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प्रश्न
\[\int\frac{\sin x}{1 + \sin x} \text{ dx }\]
योग
उत्तर
\[\int\frac{\sin x}{1 + \sin x}dx\]
\[\text{Rationalising the denominator} \]
\[ \Rightarrow \int\frac{\sin x}{1 + \sin x} \times \frac{1 - \sin x}{1 - \sin x}\text{ dx }\]
\[ \Rightarrow \int\left( \frac{\sin x - \sin^2 x}{1 - \sin^2 x} \right)dx\]
\[ \Rightarrow \int\left( \frac{\sin x}{\cos^2 x} - \tan^2 x \right)dx\]
\[ \Rightarrow \int\left\{ \frac{\sin x}{\cos x} \times \frac{1}{\cos x} - \left( \sec^2 x - 1 \right) \right\}dx\]
\[ \Rightarrow \int\left( \sec x \tan x - \sec^2 x + 1 \right)dx\]
\[ \Rightarrow \sec x - \tan x + x + C\]
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