Answer
$3^{5/2}-1$
Work Step by Step
Applying th.7, let $u=g(t)=5-4\cos t \quad $(a continuous function).
Then, $g'(t)=-4(-\sin t)=4\sin t$
Then, $\quad \left[\begin{array}{ll}
u=5-4\cos t & du=g'(t)dt\\
& du=4\sin tdt\\
& \sin tdt=\frac{du}{4}
\end{array}\right]$,
and, the borders change to $g(0)=5-4=1$ and $\quad g(\pi)=5-4(-1)=9$
$\displaystyle \int_{0}^{\pi}5(5-4\cos t)\sin tdt=\int_{1}^{9}5u^{1/4}\cdot(\frac{du}{4}) =\frac{5}{4}\int_{1}^{9}u^{-1/4}du$
$=\displaystyle \frac{5}{4}\left[ \frac{u^{1/4+1}}{1/4+1} \right]_{1}^{9}$
$=\displaystyle \frac{5}{4}\left[ \frac{u^{5/4}}{5/4} \right]_{1}^{9}$
$=\displaystyle \frac{5}{4}\cdot\frac{4}{5}\left[ u^{5/4} \right]_{1}^{9}$
$=(9^{5/4}-1)$
$=[(3^{2})^{5/4}-1]$
$=3^{5/2}-1$
