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: (D)

$4\sqrt{3}-4-\frac{\pi }{3}$

$I={\int }_0^{\pi }\sqrt{1+4{\sin }^2\left(x/2\right)-4\sin \left(x/2\right)}dx$

$I={\int }_0^{\pi }|2\sin \left(x/2\right)-1|dx$

If $0<x<\frac{\pi }{3}$, then $2\sin \left(x/2\right)-1<0$

And if $\frac{\pi }{3}<x<0$, then $2\sin \left(x/2\right)-1>0$

$\therefore I={\int }_0^{\pi /3}-\left(2\sin \right(x/2)-1)dx$$+{\int }_{\pi /3}^{\pi }\left(2\sin \right(x/2)-1)dx$

$I=\left[4cos\right(x/2)+x{]}_0^{\pi /3}+[-4\cos \left(x/2\right)-x{]}_{\pi /3}^{\pi }$

$I=\frac{4\sqrt{3}}{2}+\frac{\pi }{3}-4+(0-\pi +\frac{4\sqrt{3}}{2}+\frac{\pi }{3})$

$I=4\sqrt{3}-4-\frac{\pi }{3}$