Question

The value of ${\int }_0^{\pi /2}{\cos }^{2}xdx$ is

• A. $0$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{3}$

Answer 1

Answer: (C)

$\frac{\pi }{4}$

Let $l={\int }_0^{\pi /2}{\cos }^{2}xdx={\int }_0^{\pi /2}\left[\frac{\cos 2x+1}{2}\right]dx$
$=\frac{1}{2}[{\int }_0^{\pi /2}\cos 2xdx+{\int }_0^{\pi /2}dx]$
$=\frac{1}{2}{\left[\frac{\sin 2x}{2}\right]}_0^{\pi /2}+\frac{1}{2}[x{]}_0^{\pi /2}$
$=\frac{1}{4}[\sin \pi -\sin 0]+\frac{1}{2}[\frac{\pi }{2}-0]$
$=\frac{1}{4}[0-0]+\frac{\pi }{4}=\frac{\pi }{4}$.