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प्रश्न
उत्तर
\[\int\frac{\sin x}{\left( 1 + \cos x \right)^2}dx\]
\[\text{Let 1} + \cos x = t\]
\[ \Rightarrow - \sin x = \frac{dt}{dx}\]
\[ \Rightarrow \text{sin x dx} = - dt\]
\[Now, \int\frac{\sin x}{\left( 1 + \cos x \right)^2}dx\]
\[ = \int - \frac{dt}{t^2}\]
\[ = - \int t^{- 2} dt\]
\[ = - \left[ \frac{t^{- 2 + 1}}{- 2 + 1} \right] + C\]
\[ = \frac{1}{t} + C\]
\[ = \frac{1}{1 + \cos x} + C\]
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