Question

The value of $\int \frac {1}{1+ Cos \; 8x}dx$ is

• A. $\frac {\tan\, 8x}{8}+C$
• B. $\frac {\tan\,2x}{8}+C$
• C. $\frac {\tan\,4x}{8}+C$
• D. $\frac {\tan\,4x}{4}+C$

Answer 1

Answer: (C)

$I = \int \frac{1}{1+\cos 8x} dx $
$ = \int \frac{1}{2 \cos^{2} 4x} dx$
$ x = \frac{1}{2} \int \sec^{2} 4 x dx$
$ = \frac{1}{2} \frac{\tan 4x}{4} + c $
$ = \frac{\tan4x}{8} +c $