Question:
Evaluate the following integrals:
$\int \frac{1}{1+\cos 2 x} d x$
Solution:
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, $\int \frac{1}{1+\cos 2 x} d x=\frac{1}{2} \tan x+c$