$\cfrac{{\sin x}}{{{{\left( {1 + \cos x} \right)}^2}}}$
$\cfrac{{\sin x}}{{{{\left( {1 + \cos x} \right)}^2}}}$
Official Solution
VVidaara Team
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NCERT & Exemplar
: Let$I = \int {\cfrac{{\sin x}}{{{{\left( {1 + \cos x} \right)}^2}}}dx}$
Put $1 + \cos x = t$ $\Rightarrow$ $- \sin xdx = dt$
$\therefore$ $I = - \int {\cfrac{{dt}}{{{t^2}}} = - \cfrac{{{t^{ - 2 + 1}}}}{{ - 2 + 1}} + C = \cfrac{1}{t} + C = \cfrac{1}{{1 + \cos x}} + C}$
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