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