Answer
$\displaystyle \frac{\pi}{3}$
Work Step by Step
Applying th.7, let $u=g(\theta)=\theta^{3/2} \quad$ (a continuous function).
Then, $g'(\displaystyle \theta)=\frac{3}{2}\theta^{1/2}$
Then, $\quad \left[\begin{array}{ll}
u=y^{3}+6y^{2}-12y+9 & du=g'(\theta) dy\\
& du=\frac{3}{2}\theta^{1/2}d\theta \\
& \sqrt{\theta}d\theta=\frac{2du}{3}
\end{array}\right]$,
and, the borders change to
$g(0)=0$ and $\quad g(\pi^{2/3})=\pi$
$\displaystyle \int_{0}^{\pi^{2/3}}\sqrt{\theta}\cos^{2}(\theta^{3/2})d\theta=\int_{0}^{\pi}\cos^{2}u(\frac{2du}{3})$
$=\displaystyle \frac{2}{3}\int_{0}^{\pi}\cos^{2}udu$
Using the double angle formula for $\cos 2u$
$\cos 2u=\cos^{2}u-\sin^{2}u=\cos^{2}u-(1-\cos^{2}u)=2\cos^{2}u-1$,
We transform $\displaystyle \cos^{2}u= \frac{1}{2}(\cos 2u+1) $
$=\displaystyle \frac{2}{3}\cdot\frac{1}{2}\int_{0}^{\pi} (\cos 2u+1) du$
$=\displaystyle \frac{1}{3}\int_{0}^{\pi} (\cos 2u+1) du=\frac{1}{3}\left[\int_{0}^{\pi} \cos 2udu+ \int_{0}^{\pi}du\right]$
For the first integral, substitute: $\left[\begin{array}{ll}
t=2u, & dt=2du\\
& du=dt/2 \\
u=0\rightarrow t=0, & u=\pi\rightarrow t=2\pi
\end{array}\right]$
$=\displaystyle \frac{1}{3}\left[\int_{0}^{2\pi} \cos t(\frac{dt}{2})+ \int_{0}^{\pi}du\right]$
$=\displaystyle \frac{1}{6}\int_{0}^{2\pi} \cos tdt +\frac{1}{3}\left[ u \right]_{0}^{\pi}$
$=\displaystyle \frac{1}{6}\left[ \sin t \right]_{0}^{2\pi}+\frac{1}{3}(\pi-0)$
$=\displaystyle \frac{1}{6}(0)+\frac{\pi}{3}$
$=\displaystyle \frac{\pi}{3}$
