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Question Number 36811 by rahul 19 last updated on 05/Jun/18

$$\int \frac{\sin x}{\cos {}^{2}x.\sqrt{\cos 2x}}dx=?$$

Answered by MJS last updated on 05/Jun/18

$$\int \frac{\sin x}{{\cos }^{2}x\sqrt{\cos 2x}}dx=\int \frac{\sin x}{{\cos }^{2}x\sqrt{{\cos }^{2}x-{\sin }^{2}x}}dx=$$$$=\int \frac{\frac{\tan x}{\sec x}}{\frac{1}{{\sec }^{2}x}\sqrt{\frac{1}{{\sec }^{2}x}-\frac{{\tan }^{2}x}{{\sec }^{2}x}}}dx=$$$$=\int \frac{{\sec }^{2}x\tan x}{\sqrt{1-{\tan }^{2}x}}dx=$$$$[t=\tan x\to dx=\frac{dt}{{\sec }^{2}x}]$$$$=\int \frac{t}{\sqrt{1-t^2}}dt=$$$$[u=1-t^2\to dt=-\frac{du}{2t}]$$$$=-\frac{1}{2}\int \frac{du}{\sqrt{u}}=-\sqrt{u}=-\sqrt{1-t^2}=-\sqrt{1-{\tan }^{2}x}+C$$$$$$

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