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

KCETKCET 2007Integrals

Solution:

$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 $