Question

Evaluate:${\int }_0^{\frac{\pi }{4}}\cos 2x\sqrt{4-\sin 2x}dx$

• A. $\frac{3\sqrt{3}-8}{3}$
• B. $\frac{3\sqrt{3}+8}{3}$
• C. $\frac{3\sqrt{3}-2}{3}$
• D. $-\frac{3\sqrt{3}-8}{3}$

Answer 1

The correct option is A $\frac{3\sqrt{3}-8}{3}$

${\int }_0^{\frac{\pi }{4}}\cos 2x\sqrt{4-\sin 2x}dx$

Let $t=4-\sin 2x\Rightarrow dt=2\cos 2xdx$

$\Rightarrow \cos 2xdx=\frac{dt}{2}$

$=\frac{1}{2}{\int }_0^{\frac{\pi }{4}}\sqrt{t}dt$

$=\frac{1}{2}{\int }_0^{\frac{\pi }{4}}{t}^{\frac{1}{2}}dt$

$=\frac{1}{2}{\left[\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]}_0^{\frac{\pi }{4}}$

$=\frac{1}{2}{\left[\frac{t^{\frac{1+2}{2}}}{\frac{1+2}{2}}\right]}_0^{\frac{\pi }{4}}$

$=\frac{1}{2}{\left[\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^{\frac{\pi }{4}}$

$=\frac{1}{3}{\left[{(4-\sin 2x)}^{\frac{3}{2}}\right]}_0^{\frac{\pi }{4}}$

$=\frac{1}{3}[{(4-\sin \frac{\pi }{2})}^{\frac{3}{2}}-{(4-\sin 0)}^{\frac{3}{2}}]$

$=\frac{1}{3}[{(4-1)}^{\frac{3}{2}}-{(4-0)}^{\frac{3}{2}}]$

$=\frac{1}{3}[3^{\frac{3}{2}}-4^{\frac{3}{2}}]$

$=\frac{1}{3}(3\sqrt{3}-8)$

$=\frac{3\sqrt{3}-8}{3}$