Question

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

Answer 1

\[\text{ We  have ,} \]
\[I = \int\frac{\sin x}{\sqrt{1 + \sin x}} \text{ dx }\]
\[I = \int\frac{2 \sin\frac{x}{2}\cos\frac{x}{2}}{\sqrt{\sin^2 \frac{x}{2} + \cos^2 \frac{x}{2} + \text{ 2 }\sin\frac{x}{2}\cos\frac{x}{2}}} \text{  dx }\]
\[I = \int\frac{2 \sin\frac{x}{2}\cos\frac{x}{2}}{\sqrt{\left( \sin\frac{x}{2} + \cos\frac{x}{2} \right)^2}} \text{  dx }\]
\[I = \int\frac{2 \sin\frac{x}{2}\cos\frac{x}{2}}{\sin\frac{x}{2} + \cos\frac{x}{2}} \text{  dx }\]
\[I = \int\frac{1 + 2\sin\frac{x}{2} \cos\frac{x}{2} - 1}{\sin x + \cos x} \text{  dx }\]
\[I = \int\frac{\sin^2 \frac{x}{2} + \cos^2 \frac{x}{2} + 2\sin\frac{x}{2} \cos\frac{x}{2} - 1}{\sin x + \cos x} \text{  dx }\]
\[I = \int\frac{\left( \sin\frac{x}{2} + \cos\frac{x}{2} \right)^2 - 1}{\sin\frac{x}{2} + \cos\frac{x}{2}} \text{  dx }\]
\[I = \int\frac{\left( \sin\frac{x}{2} + \cos\frac{x}{2} \right)^2}{\sin\frac{x}{2} + \cos\frac{x}{2}} \text{  dx }- \int\frac{1}{\sin\frac{x}{2} + \cos\frac{x}{2}} \text{  dx }\]
\[I = \int\left( \sin\frac{x}{2} + \cos\frac{x}{2} \right) dx - \int\frac{1}{\sin\frac{x}{2} + \cos\frac{x}{2}} \text{  dx }\]
\[I = 2\left( - \cos\frac{x}{2} + \sin\frac{x}{2} \right) + C_1 - \frac{1}{\sqrt{2}}\int\frac{1}{\frac{1}{\sqrt{2}}\left( \sin\frac{x}{2} + \cos\frac{x}{2} \right)} \text{  dx }\]
\[I = 2\left( - \cos\frac{x}{2} + \sin\frac{x}{2} \right) + C_1 - \frac{1}{\sqrt{2}}\int\frac{1}{\sin\frac{x}{2} cos\frac{\pi}{4} + \cos\frac{x}{2} sin\frac{\pi}{4}} \text{  dx }\]
\[I = 2\left( - \cos\frac{x}{2} + \sin\frac{x}{2} \right) + C_1 - \frac{1}{\sqrt{2}}\int\frac{1}{\sin\left( \frac{x}{2} + \frac{\pi}{4} \right)} \text{  dx }\]
\[I = 2\left( - \cos\frac{x}{2} + \sin\frac{x}{2} \right) + C_1 - \frac{1}{\sqrt{2}}\int \text{ cosec} \left( \frac{x}{2} + \frac{\pi}{4} \right) \text{  dx }\]
\[I = 2\left( - \cos\frac{x}{2} + \sin\frac{x}{2} \right) - \sqrt{2}\text{ log}\left| \text{ tan}\left( \frac{x}{4} + \frac{\pi}{8} \right) \right| + C\]