If the solve the problem
Question:

$\int \frac{\sin 4 x-2}{1-\cos 4 x} e^{2 x} d x$

Solution:

Put $2 x=t$;

$2 d x=d t ; d x=d t / 2$

$=-\int \frac{\sin 4 x-2}{\cos 4 x-1} d x=-\frac{1}{2} \int \frac{e^{t}(\sin 2 t-2)}{\cos 2 t-1} d t=\frac{1}{4} \int \frac{e^{t}(2 \sin t \cos t-2)}{\cos ^{2} t} d t$

$=\frac{2}{4} \int e^{t} \cot t d t-\frac{2}{4} \int e^{t} \operatorname{cosec}^{2} t d t=\frac{1}{2}\left[\int e^{t} \operatorname{cottdt}-\int e^{t} \operatorname{cosec}^{2} t d t\right]$

$=\frac{1}{2}\left[e^{t} \cot t+\int e^{t} \operatorname{cosec}^{2} t d t-\int e^{t} \operatorname{cosec}^{2} t d t\right]$

$=\frac{1}{2}\left[\frac{e^{2 x} \cot 2 x}{2}\right]+c$

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