Question

 

\[\int\frac{\cos x}{1 - \cos x} \text{dx }or \int\frac{\cot x}{\text{cosec         } {x }- \cot x} dx\]

Answer 1

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