Question

Evaluate :

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

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{4}$
• C. $\pi$
• D. None of these

Answer 1

Answer: (B)

$\frac{\pi }{4}$

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

$I=\frac{1}{2}{\int }_0^{\pi /2}(1-\cos 2x)dx$

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

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