The value of 1 1 + cos 8 x d x is:

Question

The value of $\int \frac{1}{1 + \cos 8 x} d x$ is:

Smart Achievers · Accepted Answer

(4)

Ans. $\frac{\tan 4 x}{8} + C$

Explanation:

$\int \frac{1}{1 + \cos 8 x} d x = \int \frac{d x}{1 + \frac{(1 - \tan^{2} 4 x)}{(1 + \tan^{2} 4 x)}}$

$= \int \frac{1 + \tan^{2} 4 x}{1 + \tan^{2} 4 x + 1 - \tan^{2} 4 x} d x$

$= \int \frac{1 + \tan^{2} 4 x}{2} d x$

$= \frac{1}{2} \int \sec^{2} 4 x d x$

$= \frac{\tan 4 x}{8} + C$

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