$\cfrac{{\cos x}}{{1 + \cos x}}$
$\cfrac{{\cos x}}{{1 + \cos x}}$
Official Solution
: Let $I = \int {\cfrac{{\cos x}}{{1 + \cos x}}dx}$
$= \int {\cfrac{{\left( {1 + \cos x} \right) - 1}}{{1 + \cos x}}} dx = \int {\left( 1 \right)} dx - \int {\cfrac{1}{{1 + \cos x}}dx}$
$= x - \int {\cfrac{1}{{2{{\cos }^2}\cfrac{x}{2}}}dx + C} = x - \cfrac{1}{2}\int {{{\sec }^2}\cfrac{x}{2}dx + C}$
$= x - \cfrac{1}{2} \cdot \cfrac{{\tan \cfrac{x}{2}}}{{\left( {1/2} \right)}} + C = x - \tan \cfrac{x}{2} + C$
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