If \(\int \frac{\cos 4 x+1}{\cot x-\tan x} d x=k \cos 4 x+c\), then \(k\) is
If \(\int \frac{\cos 4 x+1}{\cot x-\tan x} d x=k \cos 4 x+c\), then \(k\) is
\(-\frac{1}{2}\)
\(-\frac{1}{4}\)
\(-\frac{1}{8}\)
-1
Solution
\(\begin{aligned}
& \int \frac{\cos 4 x+1}{\cot x-\tan x} d x \\
& =\int \frac{2 \cos ^2 2 x}{\cos ^2 x-\sin ^2 x} \times \sin x \cos x d x \\
& =\int \frac{\cos ^2 2 x}{\cos 2 x} \sin 2 x d x=\int \sin 2 x \cos 2 x d x \\
& =\frac{1}{2} \int \sin 4 x d x=-\frac{1}{8} \cos 4 x+c
\end{aligned}\)
So, \(\quad k=-\frac{1}{8}\)
Hence, option (c) is correct.