Question

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

Answer 1

\[\int \tan^{- 1} \left[ \frac{\sin \left( 2x \right)}{1 + \cos2x} \right]dx\]

`=  ∫  tan ^-1  [ (2 sin  x cos  x) / ( 2 cos^2 x)] `dx ` [∴     sin 2x  = 2  sin x cos x  &   1 + cos 2x = 2    cos ^2 x ]`

\[ = \int \tan^{- 1}  \left[ \tan x \right]\]

\[ = \int \tan^{- 1}     \left[ \tan x \right]\]

` =  ∫   x dx `

\[ = \frac{x^2}{2} + C\]