Question

Evaluate the following definite integrals :

${\int }_0^{\pi /2}{\cos }^{2}xdx$

• A. $\frac{\pi }{4}$
• B. $1$
• C. $\frac{\pi }{5}$
• D. None of these

Answer 1

The correct option is A $\frac{\pi }{4}$

$I={\int }_0^{\pi /2}{\cos }^{2}xdx$

$={\int }_0^{\pi /2}\frac{1+\cos 2x}{2}dx$........$using(\cos 2x=2co{s}^{2}x-1)$

$=\frac{1}{2}{[x+\frac{\sin 2x}{2}]}_0^{\pi /2}$

$=\frac{1}{2}[(\frac{\pi }{2}+\frac{\sin \pi }{2})-(0+\frac{\sin 0}{2})]=\frac{\pi }{4}$