Question

\[\int_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\]

Answer 1

\[I = \int_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\]
\[ = \int_0^{2\pi} \sqrt{\cos^2 \frac{x}{4} + \sin^2 \frac{x}{4} + 2\sin\frac{x}{4}\cos\frac{x}{4}}dx\]
\[ = \int_0^{2\pi} \sqrt{\left( \cos\frac{x}{4} + \sin\frac{x}{4} \right)^2}dx\]
\[ = \int_0^{2\pi} \left| \cos\frac{x}{4} + \sin\frac{x}{4} \right|dx\]

When

\[0 \leq x \leq 2\pi\]

\[0 \leq \frac{x}{4} \leq \frac{\pi}{2}\]

\[\therefore \sin\frac{x}{4} \geq 0, \cos\frac{x}{4} \geq 0\]
\[ \Rightarrow \cos\frac{x}{4} + \sin\frac{x}{4} \geq 0\]
\[ \Rightarrow \left| \cos\frac{x}{4} + \sin\frac{x}{4} \right| = \cos\frac{x}{4} + \sin\frac{x}{4}\]

\[\therefore I = \int_0^{2\pi} \left( \cos\frac{x}{4} + \sin\frac{x}{4} \right)dx\]
\[=\left.\frac{\sin\frac{x}{4}}{\frac{1}{4}}\right|_0^{2\pi} + \left.\frac{\left( - \cos\frac{x}{4} \right)}{\frac{1}{4}}\right|_0^{2\pi} \]
\[ = 4\left( \sin\frac{\pi}{2} - \sin0 \right) - 4\left( \cos\frac{\pi}{2} - \cos0 \right)\]
\[ = 4\left( 1 - 0 \right) - 4\left( 0 - 1 \right)\]
\[ = 4 + 4\]
\[ = 8\]