Question

∫cot^-1(sin 2x/1 - cos 2x) dx = ?

A. -1/(1 + x²) + C

B. -1/(1 - x²) + C

C. x²/2 + C

D. 2x² + C

Answer 1

Given:

= ∫cot -1(cot x) dx

= ∫x dx

= \(\frac{x^2}{2}+c\)