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Question Number 109016 by mathdave last updated on 20/Aug/20

Answered by Dwaipayan Shikari last updated on 20/Aug/20

$$\int \frac{1}{\sqrt{1-sinx}}dx$$$$\int \frac{1}{cos\frac{x}{2}-sin\frac{x}{2}}dx$$$$\frac{1}{\sqrt{2}}\int \frac{1}{cos(\frac{x}{2}+\frac{\pi }{4})}dx$$$$\frac{1}{\sqrt{2}}\int sec(\frac{x}{2}+\frac{\pi }{4})dx$$$$\frac{2}{\sqrt{2}}\int secudu(\frac{x}{2}+\frac{\pi }{4}=u,\frac{1}{2}=\frac{du}{dx}$$$$\sqrt{2}log(secu+tanu)+C$$$$\sqrt{2}log(sec(\frac{x}{2}+\frac{\pi }{4})+tan(\frac{x}{2}+\frac{\pi }{4}))+C$$

Commented by $@y@m last updated on 20/Aug/20

$$Wonderful!$$

Answered by mathmax by abdo last updated on 20/Aug/20

$$\mathrm{I}=\int \frac{\mathrm{dx}}{\sqrt{1-\mathrm{sinx}}}\Rightarrow \mathrm{I}=\int \frac{\mathrm{dx}}{\sqrt{1-\cos (\frac{\pi }{2}-\mathrm{x})}}=\int \frac{\mathrm{dx}}{\sqrt{{2\sin }^2(\frac{\pi }{4}-\frac{\mathrm{x}}{2})}}$$$$=\frac{1}{\sqrt{2}}\int \frac{\mathrm{dx}}{\sin (\frac{\pi }{4}-\frac{\mathrm{x}}{2})}{=}_{\frac{\pi }{4}-\frac{\mathrm{x}}{2}=\mathrm{u}}\frac{1}{\sqrt{2}}\int \frac{-2\mathrm{du}}{\mathrm{sinu}}=-\sqrt{2}\int \frac{\mathrm{du}}{\mathrm{sinu}}$$$${=}_{\tan \left(\frac{\mathrm{u}}{2}\right)=\alpha }-\sqrt{2}\int \frac{2\mathrm{d}\alpha }{(1+{\alpha }^2)\frac{2\alpha }{1+{\alpha }^2}}=-\sqrt{2}\int \frac{\mathrm{d}\alpha }{\alpha }=-\sqrt{2}\ln \mid \alpha \mid +\mathrm{C}$$$$=-\sqrt{2}\ln \mid \tan (\frac{1}{2}(\frac{\pi }{4}-\frac{\mathrm{x}}{2})\mid +\mathrm{C}=-\sqrt{2}\ln \mid \tan (\frac{\pi }{8}-\frac{\mathrm{x}}{4})\mid +\mathrm{C}$$

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