Question

If $\frac{1+\cos 8x}{\tan 2x-\cot 2x}dx=a\cos 8x+c$ , then a=

• A. $\frac{-1}{16}$
• B. $\frac{1}{8}$
• C. $\frac{1}{16}$
• D. $\frac{-1}{8}$

Answer 1

The correct option is C $\frac{1}{16}$

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

$=\int \frac{2{\cos }^{2}4x}{\frac{{\sin }^{2}2x-{\cos }^{2}2x}{\sin 2x\cos 2x}}dx$

$=\int \frac{2\sin 2x\cos 2x{\cos }^{2}4x}{-\cos 4x}dx$

$=-\int {\sin }^{4}x\cos 4xdx$

$=-\int \frac{\sin 8x}{2}dx$

$=\frac{\cos 8x}{16}+C$