Question

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

• A. $\mathrm{\pi }-4$
• B. $\left(2\mathrm{\pi }/3\right)-4-4\sqrt{3}$
• C. $4\sqrt{3}-4$
• D. $4\sqrt{3}-4-\left(\mathrm{\pi }/3\right)$

Answer 1

The correct option is D

$4\sqrt{3}-4-\left(\mathrm{\pi }/3\right)$

Step 1: Simplify the Equation :

$I={\int }_0^{\mathrm{\pi }}\sqrt{1+4{\sin }^2\frac{x}{2}-4\sin \frac{x}{2}}dx$

$\begin{array}{rcl}I& =& {\int }_0^{\mathrm{\pi }}\sqrt{1+4\frac{{\sin }^{2}x}{2}-4\frac{\sin x}{2}}dx\\ I& =& {\int }_0^{\mathrm{\pi }}\sqrt{{\left(1\right)}^2+{\left(2\sin \left(\frac{x}{2}\right)\right)}^2-2\times 1\times 2\sin \left(\frac{x}{2}\right)}dx\\ I& =& {\int }_0^{\mathrm{\pi }}\sqrt{{\left(1-2\sin \left(\frac{x}{2}\right)\right)}^2}dx\\ I& =& \left|1-2\sin \left(\frac{x}{2}\right)\right|dx\end{array}$

Step 2 : Find the value of $x$:

$\begin{array}{rcl}\sin \left(\frac{x}{2}\right)& =& \frac{1}{2}\\ \frac{x}{2}& =& \frac{\mathrm{\pi }}{6}\\ x& =& \frac{\mathrm{\pi }}{3}\end{array}$

Integrate the $I$ with respect to the limit $0$ to$\frac{\mathrm{\pi }}{3}$to $\mathrm{\pi }$.

$\begin{array}{rcl}I& =& {\int }_0^{\frac{\mathrm{\pi }}{3}}\left(1-2\sin \left(\frac{x}{2}\right)\right)dx+{\int }_{\frac{\mathrm{\pi }}{3}}^{\mathrm{\pi }}\left(2\sin \left(\frac{x}{2}\right)-1\right)dx\\ I& =& \left(x+4\cos \frac{x}{2}\right)+{\left|\left(-4\cos \left(\frac{x}{2}\right)-x\right)\right|}_{\frac{\mathrm{\pi }}{3}}^{\mathrm{\pi }}\\ I& =& \left[\left(\frac{x}{3}+4\cos \left(\frac{\mathrm{\pi }}{6}\right)\right)-\left(0+4\cos 0\right)\right]+\left[\left(-4\cos \left(\frac{\mathrm{\pi }}{2}\right)-\mathrm{\pi }\right)-\left(-4\cos \left(\frac{\mathrm{\pi }}{6}\right)\right)\right]\\ I& =& \left[\frac{\mathrm{\pi }}{3}+4\times \frac{\sqrt{3}}{2}-4\right]+\left[-\mathrm{\pi }+4\frac{\sqrt{3}}{2}+\frac{\mathrm{\pi }}{3}\right]\\ I& =& \frac{\mathrm{\pi }}{3}+4\times \frac{\sqrt{3}}{2}-4-\mathrm{\pi }+4\frac{\sqrt{3}}{2}+\frac{\mathrm{\pi }}{3}\\ \mathrm{I}& =& 4\sqrt{3}-4-\frac{\mathrm{\pi }}{3}\end{array}$

Therefore, the correct answer is Option D.