Answer
$\displaystyle \frac{3}{4}$
Work Step by Step
Applying th.7, let $ u=g(\theta)=\cos 2\theta$ (a continuous function).
Then, $ g'(\theta)=-\sin 2\theta\cdot 2=-2\sin 2\theta$
Then, $\quad \left[\begin{array}{ll}
u=\cos 2\theta, & du=g'(\theta) d\theta\\
& du=-2\sin 2\theta d\theta\\
& \sin 2\theta d\theta=\dfrac{du}{-2}
\end{array}\right]$,
and, the borders change to $g(0)=\cos 0=1$ and $\displaystyle \quad g(\pi/6)=(\cos\pi/3)=\frac{1}{2}$
$\displaystyle \int_{0}^{\pi/6}\cos^{-3}2\theta\cdot\sin 2\theta d\theta=\int_{1}^{1/2}u^{-3}\cdot\frac{du}{-2} =-\frac{1}{2}\int_{1}^{1/2}u^{-3}du$
$=-\displaystyle \frac{1}{2}\left[ \frac{u^{-3+1}}{-3+1} \right]_{1}^{1/2}$
$=-\displaystyle \frac{1}{2}\left[ \frac{u^{-2}}{-2} \right]_{1}^{1/2}$
$=\displaystyle \frac{1}{4}\left[ \frac{1}{u^{2}} \right]_{1}^{1/2}$
$=\displaystyle \frac{1}{4}\cdot(4-1)$
=$\displaystyle \frac{3}{4}$
