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Question Number 152116 by peter frank last updated on 25/Aug/21

$$\int \frac{\sin \mathrm{x}}{\sqrt{1+\sin \mathrm{x}}}\mathrm{dx}$$

Commented by puissant last updated on 26/Aug/21

$$Q=\int \frac{-cos(\frac{\pi }{2}+x)}{\sqrt{{(sin\left(\frac{x}{2}\right)+cos\left(\frac{x}{2}\right))}^2}}dx$$$$=-\int \frac{cos(\frac{\pi }{2}+x)}{sin\left(\frac{x}{2}\right)+cos\left(\frac{x}{2}\right)}dx$$$$=-\int \frac{cos2(\frac{x}{2}+\frac{\pi }{4})}{\sqrt{2}sin(\frac{x}{2}+\frac{\pi }{4})}dx$$$$=-\sqrt{2}\int \frac{1-2{sin}^2(\frac{x}{2}+\frac{\pi }{4})}{sin(\frac{x}{2}+\frac{\pi }{4})}dx$$$$=-\sqrt{2}\int cosec(\frac{x}{2}+\frac{\pi }{4})-2sin(\frac{x}{2}+\frac{\pi }{4})d(\frac{x}{2}+\frac{\pi }{4})$$$$=-\sqrt{2}(ln\mid tan\frac{1}{2}(\frac{x}{2}+\frac{\pi }{4})\mid +2cos(\frac{x}{2}+\frac{\pi }{4}))+C$$$$$$$$\therefore \because Q=-\sqrt{2}ln\mid tan(\frac{x}{4}+\frac{\pi }{8})\mid -2\sqrt{2}cos(\frac{x}{2}+\frac{\pi }{4})+C$$

Commented by peter frank last updated on 26/Aug/21

$$\mathrm{thank}\mathrm{you}$$

Answered by aleks041103 last updated on 26/Aug/21

$$\frac{\sin \mathrm{x}}{\sqrt{1+\sin \mathrm{x}}}=\frac{\sin \mathrm{x}\sqrt{1-\sin \mathrm{x}}}{\sqrt{1-{\sin }^2\mathrm{x}}}=$$$$=\frac{\sin \mathrm{x}\sqrt{1-\sin \mathrm{x}}}{{\cos }^2\mathrm{x}}\cos \mathrm{x}=\frac{\sin \mathrm{x}\sqrt{1-\sin \mathrm{x}}}{1-{\sin }^2\mathrm{x}}\cos \mathrm{x}$$$$\int \frac{\sin \mathrm{x}}{\sqrt{1+\sin \mathrm{x}}}\mathrm{dx}=$$$$=\int \frac{\sin \mathrm{x}\sqrt{1-\sin \mathrm{x}}}{1-{\sin }^2\mathrm{x}}\cos \mathrm{x}\mathrm{dx}=$$$$=\int \frac{u\sqrt{1-u}}{1-u^2}du=\int \frac{u}{(1+u)\sqrt{1-u}}du=$$$$w^2=1-u\Rightarrow u=1-w^2\Rightarrow du=-2wdw$$$$\int \frac{(1-w^2)(-2w)}{(2-w^2)w}dw=$$$$=\int -2\frac{1-w^2}{2-w^2}dw$$$$\frac{1-w^2}{2-w^2}=\frac{2-w^2}{2-w^2}-\frac{1}{2-w^2}=1-\frac{1}{2-w^2}$$$$\int -2\frac{1-w^2}{2-w^2}dw=\int (\frac{2}{2-w^2}-2)dw$$$$=2\int \frac{dw}{2-w^2}-2w$$$$\int \frac{dx}{a^2-x^2}=\int \frac{dx}{(a+x)(a-x)}=$$$$=\int \frac{1}{2a}(\frac{1}{a+x}+\frac{1}{a-x})dx=$$$$=\frac{1}{2a}(ln(a+x)-ln(a-x))=$$$$=\frac{1}{2a}ln\left(\frac{a+x}{a-x}\right)$$$$\Rightarrow integral=\frac{\sqrt{2}}{2}ln\left(\frac{\sqrt{2}+w}{\sqrt{2}-w}\right)-2w$$$$integral=\frac{\sqrt{2}}{2}ln\left(\frac{\sqrt{2}+\sqrt{1-sinx}}{\sqrt{2}-\sqrt{1-sinx}}\right)-2\sqrt{1-sinx}+C$$

Commented by peter frank last updated on 26/Aug/21

$$\mathrm{thank}\mathrm{you}$$

Answered by peter frank last updated on 26/Aug/21

$$\int \frac{2\sin \frac{\mathrm{x}}{2}\cos \frac{\mathrm{x}}{2}}{\sin \frac{\mathrm{x}}{2}+\cos \frac{\mathrm{x}}{2}}$$$$....$$

Answered by Ar Brandon last updated on 26/Aug/21

$$I=\int \frac{\sin x}{\sqrt{1+\sin x}}dx=\int (\sqrt{1+\sin x}-\frac{1}{\sqrt{1+\sin x}})dx$$$$=\int (\cos \frac{x}{2}+\sin \frac{x}{2}-\frac{1}{\cos \frac{x}{2}+\sin \frac{x}{2}})dx$$$$=2(\sin \frac{x}{2}-\cos \frac{x}{2})-\frac{1}{\sqrt{2}}\int \frac{dx}{\sin (\frac{x}{2}+\frac{\pi }{4})}$$$$=2\sqrt{2}\sin (\frac{x}{2}-\frac{\pi }{4})-\sqrt{2}\ln \left(\cot (\frac{x}{4}+\frac{\pi }{8})\right)+C$$

Commented by peter frank last updated on 26/Aug/21

$$\mathrm{thank}\mathrm{you}$$

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