Answer
$\displaystyle \frac{1}{5}$
Work Step by Step
Applying th.7, let $u=g(t)=1-\sin 2t \quad $(a continuous function).
Then, $g'(t)=-(\cos 2t\cdot 2)=-2\cos 2t$
Then, $\quad \left[\begin{array}{ll}
u=5-4\cos t & du=g'(t)dt\\
& du=-2\cos 2tdt \\
& \cos 2tdt=\frac{du}{-2}
\end{array}\right]$,
and, the borders change to $g(0)=1-1=0$ and $\quad g(\pi/4)=1-0=1$
$\displaystyle \int_{0}^{\pi}(1-\sin 2t)^{3/2}\cos 2tdt=\int_{1}^{0}u^{3/2}\cdot(\frac{du}{-2}) =-\frac{1}{2}\int_{1}^{0}u^{3/2}du$
$=-\displaystyle \frac{1}{2}\left[ \frac{u^{3/2+1}}{3/2+1} \right]_{1}^{0}$
$=-\displaystyle \frac{1}{2}\left[ \frac{u^{5/2}}{5/2} \right]_{1}^{0}$
$=-\displaystyle \frac{1}{2}\cdot\frac{2}{5}\left[ u^{5/2} \right]_{1}^{0}$
$=-\displaystyle \frac{1}{5}\left[ u^{5/2} \right]_{1}^{0}$
$=-\displaystyle \frac{1}{5}(0-1)$
$=\displaystyle \frac{1}{5}$
