Question

Solve the integral $I={\int }_0^{\pi }{\sin }^{2}xdx$

• A. $\pi$
• B. $\frac{\pi }{2}$
• C. $0$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is B $\frac{\pi }{2}$

We have, $I={\int }_0^{\pi }{\sin }^{2}xdx$
Using trigonometric identity ${\sin }^{2}x=\frac{1-\cos 2x}{2}$
$I={\int }_0^{\pi }\frac{[1-\cos 2x]}{2}dx$
$=\frac{1}{2}[{\int }_0^{\pi }dx-{\int }_0^{\pi }\cos 2xdx]$
$=\frac{1}{2}{[x-\frac{\sin 2x}{2}]}_0^{\pi }$
$=\frac{1}{2}\left[\right(\pi -0)-\frac{(\sin 2\pi -\sin 0)}{2}]$
$\therefore I=\frac{\pi }{2}$