Question
Find the integral of the function $\frac{1-\cos x}{1+\cos x}$

Answer

We have $\frac{1-\cos x}{1+\cos x}=\frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}=2 \tan ^{2} \frac{x}{2}=\left(\sec ^{2} \frac{x}{2}-1\right)$ 
$\Rightarrow \int \frac{1-\cos x}{1+\cos x} d x=\int\left(\sec ^{2} \frac{x}{2}-1\right) d x = \int \sec ^{2} \frac{x}{2} d x-\int 1 d x$ 
$=\frac{\tan \frac{x}{2}}{\frac{1}{2}}-x+C~~ \begin{equation} \left(\because \int \sec ^{2}(a x+b) d x=\frac{\tan (a x+b)}{a}+C\right) \end{equation}$
$= 2 \tan \frac{\mathrm{x}}{2}-\mathrm{x}+\mathrm{C}$

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