Đáp án:
$\begin{array}{l}
x = {\mathop{\rm sinu}\nolimits} \\
\Rightarrow dx = \cos udu\\
\Rightarrow \left\{ \begin{array}{l}
x = 0 \Rightarrow u = 0\\
x = 1 \Rightarrow u = \frac{\pi }{2}
\end{array} \right.\\
I = \int\limits_0^{\pi /2} {{{\sin }^2}u.\sqrt {1 - {{\sin }^2}u} } .\cos udu\\
= \int\limits_0^{\pi /2} {{{\sin }^2}u.\cos u} .\cos udu\\
= \int\limits_0^{\pi /2} {\frac{1}{4}.4.{{\sin }^2}u.co{s^2}u} du\\
= \frac{1}{4}.\int\limits_0^{\pi /2} {{{\sin }^2}2u} .du\\
= \frac{1}{{16}}\int\limits_0^{\pi /2} {\frac{{1 - \cos 4u}}{2}} .d4u\\
= \frac{1}{{16}}.\left( {\frac{1}{2}.4u - \frac{1}{2}\sin 4u} \right)_0^{\pi /2}\\
= \frac{\pi }{{16}}
\end{array}$
