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Question

\[\int\frac{\cos x}{1 + \cos x} dx\]
Sum
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Solution

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

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Chapter 19: Indefinite Integrals - Exercise 19.02 [Page 15]

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RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
Exercise 19.02 | Q 42 | Page 15

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