${\int }_0^{\pi /2}{\sin }^2(x)dx$

Solution

1. Apply the trigonometric identity ${\sin }^2(x)=\frac{1-\cos (2x)}{2}$ to the integral.
2. Rewrite the integral as ${\int }_0^{\pi /2}\frac{1}{2}dx-{\int }_0^{\pi /2}\frac{\cos (2x)}{2}dx$.
3. Evaluate the antiderivative to get ${\left[\frac{x}{2}-\frac{\sin (2x)}{4}\right]}_0^{\pi /2}$.
4. Substitute the limits to calculate $\left(\frac{\pi }{4}-\frac{\sin (\pi )}{4}\right)-(0-0)=\frac{\pi }{4}$.
5. Hence the answer is (B).