If $\int \frac{1+\cos 8 x}{\tan 2 x-\cot 2 x} d x=a \cos 8 x+C$, then $a = $

$\frac{1}{16}$

We have,

$I=\int \frac{1+\cos 8 x}{\tan 2 x-\cot 2 x} d x$

$\Rightarrow I=\int \frac{2 \cos ^2 4 x}{-2 \cot 4 x} d x$          $[∵ \cot \theta-\tan \theta=2 \cot 2 \theta]$

$\Rightarrow I=-\int \cos 4 x \sin 4 x d x=-\frac{1}{2} \int \sin 8 x d x=\frac{1}{16} \cos 8 x+C$

∴  $a \cos 8 x+C=\frac{1}{16} \cos 8 x+C \Rightarrow a=\frac{1}{16}$