Question

Evaluate: ${\int }_{-\pi /3}^{\pi /3}{\cos }^{2}xdx$

• A. $\sqrt{3}/4$
• B. $\pi /3$
• C. $\frac{\pi }{3}+\frac{\sqrt{3}}{4}$
• D. $\frac{\pi }{3}-\frac{\sqrt{3}}{4}$

Answer 1

The correct option is C $\frac{\pi }{3}+\frac{\sqrt{3}}{4}$

Given

${\int }_{\frac{-\pi }{3}}^{\frac{\pi }{3}}{\cos }^2\left(x\right)dx$

$={\int }_{\frac{-\pi }{3}}^{\frac{\pi }{3}}\frac{1+\cos \left(2x\right)}{2}dx$ $\because {\cos }^2\left(x\right)=\frac{1+\cos \left(2x\right)}{2}$

$={\int }_{\frac{-\pi }{3}}^{\frac{\pi }{3}}\frac{1}{2}dx+{\int }_{\frac{-\pi }{3}}^{\frac{\pi }{3}}\frac{\cos \left(2x\right)}{2}dx$

$=[\frac{x}{2}{]}_{\frac{-\pi }{3}}^{\frac{\pi }{3}}+\frac{1}{2}[\frac{\sin \left(2x\right)}{2}{]}_{\frac{-\pi }{3}}^{\frac{\pi }{3}}$

$=[\frac{\pi }{6}+\frac{\pi }{6}]+\frac{1}{2}[\frac{\pi }{4}+\frac{\pi }{4}]$

$=\frac{\pi }{3}+\frac{1}{2}\frac{2\sqrt{3}}{4}$

$=\frac{\pi }{3}+\frac{\sqrt{3}}{4}$