Question

\[\int \cot^{- 1} \left( \frac{\sin 2x}{1 - \cos 2x} \right) dx\]

Answer 1

\[\int \cot^{- 1} \left( \frac{\sin 2x}{1 - \cos 2x} \right)dx\]
` = ∫    cot ^-1  (( 2 sin x  cos x) /( 2 sin^2 x))` dx ` [∴ sin  2x = 2   sin x cos x  & 1 - cos   2x = 2   sin^2 x ]`
\[ = \int \cot^{- 1} \left( \cot x \right)dx\]
` = ∫  x   dx `
\[ = \frac{x^2}{2} + C\]