Evaluate : $\int e^{2x}\sin\: x\: dx$

Toolbox:
• If there are two functions u and v, such that the integral function is of the form $\int vdv,$ then we can solve it by the method of integration by parets which states $\int udv=uv-\int vdu.$
• $\int \sin x dx =-\cos x +c$
• $\int \cos x dx =\sin x +c$
Step 1:
$I=\int e^{2x}\sin x dx$
Clearly the integral function is of the form $\int udv=uv-\int vdu$
Let us take $a=e^{2x}$
on differentiating w.r.t.x $du=2e^{2x}$
Let $dv=\sin x dx$
on integrating,$v=-\cos x$
Now substituting for u,v,du and dv we get,
$\int e^{2x}sin x dx=e^{2x}(-\cos x)-2\int (-\cos x)e^{2x}dx$
$\qquad\;\;\;\;\;\;\;=(-e^{2x}-\cos x+2\int e^{2x} (-\cos x)dx$ --------(1)
Step 2:
Again this is of the form $I_1=\int udv.\;where\; I_1= \int e^{2x}\cos x dx$
Let $e^{2x}=u$
on differentiating w.r.t.x $2e^{2x}=du$
Let $dv=\cos x dx$
On integrating ,$v=\sin x$
Now sustituting for u,v,du and dv we get,
$I_1=2[(e^{2x}\sin x)-\int \sin x.(2e^{2x})]dx$
$\qquad=2[e^{2x}\sin x-2\int e^{2x}\sin x.dx]$
But $\int e^2x\sin x dx=I$
Hence $I_1=e^{2x}\sin x -2I$
Step 3:
Now substituting this value for $I_1$ in equ(1)
$I=-e^{2x}\cos x +2 e^{2x}\sin x -4I.$
=>$I=2e^{2x}\sin x-e^{2x}\cos x$
Taking $e^{2x}$ as the common factor,