## Calculus, 10th Edition (Anton)

$$\frac{\pi }{{16}}$$
\eqalign{ & \int_0^{\pi /2} {{{\sin }^2}\frac{x}{2}{{\cos }^2}\frac{x}{2}} dx \cr & = \frac{1}{4}\int_0^{\pi /2} {4{{\sin }^2}\frac{x}{2}{{\cos }^2}\frac{x}{2}} dx \cr & {\text{Use }}{\left( {ab} \right)^n} = {a^n}{b^n} \cr & = \frac{1}{4}\int_0^{\pi /2} {{{\left( {2\sin \frac{x}{2}\cos \frac{x}{2}} \right)}^2}} dx \cr & {\text{By the trigonometric identity }}\sin 2\theta = 2\sin \theta \cos \theta \cr & = \frac{1}{4}\int_0^{\pi /2} {{{\left( {\sin x} \right)}^2}} dx \cr & {\text{Use }}{\sin ^2}\theta = \frac{1}{2} - \frac{1}{2}\cos 2\theta \cr & = \frac{1}{4}\int_0^{\pi /2} {\left( {\frac{1}{2} - \frac{1}{2}\cos 2x} \right)} dx \cr & {\text{Integrating}} \cr & = \frac{1}{4}\left( {\frac{1}{2}x - \frac{1}{4}\sin 2x} \right)_0^{\pi /2} \cr & = \frac{1}{4}\left( {\frac{1}{2}\left( {\frac{\pi }{2}} \right) - \frac{1}{4}\sin 2\left( {\frac{\pi }{2}} \right)} \right) - \frac{1}{8}\left( {\frac{1}{2}\left( 0 \right) - \frac{1}{4}\sin 2\left( 0 \right)} \right) \cr & = \frac{1}{4}\left( {\frac{\pi }{4}} \right) - \frac{1}{8}\left( 0 \right) \cr & = \frac{\pi }{{16}} \cr}