integration
$$\int e^{2x}\sin(3x)dx$$
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:
$$-\frac{e^{2x}}{3}\cos(3x) + \frac{2}{3}\left(\frac{e^{2x}}{3}\sin(3x) -\frac{2}{3}\int e^{2x}\sin(3x)dx\right)$$
And it just keeps going on and on and I'm stuck. How can I solve this?
For integration by parts, you will need to do it twice to get the same integral that you started with. When that happens, you substitute it for L, M, or some other letter. So we start by taking your original integral and begin the process as shown below.
Now using the formula
We get the following sequence, and we do another round of integration by parts.
So we end up the integral part that is the same as your original equation like so:
Now we substitute your original integral with the letter L like so...
...and replace every occurrence so the equation looks like the following:
Then we solve for L as shown below.
So your complete solution is this:
Hope this helps.
First, they used the double angle formula:
$$\sin u = \sin\left(2\frac{u}{2}\right) = 2\sin\frac{u}{2}\cos\frac{u}{2}.$$
Then they replaced the $1$ with $\sin^2\frac{u}{2}+\cos^2\frac{u}{2}.$
Best Answer
You almost have it! Put
$$I:=\int e^{2x}\sin 3x\,dx$$
and now take your last line:
$$I=-\frac{e^{2x}}{3}\cos3x + \frac{2}{3}\left(\frac{e^{2x}}{3}\sin3x -\frac{2}{3}\int e^{2x}\sin3x\,dx\right)=\frac{e^{2x}}{3}\cos3x + \frac{2}{9}e^{2x}\sin3x -\color{red}{\frac{4}{9}\int e^{2x}sin3x}\implies$$
$$\implies\frac{13}9I=\frac{e^{2x}}{3}\cos3x + \frac{2}{9}e^{2x}\sin3x$$
and now end the exercise.