$\frac{d}{dx}\left(\frac{1}{\sin\left(2x\right)^2\cos\left(2x\right)^4}\right)=\frac{-\left(4\cos\left(2x\right)^{5}\sin\left(2x\right)-8\sin\left(2x\right)^{3}\cos\left(2x\right)^{3}\right)}{\sin\left(2x\right)^{4}\cos\left(2x\right)^{8}}$
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Integral
$\int\frac{1}{\sin\left(2x\right)^2\cos\left(2x\right)^4}dx=\frac{5}{6}\tan\left(2x\right)+\frac{\tan\left(2x\right)\sec\left(2x\right)^{2}}{6}-\frac{1}{2}\cot\left(2x\right)+C_0$
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