# Find the integrals of the functions$\frac{1-\cos x}{1+\cos x}$

$\begin{array}{1 1} 2\tan \frac{x}{2}-x+c. \\ 2\tan \frac{x}{2}+x+c \\ 2\cot \frac{x}{2}-x+c \\ 2\cot \frac{x}{2}+x+c. \end{array}$

Toolbox:
• (i)$1-\cos x=2\sin^2\frac{x}{2}$.
• (ii)$1+\cos x=2\cos^2\frac{x}{2}$.
• (iii)$\int \tan xdx=log|\sec x|+c.$
• (iv)$\sec^2xdx=\tan x+c.$
Given $I=\int\frac{1-\cos x}{1+\cos x}dx.$

Using the information in the tool box we get

$1-\cos x=2\sin^2\frac{x}{2}$ and $1+\cos x=2\cos^2\frac{x}{2}$.

$I=\int\frac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}}=\int\tan^2\frac{x}{2}.$

But $\tan^2\frac{x}{2}=\sec^2\frac{x}{2}-1$.

Substituting this,

$I=\int(\sec^2\frac{x}{2}-1)dx.$

On separating the terms,

$\;\;\;=\int\sec^2\frac{x}{2}dx-\int dx.$

On integrating we get,

$\;\;\;=\frac{\tan\frac{x}{2}}{\frac{1}{2}}-x+c.$

$\;\;\;=2\tan\frac{x}{2}-x+c.$