$$\int e^{ax}\sin(bx)dx=\frac{1}{a}e^{ax}\sin(bx)-\frac{1}{a}\int be^{ax}\cos(bx)dx$$ Now, we do it again.
$$\frac{b}{a} \int e^{ax}\cos(bx)dx=\frac{b}{a}\left(\frac{b}{a^2}e^{ax}\cos(x)-\frac{b^2}{a^2}\int e^{ax}[-\sin(bx)]\right)dx= \dots$$
Now, you take it from here, noticing that that last integral is your original one (with a negative). Set $\displaystyle I=\int e^{ax}\sin(bx)dx$, and solve for $I$ after substituting the above expression into the original one.
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