If \(\int \frac{\cos 8 x+1}{\cot 2 x-\tan 2 x} \mathrm{~d} x=\mathrm{A} \cos 8 x+\mathrm{c}\), where \(\mathrm{c}\) is an arbitrary constant, then the value of \(\mathrm{A}\) is

\(\begin{aligned} I & =\int \frac{\cos 8 x+1}{\cot 2 x-\tan 2 x} d x \\ & =\int \frac{2 \cos ^2\left(\frac{8 x}{2}\right)}{\frac{\cos 2 x}{\sin 2 x}-\frac{\sin 2 x}{\cos 2 x}} d x\end{aligned}\)
\(\ldots\left[\because 1+\cos \theta=2 \cos ^2 \frac{\theta}{2}\right]\)
\(\begin{aligned}
& =\int \frac{2 \cos ^2(4 x) \times \sin 2 x \times \cos 2 x}{\cos ^2 2 x-\sin ^2 2 x} \mathrm{~d} x \\
& =\int \frac{\cos ^2(4 x) \sin (4 x)}{\cos (4 x)} \mathrm{d} x \\
& \quad \cdots\left[\begin{array}{r}
\because \sin 2 \theta=2 \sin \theta \cos \theta \text { and } \\
\cos 2 \theta=\cos ^2 \theta-\sin ^2 \theta
\end{array}\right]
\end{aligned}\)
\(\begin{aligned}
& =\frac{1}{2} \int 2 \sin (4 x) \cos (4 x) \mathrm{d} x \\
& =\frac{1}{2} \int \sin 8 x \mathrm{~d} x \\
& =\frac{-\cos 8 x}{2 \times 8}+c \\
& =\frac{-\cos 8 x}{16}+c
\end{aligned}\)
Comparing with ' \(\mathrm{A} \cos 8 x+\mathrm{c}\) ', we get \(\mathrm{A}=\frac{-1}{16}\)