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.
 A: Solution without complex numbers:
Let $I=\int e^x x\sin xdx$ Integrating by parts:
$$I=e^x x\sin x-\int e^x x\cos xdx-\int e^x\sin xdx$$
Then one more time by parts:
$$\int e^x x\cos xdx=e^x x\cos x + I-\int e^x\cos xdx$$
So:
$$2I=e^x x\sin x-e^x x\cos x+\int e^x(\cos x-\sin x)dx$$
Now (by parts again or by direct observation):
$$\int e^x(\cos x-\sin x)dx=e^x \cos x$$
So:
$$I=\frac{e^x x\sin x-e^x x\cos x+e^x \cos x}{2}$$
A: HINTS: 
Use Euler's Formula to write $$x\sin(x)e^x=\text{Im}(xe^{(1+i)x})$$Integrate $\int xe^{(1+i)x}\,dx$ by parts with $u=x$ and $v=\frac{e^{(1+i)x}}{1+i}$ and finish by taking the imaginary part.
A: This solution doesn't use integration by parts. We start with
$$\int\exp(x) dx = \exp(x)$$
Substituting  $x = \lambda t$ yields:
$$\int\exp(\lambda t) dt = \frac{\exp(\lambda t)}{\lambda}$$
Substitute $\lambda = 1+\epsilon + i$ and expand both sides to first order in $\epsilon$. Equating the coefficient of $\epsilon$ of both sides yields:
$$\begin{split}\int t\exp(t)\exp(i t) dt &= \exp(t)\exp(it)\left[\frac{1-i}{2}t + \frac{i}{2}\right]\\
& = \exp(t)\left[\frac{\exp(i(t-\frac{\pi}{4}))}{\sqrt{2}}t + \frac{\exp(i(t+\frac{\pi}{2}))}{2}\right]
\end{split}
$$
Finally, take the imaginary part of both sides:
$$\int t\exp(t)\sin(t) dt = \exp(t)\left[\frac{\sin\left(t-\frac{\pi}{4}\right)}{\sqrt{2}}t + \frac{\cos(t)}{2}\right]$$
A: 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:
$$
I = \frac {xe^x\sin (x) - xe^x\cos (x) +e^x\cos (x)}{2} + C
$$
A: If one may recall that $\sin(x)=\Im e^{ix}$, then
$$\int x\sin(x)e^x\ dx=\Im\int xe^{(1+i)x}\ dx$$
With a quick integration by parts, we have
$$=\Im\left(\frac1{1+i}xe^{(1+i)x}-\frac1{1+i}\int e^{(1+i)x}\ dx\right)\\=\Im\left(\frac1{1+i}xe^{(1+i)x}-\frac1{(1+i)^2}e^{(1+i)x}+C\right)\\=\frac12\left(x\sin(x)e^x-x\cos(x)e^x+\cos(x)e^x\right)$$
A: By indeterminate coefficients:
A term like $xe^x\sin x$ can be generated by the derivative of itself (due to $e^x$), which will also generate $xe^x\cos x$ and $e^x\sin x$.
Then we are tempted to try
$$f(x)=e^x(x(A\sin x+B\cos x)+(C\sin x+D\cos x)),$$
and
$$f'(x)=e^x(x(A\sin x+B\cos x)+(C\sin x+D\cos x)+(A\sin x+B\cos x)+x(A\cos x-B\sin x)+(C\cos x-D\sin x)).$$
We identify,
$$A-B=1,\\A+B=0,\\C+A-D=0,\\D+B+C=0$$
and obtain
$$\frac12xe^x(\sin x-\cos x)+\frac12\cos x.$$
A: First integrate $\sin(x) \exp(x),$ by parts, with the exponential as $d v.$ (you will need to do the integration by parts twice), then use the same method for the original integral.
