Solution:

let $\quad I=\int_{0}^{\pi / 6} \cos ^{6} 3 \theta \cdot \sin ^{2} 6 \theta d \theta$

Put $\quad 3 \theta=t$

Diff. w.r.t $\theta$

$\quad d \theta=\frac{d t}{3} \quad$ limits: $\left[0, \frac{\pi}{2}\right]$

$\begin{aligned} \therefore \mathrm{I} &=\frac{1}{3} \int_{0}^{\pi / 2} \cos ^{6} t \cdot \sin ^{2} 2 t d t \\ &=\frac{4}{3} \int_{0}^{\pi / 2} \cos ^{3} t(\sin t \cdot \cos t)^{2} d t \\ &=\frac{4}{3} \int_{0}^{\pi / 2} \cos ^{5} t \cdot \sin ^{2} t d t \end{aligned}$

$\begin{aligned}=& \frac{4}{3} \times \frac{1}{2} \times \beta\left(3, \frac{3}{2}\right) \\ \therefore I=& \frac{32}{315} \end{aligned}$