Integrals — Class 12 Maths Solution

: Let $I = \int {\cfrac{{1 - \cos x}}{{1 + \cos x}}} dx = \int {\cfrac{{2{{\sin }^2}\cfrac{x}{2}}}{{2{{\cos }^2}\cfrac{x}{2}}}dx} = \int {{{\tan }^2}\cfrac{x}{2}dx}$

$= \int {\left( {{{\sec }^2}\cfrac{x}{2} - 1} \right)} dx = \cfrac{{\tan \cfrac{x}{2}}}{{\left( {1/2} \right)}} - x + C = 2\tan \cfrac{x}{2} - x + C$