First of all, sorry for my poor English. Can I integrate the following using integration by parts, because I've tried a lot and it didn't work. (And sorry if it's too easy for you, guys! I'm such a fool with numbers lol)
$$ \int e^{-2x}\sin(x) \, dx $$
Thank you!
Integration by parts (one of your "parts" in your comment below your answer is a little off):
$$u = \sin x,\quad du= \cos x\,dx\quad\quad dv=e^{-2x} dx, \quad v= -\dfrac{e^{-2x}}{2}$$
Set this up (perform integration by parts) and then, with the resultant right-hand side (RHS), you'll want to do integration by parts again.
You'll have an equation with left-hand side equal the original integral = $\left(I = \int e^{-2x}\sin x\,dx\right)$, and the RHS which includes in its expansion, "blah" and "blah" and the original integral $I.\;$ Then solve that equation for $I$.
Take a look at the "general case" of an integral such as yours: $$\int e^{ax}\cos(bx)\,dx$$ to see (in action) the general strategy of "taking integration by parts, twice, and solving for the original integral", and how this is carried out. It can easily be modeled for evaluating your integral.
Such a strategy is something to make note of, as its need reappears, and it is a tremendously useful when it can be applied (to circumvent what seems to amount to simply "running in circles".)
Hint: Think complex exponentials, they are often simpler to manipulate than the sine and cosine functions.
More precisely, note that $\mathrm e^{-2x}\sin x=\Im(\mathrm e^{ax})$ for some $a\ne0$ in $\mathbb C$ and that a primitive of $x\mapsto\mathrm e^{ax}$ is $x\mapsto a^{-1}\mathrm e^{ax}$. Thus, a primitive of $x\mapsto\mathrm e^{-2x}\sin x$ is $x\mapsto\Im(a^{-1}\mathrm e^{ax})$ (in words, the imaginary part of any primitive is a primitive of the imaginary part). Now, $a=$ $\ldots$ hence $a^{-1}=$ $\ldots$, which leads to...
The integration by parts approach has been very well described by amWhy. We describe another approach.
Guess that one antiderivative of $e^{-2x}\sin x$ has shape $$Ae^{-2x}\cos x+Be^{-2x}\sin x.$$ Differentiate. We get $$A\left(-e^{-2x}\sin x-2e^{-2x}\cos x\right)+B\left(e^{-2x}\cos x-2e^{-2x}\sin x \right).$$ This is equal to $$(-A-2B)e^{-2x}\sin x+(-2A+B)e^{-2x}\cos x.\tag{1}$$ In order to make (1) identically equal to $e^{-2x}\sin x$, we want $$-A-2B=1 \quad\text{and}\quad -2A+B=0.$$ Solve the above system of linear equations for $A$ and $B$. We get $A=-\frac{1}{5}$ and $B=-\frac{2}{5}$. Thus our integral is $$-\frac{1}{5}e^{-2x}\cos x -\frac{2}{5}e^{-2x}\sin x +C.$$
Remark: Variants of the above process are omnipresent in beginning differential equations. We can often use the idea to bypass integration by parts.
However, integration by parts has to be mastered. In addition to its uses in finding antiderivatives, it can be a very handy tool in producing estimates.
Complex functions of a real variable can be formally integrated using the rules for real functions. So, remember that $$\sin x=\frac{e^{ix}-e^{-ix}}{2i}$$ Therefore $$e^{-2x}\sin x=\frac{1}{2i}(e^{x(i-2)}-e^{-x(i+2)})$$ Hence $$\int e^{-2x}\sin x=\int\frac{1}{2i}(e^{x(i-2)}-e^{-x(i+2)})\;dx$$ $$\frac{1}{2i}\left(\dfrac {e^{x(i-2)}}{i-2}-\dfrac {e^{-x(i+2)})}{-(i+2)}\right)$$ $$\frac{e^{-2x}}{2i}\left(\dfrac {e^{ix}}{i-2}+\dfrac {e^{-ix}}{i+2}\right)$$ $$\frac{e^{-2x}}{2i}\left(\dfrac {\cos x+i\sin x}{i-2}+\dfrac {\cos x-i\sin x}{i+2}\right)$$ $$\frac{e^{-2x}}{2i}\left(\dfrac {(\cos x+i\sin x)(i+2)+(\cos x-i\sin x)(i-2)}{i^2-4}\right)$$ $$\dfrac {e^{-2x}(4i\sin x+2i\cos x)}{-10i}$$ $$-\dfrac {e^{-2x}(2\sin x+\cos x)}{5}+C$$
So generalized form :$$\int e^{ax}\sin bx=\dfrac {e^{ax}(a\sin bx-b\cos bx)}{a^2+b^2}+C$$
