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Question Number 91377 by A8;15: last updated on 30/Apr/20

Commented by Tony Lin last updated on 30/Apr/20

$$\int \frac{1}{\sqrt{sinx}}dx$$$$=\int \frac{1}{\sqrt{cos(x-\frac{\pi }{2})}}dx$$$$=\int \frac{1}{\sqrt{1-2{sin}^2\left(\frac{2x-\pi }{4}\right)}}dx$$$$let\frac{2x-\pi }{4}=u,\frac{du}{dx}=\frac{1}{2}$$$$2\int \frac{1}{\sqrt{1-2{sin}^{2}u}}du$$$$=F(u\mid 2)+c$$$$=F(\frac{2x-\pi }{4}\mid 2)+c$$

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