integrals involving tan and sec; cot and cos

Question

evaluate pls

∫cot⁶2xdx

Answer 1

∫cot⁶2xdx = ∫(csc²2x -1)³dx = ∫(csc⁶2x - 3csc⁴2x + 3csc²2x - 1)dx = - 1/2∫(csc⁴2x - 3csc²2x + 3)d(cot2x) - x + C = -1/2∫(cot⁴2x + 2cot²2x + 1 - 3cot²2x )d(cot2x) - x + C = - cot⁵2x/10 + cot³2x/6 - cot2x/2 - x + C