Question

The integral ${\int }_0^{\pi }\sqrt{1+4{\sin }^2\frac{x}{2}-4\sin \frac{x}{2}}dx$ equals.

• 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

Answer: (A)

$\pi -4$

${\int }_0^{\pi }\sqrt{1+4{\sin }^2\frac{x}{2}-4\sin \frac{x}{2}}dx={\int }_0^{\pi }\sqrt{{(1-2\sin \frac{x}{2})}^2}dx={\int }_0^{\pi }(1-2\sin \frac{x}{2})dx={[x+2\left(2\cos \frac{x}{2}\right)]}_0^{\pi }=(\pi +4\cos \frac{\pi }{2})-(0+4\cos 0)=\pi -4$