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Integral 1+cos8x/tan2x-cot2x
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Integral 1+cos8x/tan2x-cot2x

User Kishma
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1 Answer

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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))].


(1)/(tan(2x)-cot(2x))+ (cos(8x))/(tan(2x)-cot(2x))
User Michael Hewson
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