Evaluate $\int_0^5 e^{-2t}\sin(t)\, \mathrm{d}t$. 
$$\int_0^5 e^{-2t}\sin(t) \,\mathrm{d}t$$

I know I should be able to integrate this by parts but I can't seem to get the parts I choose to make the result any easier to integrate. 
 A: In fact, you're not meant to find something easier to integrate.
After integrating by parts twice (keep integrating $e^{-2t}$ and deriving the trigonometric part), you'll end up with something like $$\int_0^5 e^{-2t}\sin t\,dt=\text{something }+k\int_0^5e^{-2t}\sin t\,dt$$ with $k\ne 1$.
At that point, you just need to do this $$(1-k)\int_0^5 e^{-2t}\sin t\,dt=\text{something }\\\int_0^5 e^{-2t}\sin t\,dt=\frac{\text{something}}{1-k}$$
A: This integral is the imaginary part of 
\begin{align*}\int_0^5\mathrm e^{(-2+\mathrm i)t}\,\mathrm d \mkern1mu t&=\frac1{-2+\mathrm i}\mathrm e^{(-2+\mathrm i)t}\biggr\rvert_0^5=-\frac{2+ \mathrm i}{5}\bigl(\mathrm e^{-10+5\mathrm i}-1)\\
&=-\frac15\bigl(2(\mathrm e^{-10}\cos 5-1)-\mathrm e^{-10}\sin 5+\mathrm i(\mathrm e^{-10}\cos 5-1+2\mathrm e^{-10}\sin 5)\bigr),
\end{align*}
so the answer is $$\frac{1-\mathrm e^{-10}(\cos 5+2\sin 5)}5.$$
A: By indeterminate coefficients:
By educated guess, you try a solution of the form
$$e^{-2t}(a\cos(t)+b\sin(t)).$$
Deriving, you get
$$e^{-2t}(-2a\cos(t)-2b\sin(t)-a\sin(t)+b\cos(t)).$$
The unknown coefficients are obtained by solving
$$-2a+b=0,\\-2b-a=1,$$ giving
$$a=-\frac15,b=-\frac25.$$
The definite integral easily follows.
A: Hint. Integrating by parts twice gives
$$
\begin{align}
\int_0^5e^{-2t}\sin t\:dt&=\left. -\frac12e^{-2t}\sin t\right|_0^5+\frac{1}{2}\int_0^5e^{-2t}\cos t\:dt\
\end{align}
$$ and
$$
\begin{align}
\int_0^5e^{-2t}\cos t\:dt&=\left. -\frac12e^{-2t}\cos t\right|_0^5-\frac{1}{2}\int_0^5e^{-2t}\sin t\:dt\
\end{align}
$$
Can you take it from here?
A: Since we can do integrals similarly to $\mathbb{R}\to\mathbb{C}$ as well:
$$
\sin x=\frac{e^{xi}-e^{-xi}}{2i}\qquad e^{a+bi}=e^a\left(\cos b+i\sin b\right)
$$
So with a more direct approach:
$$
\int_0^5e^{-2t}\sin t\mathrm{d}t=\frac{1}{2i}\int_0^5e^{(-2+i)t}-e^{(-2-i)t}\mathrm{d}t=\frac{1}{2i}\left.\left(\frac{e^{(-2+i)t}}{-2+i}-\frac{e^{(-2-i)t}}{2-i}\right)\right|_0^5=
$$
$$
=\frac{(-2-i)e^{(-2+i)t}-(-2+i)e^{(-2-i)t}}{10i}=\frac{1-\mathrm e^{-10}(\cos 5+2\sin 5)}5.
$$
