Question
Integrate the function: $\frac{{\sin x}}{{1 + \cos x}}$
$= - \int {\frac{{ - \sin x}}{{1 + \cos x}}dx} $ ...(i)
Putting 1 + cos x = t
$\Rightarrow - \sin x = \frac{{dt}}{{dx}}$
$\Rightarrow - \sin xdx = dt$
$\therefore$ From eq. (i), $I = - \int {\frac{{dt}}{t}}$
= - log |t| + c
= -log |1 + cos x| + c
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