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Question Number 42579 by Raj Singh last updated on 28/Aug/18

$$\int sinx/\sqrt{1+sinx}$$

Commented by maxmathsup by imad last updated on 28/Aug/18

$$letA=\int \frac{sinx}{\sqrt{1+sinx}}dxwehave1+sinx=1+2sin\left(\frac{x}{2}\right)cos\left(\frac{x}{2}\right)$$$$={(sin\left(\frac{x}{2}\right)+cos\left(\frac{x}{2}\right))}^2\Rightarrow \sqrt{1+sinx}=sin\left(\frac{x}{2}\right)+cos\left(\frac{x}{2}\right)$$$$A=\int \frac{1+sinx-1}{\sqrt{1+sinx}}dx=\int \sqrt{1+sinx}dx-\int \frac{dx}{\sqrt{1+sinx}}$$$$=\int (sin\left(\frac{x}{2}\right)+cos\left(\frac{x}{2}\right))dx-\int \frac{dx}{sin\left(\frac{x}{2}\right)+cos\left(\frac{x}{2}\right)}$$$$=-2cos\left(\frac{x}{2}\right)+2sin\left(\frac{x}{2}\right)-\int \frac{dx}{\sqrt{2}cos(\frac{x}{2}-\frac{\pi }{4})}changement\frac{x}{2}-\frac{\pi }{4}=tgive$$$$\int \frac{dx}{\sqrt{2}cos(\frac{x}{2}-\frac{\pi }{4})}=\int \frac{2dt}{\sqrt{2}cos\left(t\right)}=\sqrt{2}\int \frac{dt}{cos\left(t\right)}$$$${=}_{tan\left(\frac{t}{2}\right)=u}\sqrt{2}\int \frac{1}{\frac{1-u^2}{1+u^2}}\frac{2du}{1+u^2}=2\sqrt{2}\int \frac{du}{1-u^2}$$$$=\sqrt{2}\int (\frac{1}{1+u}+\frac{1}{1-u})du=\sqrt{2}ln\mid \frac{1+u}{1-u}\mid =\sqrt{2}ln\mid \frac{1+tan\left(\frac{t}{2}\right)}{1-tan\left(\frac{t}{2}\right)}\mid$$$$=\sqrt{2}ln\mid tan(\frac{t}{2}+\frac{\pi }{4})\mid =\sqrt{2}ln\mid tan(\frac{x}{4}-\frac{\pi }{8}+\frac{\pi }{4})\mid =\sqrt{2}ln\mid tan(\frac{x}{4}+\frac{\pi }{8})\mid \Rightarrow$$$$A=2sin\left(\frac{x}{2}\right)-2cos\left(\frac{x}{2}\right)-\sqrt{2}ln\mid tan(\frac{x}{4}+\frac{\pi }{8})\mid +c.$$$$$$

Answered by tanmay.chaudhury50@gmail.com last updated on 28/Aug/18

$$\int \frac{sinx}{\sqrt{1+sinx}}dx$$$$\int \frac{1+sinx-1}{\sqrt{1+sinx}}dx$$$$\int \sqrt{1+sinx}dx-\int \frac{dx}{\sqrt{1+sinx}}dx$$$$\int (sin\frac{x}{2}+cos\frac{x}{2})-\int \frac{dx}{sin\frac{x}{2}+cos\frac{x}{2}}$$$$\int sin\frac{x}{2}+cos\frac{x}{2}dx-\frac{1}{\sqrt{2}}\int \frac{dx}{(\frac{1}{\sqrt{2}}sin\frac{x}{2}+\frac{1}{\sqrt{2}}cos\frac{x}{2})}$$$$\int (sin\frac{x}{2}+cos\frac{x}{2})dx-\frac{1}{\sqrt{2}}\int cosec(\frac{\mathrm{\Pi }}{4}+\frac{x}{2})dx$$$$=-\frac{cos\frac{x}{2}}{\frac{1}{2}}+\frac{sin\frac{x}{2}}{\frac{1}{2}}-\frac{1}{\sqrt{2}}lntan\left(\frac{\frac{\mathrm{\Pi }}{4}+\frac{x}{2}}{2}\right)+c$$$$=2(-cos\frac{x}{2}+sin\frac{x}{2})-\frac{1}{\sqrt{2}}lntan(\frac{\mathrm{\Pi }}{8}+\frac{x}{4})+c$$$$$$

Commented by maxmathsup by imad last updated on 28/Aug/18

$$youransweriscorrectsirTanmay.$$

Commented by tanmay.chaudhury50@gmail.com last updated on 28/Aug/18

$$thankyousir...$$

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