Answer
$a.\quad \displaystyle \frac{1}{6}$
$ b.\quad \displaystyle \frac{1}{2}$
Work Step by Step
$a.$
Applying th.7, let $u=g(t)=1-\cos 3t$ (a continuous function).
Then, $\quad du=g'(t)dt=-[(-\sin 3t)\cdot 3]\cdot dt=(3\sin 3t)dt $
and, the borders change to $g(0)=1-1=0$ and $g(\displaystyle \frac{\pi}{6})=1-\cos\frac{\pi}{2}=1.$
$\displaystyle \int_{0}^{\pi/6}(1-\cos 3t)\sin 3tdt=\int_{0}^{1}u\cdot\frac{1}{3}du =\frac{1}{3} \int_{0}^{1}udu$
$=\displaystyle \frac{1}{3}\left[ \frac{u^{2}}{2} \right]_{0}^{1}$
$=\displaystyle \frac{1}{3}(\frac{1}{2}-0)$
$=\displaystyle \frac{1}{6}$
$a.$
Applying the same substitution,
the borders here change to $g(\displaystyle \frac{\pi}{6})=1-\cos\frac{\pi}{2}=1$ and $g(\displaystyle \frac{\pi}{3})=1-\cos\pi=2$
$\displaystyle \int_{\pi/6}^{\pi/3}(1-\cos 3t)\sin 3tdt=\int_{1}^{2}u\cdot\frac{1}{3}du =\frac{1}{3} \int_{1}^{2}udu$
$=\displaystyle \frac{1}{3}\left[ \frac{u^{2}}{2} \right]_{1}^{2}$
$=\displaystyle \frac{1}{3}(\frac{4}{2}-\frac{1}{2})$
$=\displaystyle \frac{1}{3}(\frac{3}{2})$
$=\displaystyle \frac{1}{2}$
