Integrate $\int x e^{x} \sin x dx$ 
Evaluate:
$$\int x e^{x} \sin x dx$$

Have you ever come across such an integral? I have no idea how to start with the calculation.

This can also be solved another way (a little longer but correct nevertheless and could be more basic and readable for people who just started learning calculus):
We will use the results of following(easy - just substitute):
$$
∫ e^x \sin (x) dx = \frac{e^x \sin (x) - e^x \cos (x)}{2} + C
$$
$$
∫ e^{\ln (x)+x}dx = ∫ xe^xdx = xe^x-e^x + C
$$
and
$$
∫ e^x \cos (x) dx = \frac{e^x \sin (x) + e^x \cos (x)}{2} + C
$$
Now, back to the main integral. The main problem is that the formula contains not the usual two but three factors($x$, $e^x$ and $\sin (x)$). However, using basic logarithm and $\exp $ properties we can transform the \exp ression into one with just two factors:
$$
x e^x = e^{\ln (x)} e^x = e^{\ln (x) + x}
$$
this leaves us with
\begin{equation*}
I =
\int xe^xsinx dx =
\int e^{ln(x) + x} \sin (x) dx =
\end{equation*}
integrate by parts with substituting:
$$
\begin{equation*}
\left[
 \begin{alignedat}{2}
 u &= \sin (x) \quad & du &=\cos (x) \\
 dv &= e^{\ln (x)+x} \quad & v &= xe^x-e^x
 \end{alignedat}\,
\right] 
\end{equation*}
$$
$$
= e^xx\sin (x)-e^x\sin (x)-∫ xe^x\cos (x) -e^x\cos (x)dx 
$$
$$
= e^xx\sin (x)-e^x\sin (x)-∫ xe^x\cos (x)dx + ∫ e^x\cos (x)dx
$$
Now, another three factor integral appears, which can also be integrated by parts using the previous trick.
$$
∫ xe^x\cos (x)dx = 
$$
$$
\begin{equation*}
= ∫ e^{\ln (x)+x}\cos (x)dx = 
\left[
 \begin{alignedat}{2}
 u &= \cos (x) \quad & du &=-\sin (x) \\
 dv &= e^{\ln (x)+x} \quad & v &= xe^x-e^x
 \end{alignedat}\,
\right] =
\end{equation*}
$$
$$
= xe^x\cos (x)-e^x\cos (x)+ ∫ x e^x \sin (x) dx - ∫ e^x \sin (x) dx =
$$
$$
= xe^x\cos (x)-e^x\cos (x)+ I - ∫ e^x \sin (x) dx 
$$
Back to the previous equation, plug in the result above flipping the signs accordingly:
$$
I = e^xx\sin (x)-e^x\sin (x) + ∫ e^x\cos (x)dx - xe^x\cos (x)+e^x\cos (x)- I + ∫ e^x \sin (x) dx
$$
Great! We obtained the wanted integral with negative sign. Let's move it to the other side of equation:
$$
2I = e^xx\sin (x)-e^x\sin (x) + ∫ e^x\cos (x)dx - xe^x\cos (x)+e^x\cos (x) + ∫ e^x \sin (x) dx
$$
and plug in the results of simpler integrals:
$$
2I = e^xx\sin (x)-e^x\sin (x) + \frac{e^x \sin (x) + e^x \cos (x)}{2} - xe^x\cos (x)+e^x\cos (x) + \frac{e^x \sin (x) - e^x \cos (x)}{2} + C
$$
now just simplify the right side and divide by 2 to obtain the final result:
$$
I = \frac {xe^x\sin (x) - xe^x\cos (x) +e^x\cos (x)}{2} + C
$$
