Question

Integral 1+cos8x/tan2x-cot2x

Answer 1

I believe it's 8cos(x)⁸ - 16cos(x)⁶ + 10cos(x)⁴ - 2cos(x)².

$8cosx^8 - 16cosx^6+10cosx^4-2cosx^2$

Alternately, you can write [1 / (tan(2x) - cot(2x))] + [cos(8x) / (tan(2x) - cot(2x))].

$\dfrac{1}{tan(2x)-cot(2x)}+ \dfrac{cos(8x)}{tan(2x)-cot(2x)}$