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Question Number 79413 by rajesh4661kumar@gmail.com last updated on 25/Jan/20

Commented by john santu last updated on 25/Jan/20

$${\left(\frac{\cos x\sin x}{\cos 2x({\cos }^{2}x-\sin {}^{2}x)}\right)}^2=$$$$\frac{1}{4}.\frac{\sin {}^{2}2x}{\cos {}^{4}2x}=\frac{1}{4}\tan {}^{2}2x\sec {}^{2}2x$$$$\int \frac{1}{8}{\tan }^2\left(2x\right)d\left(\tan \left(2x\right)\right)=$$$$\frac{1}{24}\tan {}^3\left(2x\right)+c$$

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