Let $I=\int_0^{\pi / 2} \sqrt{\sin \theta} \cdot \cos ^3 \theta d \theta$
Let $\sin \theta=t \Rightarrow \cos \theta d \theta=d t$
When $\quad \theta=0 \Rightarrow t=0$

When
$$
\begin{aligned}
& \theta=\frac{\pi}{2} \Rightarrow 1=1 \\
& I=\int_0^1 \sqrt{t}\left(\left(-t^2\right) d t=\int_0^1\left(t^{1 / 2}-t^{5 / 2}\right) \cdot d t\right. \\
& =\left[\begin{array}{r}
3 \\
t^2 \\
3 \\
2
\end{array}\right]_0^1-\left[\begin{array}{r}
7 \\
t^2 \\
7 \\
2
\end{array}\right]_0^1=\frac{2}{3}[1-0]-\frac{2}{7}[1-0] \\
& =\frac{2}{3}-\frac{2}{7}=\frac{14-6}{21}=\frac{8}{21} \\
&
\end{aligned}
$$