question_answer

Evaluate \[\int{\frac{\sin x}{\sqrt{1+\sin x}}}\,dx.\]

Answer:

\[\int{\frac{\sin x}{\sqrt{1+\sin x}}d\,\,x}=\int{\frac{(1+\sin \,x)-1}{\sqrt{1+\sin \,x}}\,dx}\] \[=\int{\sqrt{1+\sin \,xdx}-\int{\frac{dx}{\sqrt{1+\sin x}}}}\] \[=\int{\sqrt{{{\cos }^{2}}\frac{x}{2}+{{\sin }^{2}}\frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2}dx}}\] \[-\int{\frac{dx}{\sqrt{{{\cos }^{2}}\left( \frac{x}{2} \right)+{{\sin }^{2}}\left( \frac{x}{2} \right)+2\sin \left( \frac{x}{2} \right)\cos \left( \frac{x}{2} \right)}}}\] \[=\int{\left[ \cos \left( \frac{x}{2} \right)+\sin \left( \frac{x}{2} \right) \right]}dx-\int{\frac{dx}{\left[ \cos \left( \frac{x}{2} \right)+\sin \left( \frac{x}{2} \right) \right]}}\] \[=\left( 2\sin \frac{x}{2}-2\cos \frac{x}{2} \right)-\frac{1}{\sqrt{2}}\cdot \int{\frac{dx}{\sin \left( \frac{x}{2}+\frac{\pi }{4} \right)}}\]  \[=\left( 2\sin \frac{x}{2}-2\cos \frac{x}{2} \right)-\frac{1}{\sqrt{2}}\int{\text{cosec}\left( \frac{x}{2}+\frac{\pi }{4} \right)}\,dx\] \[=2\left( \sin \frac{x}{2}-\cos \frac{x}{2} \right)-\frac{1}{\sqrt{2}}\times 2\,\,\log \left| \tan \left( \frac{x}{4}+\frac{\pi }{8} \right) \right|+C\]\[=2\left( \sin \frac{x}{2}-\cos \frac{x}{2} \right)-\sqrt{2}\,\,\log \left| \tan \left( \frac{x}{4}+\frac{\pi }{8} \right) \right|+C\]