Question
$\int\frac{1-\cos\text{x}}{1+\cos\text{x}}\text{dx}$

Answer

Let $\text{I}=\int\frac{1-\cos\text{x}}{1+\cos\text{x}}\times\text{dx}.$ Then,
$\text{I}=\int\frac{2\sin^2\frac{\text{x}}{2}}{2\cos^2\frac{\text{x}}{2}}\times\text{dx}$
$=\int\frac{\sin^2\frac{\text{x}}{2}}{\cos^2\frac{\text{x}}{2}}\times\text{dx}$
$=\int\tan^2\frac{\text{x}}{2}\text{dx}$
$=\int\Big(\sec^2\frac{\text{x}}{2}-1\Big)\text{dx}$
$=\frac{\tan\frac{\text{x}}{2}}{\frac{1}2{}}-\text{x+c}$
$=2\tan\frac{\text{x}}{2}-\text{x+c}$

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