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Question
Solution
\[\int\left( \frac{1 + \cos 4x}{\cot x - \tan x} \right) dx\]
\[ = \int\frac{\left( 1 + \cos 4x \right)}{\left( \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} \right)} dx\]
\[ = \int\frac{2 \cos^2 2x \times \sin x \cos x}{\left( \cos^2 x - \sin^2 x \right)}dx\]
\[ = \int\frac{\cos^2 2x \times 2 \sin x \cos x}{\cos 2x}dx\]
\[ = \int\cos 2x \sin 2xdx\]
\[ = \frac{1}{2}\int2 \sin 2x \cos 2xdx\]
\[ = \frac{1}{2}\int\sin 4xdx\]
\[ = \frac{1}{2}\left[ - \frac{\cos 4x}{4} \right] + C\]
\[ = - \frac{1}{8}\cos 4x + C\]
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