Question

${\int }_0^{32\pi /3}\sqrt{1+\cos 2x}dx$ is equal to

• A. $\sqrt{\frac{3}{2}}$
• B. $\sqrt{2}+\sqrt{\frac{3}{2}}$
• C. $\sqrt{\frac{2}{3}}$
• D. none

Answer 1

The correct option is A $\sqrt{\frac{3}{2}}$

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

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

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

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

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

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

$=\sqrt{2}\sin \frac{32\pi }{3}$

$=\sqrt{2}\sin (11\pi -\frac{\pi }{3})$ is in second quadrant.

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

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

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

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

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