How to integrate $\int_{0}^{\pi /2} e^{−2x}\sin(3x)\rm dx $? 
How to integrate $$\int_{0}^{\pi/2} e^{−2x}\sin(3x)\rm dx $$

I have attempted to this question with integration by parts, but I'm hitting a lot of walls. I have a feeling it might have to do with representing repeating integrals with I, but I'm not quite sure as to how to approach this question. Sorry for the terrible formatting, it's my first time on this site.
 A: Letting $I$ be your definite integral, 
$$\begin{align}I=\int_{0}^{\pi/2}e^{-2x}\sin(3x)dx&=\int_{0}^{\pi/2}\left(-\frac 12e^{-2x}\right)^\prime\sin (3x)dx\\&=(-1/2)e^{-2x} \sin(3x) - \int_{0}^{\pi/2}(-1/2)e^{-2x} \cdot 3\cdot \cos(3x) dx\\&=(-1/2)e^{-2x} \sin(3x) -(-3/2)\int_{0}^{\pi/2} e^{-2x} \cos(3x) dx.\end{align}$$
Also, since
$$\begin{align}\int_{0}^{\pi/2} e^{-2x} \cos(3x) dx　&= (-1/2)e^{-2x} \cos(3x) - \int_{0}^{\pi/2} (-1/2)e^{-2x} (-3) \sin(3x) dx\\&=(-1/2)e^{-2x} \cos(3x) -(3/2)I,\end{align}$$
you'll get
$$I= (-1/2)e^{-2x} \sin(3x) - (-3/2)\{ (-1/2)e^{-2x} \cos(3x) - (3/2)I \}.$$
Then, you can solve this equation of $I$.
A: Hint: Use the fact that $\sin x=\Im(e^{ix})$, or $\sin x=\dfrac{e^{ix}-e^{-ix}}2$ .
A: First you want to find a function $y(x)$ such that $y'(x)=e^{-2x}\sin3x$. For this you can use the method of undetermined coefficients. To start with, you guess (or know) that there is a solution of the form $y=Ae^{-2x}\cos3x+Be^{-2x}\sin3x$ where $A$ and $B$ are constants to be determined. Differentiate and equate coefficients:
$$y=e^{-2x}(A\cos3x+B\sin3x);$$
$$y'=-2e^{-2x}(A\cos3x+B\sin3x)+e^{-2x}(-3A\sin3x+3B\cos3x)=e^{-2x}(-2A+3B)\cos3x+e^{-2x}(-3A-2B)\sin3x=e^{-2x}\sin3x;$$
$$-2A+3B=0,\ -3A-2B=1.$$
The solution of the linear equations is $A=\frac{-3}{13}$ and $B=\frac{-2}{13}$ (if I did all that arithmetic right, which is not very likely), so your antiderivative is
$$y(x)=\frac1{13}e^{-2x}(-3\cos3x-2\sin3x)$$
and the value of your definite integral is
$$\int_0^\frac\pi2e^{-2x}\sin3xdx=y(\frac\pi2)-y(0)=\frac{2e^{-\pi}+3}{13}$$
or something like that. Alternatively, you could just look up the formula for $\int e^{ax}\sin bxdx$.
A: You have to be careful using integration by parts:
$$\int e^{-2x} \sin (3x) dx= -\frac{1}{2} e^{-2x} \sin (3x) +\frac{3}{2} \int e^{-2x} \cos (3x) dx = -\frac{1}{2} e^{-2x} \sin (3x)
-\frac{3}{4} e^{-2x} \cos (3x)-\frac{9}{4} \int e^{-2x} \sin (3x) dx$$
note that in the latter part you recovered the initial integral (but with a different coefficient). Thus 
$$\int e^{-2x} \sin (3x) dx= -\frac{2}{13} e^{-2x} \sin (3x)-\frac{3}{13} e^{-2x} \cos (3x)$$ and 
$$\int_{0}^{\frac{\pi}{2}} e^{-2x} \sin (3x) dx= -\frac{2}{13}(e^{-\pi}\sin(\frac{3\pi}{2})-e^{0} \sin(0)) -\frac{3}{13}(e^{-\pi}\cos(\frac{3\pi}{2})-e^{0} cos(0))=\frac{2}{13}e^{-\pi}+\frac{3}{13}$$
