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$\int \frac{1+\cos 4 x}{\cot x-\tan x} d x$ is equal to
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$-\frac{1}{8} \cos 4 x+C$
$\begin{aligned} & \text { Let } I=\int \frac{1+\cos 4 x}{\cot x-\tan x} d x \\ & =\int \frac{\frac{2 \cos ^2 2 x}{\cos ^2 x-\sin ^2 x}}{\sin x \cos x} d x \\ & =\int \frac{\sin 2 x \cos ^2 2 x}{\cos 2 x} d x \\ & =\frac{1}{2} \int \sin 4 x d x=-\frac{\cos 4 x}{8}+C \\ & \end{aligned}$
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