Evaluate the following integrals:


Evaluate the following integrals:

$\int \frac{1}{1+\cos 2 x} d x$


Let $I=\int \frac{1}{1+\cos 2 x} d x$

We know $\cos 2 \theta=2 \cos ^{2} \theta-1$

Hence, in the denominator, we can write $1+\cos 2 x=2 \cos ^{2} x$

Therefore, we can write the integral as

$I=\int \frac{1}{2 \cos ^{2} x} d x$

$\Rightarrow I=\frac{1}{2} \int \frac{1}{\cos ^{2} x} d x$

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

Recall $\int \sec ^{2} x d x=\tan x+c$

$\therefore I=\frac{1}{2} \tan x+c$

Thus, $\int \frac{1}{1+\cos 2 x} d x=\frac{1}{2} \tan x+c$

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