Question

$\int \frac{e^{2x}\left(\sin 4x-2\right)}{1-\cos 4x}dx$

Answer 1

$I=\int \frac{{\theta }^2[\sin 4x-2]}{[1-\sin 4x]}dx$

$2x=t\Rightarrow dt=dx/2$

$\Rightarrow I=\frac{1}{2}\int \frac{e^t[\sin 2t-2]}{[1-\cos 2t]}dt$

by $\sin 2t=2\sin t\cos t1-\cos 2t=2{\sin }^{2}2t$

$\Rightarrow I=\frac{1}{2}\int \frac{e^t[2\sin t\cos t-2]}{2{\sin }^{2}tdt}$

$=\frac{1}{2}\int \frac{e^t[\sin t\cos t-1]}{{\sin }^{2}t}dt$

$=\frac{1}{2}\left[e^t\cot t-\csc t\right]dt$

$\int e^x\left[f\right(x)+f(x\left)\right]=e^{x}f\left(x\right)+c$

$I=\frac{1}{2}{e}^t\cot t+c$

$I=\frac{1}{2}{e}^{2x}\cot 2x+c$