# Evaluate the following$\int\frac{dx}{1+\cos x}$

$\begin{array}{1 1} (A)\;\tan^2 x/2+c \\ (B)\;\tan x/2+c \\ (C)\;\tan x+c \\ (D)\;\tan^2x+c \end{array}$

Toolbox:
• $1+\cos x=2 \cos ^2 x/2$
• $\int \sec^2 x=\tan x+c$
Let $I=\large\int\frac{dx}{1+\cos x}$
But $1+\cos x=2 \cos ^2 x/2$
Therefore $I=\large\int\frac{dx}{2\cos^2 x/2}=\frac{1}{2}\int\frac{dx}{\cos^2 x/2}$
But $\large\int\frac{1}{1+\cos ^2 x/2}=\sec^2 x/2$
Therefore $I=\frac{1}{2} \int sec^2 x/2.dx$
on integrating we get
$I=\frac{1}{2} \times 2(\tan x/2)+c$
Therefore $I=\tan x/2+c$