Question

Prove that $\underset{0}{\overset{\frac{32\pi }{3}}{\int }}\sqrt{1+\cos 2x}dx=\sqrt{\frac{3}{2}}$

Answer 1

${\int }_0^{\frac{32\pi }{3}}\sqrt{1+\cos 2x}dx$

$={\int }_0^{\frac{32\pi }{3}}\sqrt{2{\cos }^{2}x}dx$

$=\sqrt{2}{\int }_0^{\frac{32\pi }{3}}\cos xdx$

$=\sqrt{2}{\left[\sin x\right]}_0^{\frac{32\pi }{3}}$

$=\sqrt{2}[\sin \frac{32\pi }{3}-\sin 0]$

$=\sqrt{2}[\sin (10\pi +\frac{2\pi }{3})-0]$

$=\sqrt{2}\sin \frac{2\pi }{3}$

$=\sqrt{2}\sin (\pi -\frac{\pi }{3})$

$=\sqrt{2}\sin \frac{\pi }{3}$

$=\sqrt{2}\times \frac{\sqrt{3}}{2}$

$=\sqrt{2}\times \frac{\sqrt{3}}{2}\times \frac{\sqrt{2}}{\sqrt{2}}$

$=\sqrt{\frac{3}{2}}$