Integrate the function: x ( 1 + x ) ^2 - Vidyadip

Question

Integrate the function: $\frac{{\sin x}}{{{{\left( {1 + \cos x} \right)}^2}}}$

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Answer

Let $I = \int {\frac{{\sin x}}{{{{\left( {1 + \cos x} \right)}^2}}}dx} $
$= - \int {\frac{{ - \sin x}}{{{{\left( {1 + \cos x} \right)}^2}}}dx} $…(i)
Putting 1 + cos x = t
$ \Rightarrow - \sin x = \frac{{dt}}{{dx}}$
$\Rightarrow $ -sin xdx = dt
$\therefore$ From eq. (i), $I = - \int {\frac{{dt}}{{{t^2}}}} $
$ = - \int {{t^{-2}}dt} $
$= \frac{{ - {t^{ - 1}}}}{{ - 1}} + c$
$= \frac{1}{t} + c$
$= \frac{1}{{1 + \cos x}} + c$

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