Question

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

Answer: (D)

PLAN Use the formula, $|x-a|=\{\begin{array}{ll}x-a,& x\ge a\\ -(x-a),& x<a\end{array}$
to break given integral in two parts and then integrate separately.
$\underset{0}{\overset{\pi }{\int }}\sqrt{{\left(1-2\sin \frac{x}{2}\right)}^2}dx=\underset{0}{\overset{\pi }{\int }}\left|1-2\sin \frac{x}{2}\right|dx$
$=\underset{0}{\overset{\frac{\pi }{3}}{\int }}\left(1-2\sin \frac{x}{2}\right)dx-\underset{\frac{\pi }{3}}{\overset{\pi }{\int }}\left(1-2\sin \frac{x}{2}\right)dx$
$={\left(x+4\cos \frac{x}{2}\right)}_0^{\frac{\pi }{3}}-{\left(x+4\cos \frac{x}{2}\right)}_{\frac{\pi }{3}}^{\pi }$
$=4\sqrt{3}-4-\frac{\pi }{3}$