Question

# Evaluate : ${\int }_{0}^{\mathrm{\pi }}\sqrt{1+4{\mathrm{sin}}^{2}\left(x/2\right)-4\mathrm{sin}\left(x/2\right)}dx$

A

$\mathrm{\pi }-4$

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

$\left(2\mathrm{\pi }/3\right)-4-4\sqrt{3}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

$4\sqrt{3}-4$

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

## 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{\mathrm{sin}}^{2}\frac{x}{2}-4\mathrm{sin}\frac{x}{2}}dx$$\begin{array}{rcl}I& =& {\int }_{0}^{\mathrm{\pi }}\sqrt{1+4\frac{{\mathrm{sin}}^{2}x}{2}-4\frac{\mathrm{sin}x}{2}}dx\\ I& =& {\int }_{0}^{\mathrm{\pi }}\sqrt{{\left(1\right)}^{2}+{\left(2\mathrm{sin}\left(\frac{x}{2}\right)\right)}^{2}-2×1×2\mathrm{sin}\left(\frac{x}{2}\right)}dx\\ I& =& {\int }_{0}^{\mathrm{\pi }}\sqrt{{\left(1-2\mathrm{sin}\left(\frac{x}{2}\right)\right)}^{2}}dx\\ I& =& \left|1-2\mathrm{sin}\left(\frac{x}{2}\right)\right|dx\end{array}$Step 2 : Find the value of $x$:$\begin{array}{rcl}\mathrm{sin}\left(\frac{x}{2}\right)& =& \frac{1}{2}\\ \frac{x}{2}& =& \frac{\mathrm{\pi }}{6}\\ x& =& \frac{\mathrm{\pi }}{3}\end{array}$Integrate the $I$ with respect to the limit $0$ to$\frac{\mathrm{\pi }}{3}$to $\mathrm{\pi }$.$\begin{array}{rcl}I& =& {\int }_{0}^{\frac{\mathrm{\pi }}{3}}\left(1-2\mathrm{sin}\left(\frac{x}{2}\right)\right)dx+{\int }_{\frac{\mathrm{\pi }}{3}}^{\mathrm{\pi }}\left(2\mathrm{sin}\left(\frac{x}{2}\right)-1\right)dx\\ I& =& \left(x+4\mathrm{cos}\frac{x}{2}\right)+{\left|\left(-4\mathrm{cos}\left(\frac{x}{2}\right)-x\right)\right|}_{\frac{\mathrm{\pi }}{3}}^{\mathrm{\pi }}\\ I& =& \left[\left(\frac{x}{3}+4\mathrm{cos}\left(\frac{\mathrm{\pi }}{6}\right)\right)-\left(0+4\mathrm{cos}0\right)\right]+\left[\left(-4\mathrm{cos}\left(\frac{\mathrm{\pi }}{2}\right)-\mathrm{\pi }\right)-\left(-4\mathrm{cos}\left(\frac{\mathrm{\pi }}{6}\right)\right)\right]\\ I& =& \left[\frac{\mathrm{\pi }}{3}+4×\frac{\sqrt{3}}{2}-4\right]+\left[-\mathrm{\pi }+4\frac{\sqrt{3}}{2}+\frac{\mathrm{\pi }}{3}\right]\\ I& =& \frac{\mathrm{\pi }}{3}+4×\frac{\sqrt{3}}{2}-4-\mathrm{\pi }+4\frac{\sqrt{3}}{2}+\frac{\mathrm{\pi }}{3}\\ \mathrm{I}& =& 4\sqrt{3}-4-\frac{\mathrm{\pi }}{3}\end{array}$Therefore, the correct answer is Option D.

Suggest Corrections
3
Join BYJU'S Learning Program
Related Videos
Integration of Piecewise Functions
MATHEMATICS
Watch in App
Join BYJU'S Learning Program