Question

The integral $\underset{0}{\overset{\pi }{\int }}\sqrt{1+4{\sin }^2\frac{x}{2}-4\sin \frac{x}{2}}dx$ equals to:

• A. $\pi -4$
• B. $\frac{2\pi }{3}-4-4\sqrt{3}$
• C. $4\sqrt{3}-4$
• D. $4\sqrt{3}-4-\frac{\pi }{3}$

Answer 1

The correct option is D $4\sqrt{3}-4-\frac{\pi }{3}$

$\underset{0}{\overset{\pi }{\int }}\sqrt{1+4{\sin }^2\frac{x}{2}-4\sin \frac{x}{2}}dx=\underset{0}{\overset{\pi }{\int }}\sqrt{{(1-2\sin \frac{x}{2})}^2}dx$
$=\underset{0}{\overset{\pi }{\int }}|1-2\sin \frac{x}{2}|dx$
Put $\frac{x}{2}=t\therefore dx=2dt$ and $t=0,t=\frac{\pi }{2}$
$\therefore \underset{0}{\overset{\pi }{\int }}|1-2\sin \frac{x}{2}|dx=2\underset{0}{\overset{\frac{\pi }{2}}{\int }}|1-2\sin t|dt$
$=2\left[\underset{0}{\overset{\frac{\pi }{6}}{\int }}(1-2\sin t)dt+\underset{\frac{\pi }{6}}{\overset{\frac{\pi }{2}}{\int }}(2\sin t-1)dt\right]$
$=2([t+2\cos t{]}_0^{\pi /6}+[-2\cos t-t{]}_{\pi /6}^{\pi /2})$
$=2(\frac{\pi }{6}+2\frac{\sqrt{3}}{2}-2-\frac{\pi }{2}+2\frac{\sqrt{3}}{2}+\frac{\pi }{6})$
$=2(4\frac{\sqrt{3}}{2}-2-\frac{\pi }{6})$
$=4\sqrt{3}-4-\frac{\pi }{3}$