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प्रश्न
\[\int\frac{\cos x}{1 + \cos x} dx\]
योग
उत्तर
\[\int\frac{\cos x}{1 + \cos x}dx\]
\[ = \int\frac{\cos x\left( 1 - \cos x \right)}{\left( 1 + \cos x \right)\left( 1 - \cos x \right)}dx\]
\[ = \int\frac{\cos x - \cos^2 x}{1 - \cos^2 x}dx\]
\[ = \int\frac{\cos x - \cos^2 x}{\sin^2 x}dx\]
\[ = \int\frac{\cos x}{\sin^2 x} - \frac{\cos^2 x}{\sin^2 x}dx\]
\[ = \int\left( \text{cot x cosec x} - \cot^2 x \right)dx\]
\[ = \int\left( \text{cot x cosec x} - cosec^2 x + 1 \right)dx\]
\[ = \int\text{cot x cosec x dx} - \ ∫ co \sec^2 x dx + \int1dx\]
\[ =\text{ - cosec x }+ \cot x + x + C\]
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