Integral 1+cos8x/tan2x-cot2x

1 answer:

5 0

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))].

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

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