Question

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

Answer 1

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

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

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

$=\int e^x(\cot 2x-2{\csc }^{2}2x)dx$

$=\int e^x\cot 2xdx-2\int e^x{\csc }^{2}2xdx$

Let $u=e^x\Rightarrow du=e^{x}dx$

$dv=-{\csc }^{2}2xdx\Rightarrow v=\frac{-\cot 2x}{2}$

$=\int e^x\cot 2xdx-2[\frac{-e^x\cot 2x}{2}+\frac{1}{2}\int e^x\cot 2xdx]$

$=\int e^x\cot 2xdx+2\frac{e^x\cot 2x}{2}-\frac{1}{2}\int e^x\cot 2xdx+c$

$=e^x\cot 2x+c$