Verified answer

I will try to explain it to you. But I can only write in English.

You use the integration by parts technique.

u = sin(x)

dv = e^(2x) dx

du = cos(x)

v = e^(2x)/2

∫ e^(2x) sin(x) dx =

sin(x) e^(2x)/2 − ∫ e^(2x) cos(x) / 2 dx

But then you do it again. Use integration by parts on the second integral

sin(x) e^(2x)/2 − ∫ e^(2x) cos(x) / 2 dx

u = cos(x)

dv = e^(2x)/2 dx

du = −sin(x) dx

v = e^(2x) / 4

sin(x) e^(2x)/2 − [ cos(x) e^(2x)/4 − ∫ −e^(2x) sin(x) / 4 dx ]

I think I did that right. I just woke up, I might have made an error somewhere...

Simplify

sin(x) e^(2x)/2 − cos(x) e^(2x)/4 − ∫ e^(2x) sin(x) / 4 dx

This is what youre left with. Which is equal to what we started with...

∫ e^(2x) sin(x) dx = sin(x) e^(2x)/2 − cos(x) e^(2x)/4 − ∫ e^(2x) sin(x) / 4 dx

The integral on the right is the same as the integral on the left. They are like terms.

(5/4) ∫ e^(2x) sin(x) dx = sin(x) e^(2x)/2 − cos(x) e^(2x)/4

∫ e^(2x) sin(x) dx = 2sin(x) e^(2x)/5 − cos(x) e^(2x)/5