Question

Solve ${\int }_{\pi /6}^{\pi /3}{\cos }^{2}xdx$

Answer 1

$\begin{array}{c}{\int }_{\pi /6}^{\pi /3}{\cos }^{2}xdx\\ =\frac{1}{2}{\int }_{\pi /6}^{\pi /3}{\cos }^{2}xdx\\ =\frac{1}{2}{\int }_{\pi /6}^{\pi /3}\left(1+\cos 2x\right)dx\\ =\frac{1}{2}{{\left[x+\frac{\sin 2x}{2}\right]}_{\pi /6}}^{\pi /3}\\ =\frac{1}{2}\left[\left(\frac{\pi }{3}+\frac{1}{2}\sin \frac{2\pi }{3}\right)-\left(\frac{\pi }{6}+\frac{1}{2}+\sin \frac{\pi }{3}\right)\right]\\ =\frac{1}{2}\left[\left(\frac{\pi }{3}+\frac{1}{2}\times \sin \frac{\pi }{3}\right)-\left(\frac{\pi }{2}+\frac{1}{2}\times \frac{\sqrt{3}}{2}\right)\right]\\ =\frac{1}{2}\left[\left(\frac{\pi }{3}+\frac{1}{2}\times \frac{\sqrt{3}}{3}\right)-\left(\frac{\pi }{2}+\frac{\sqrt{3}}{4}\right)\right]\\ =\frac{1}{2}\left[\frac{\pi }{3}+\frac{\sqrt{3}}{4}-\frac{\pi }{2}-\frac{\sqrt{3}}{4}\right]\\ =\frac{1}{2}\left[\frac{-\pi }{6}\right]\\ =\frac{-\pi }{12}\end{array}$