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How do you integrate $$ 1/(\sin2x(\tan^5x+\cot^5x)) $$ with respect to $x$?
I tried writing tan and cot in terms of sin and cos but when I take $\mathrm{LCM}$ I get powers of $10$ for $\sin$ and $\cos$ in the denominator. I can't think of how I would simplify this without the use of binomial expansion but that might make things more complicated.

Is there another method I could use to approach this problem?

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Hint

I would say:

$\displaystyle \int \frac{dx}{\sin2x\cdot(tan^5x+cot^5x)} = \frac12 \int \frac{\cot x \,dx}{\cos^2x\cdot(\tan^5x+\cot^5x)}$

and substitution:

$\displaystyle \tan x = t, \frac{dx}{\cos^2x}=dt$

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