Answer
\begin{aligned}
\int_{0}^{\pi / 6} 3 \cos ^{5} 3 x d x =\frac{8}{15}
\end{aligned}
Work Step by Step
Given $$ \int_{0}^{\pi / 6} 3 \cos ^{5} 3 x d x $$
So, we have
\begin{aligned}
I&= \int_{0}^{\pi / 6} 3 \cos ^{5} 3 x d x\\
&=\int_{0}^{\pi / 6}\left(\cos ^{2} 3 x\right)^{2} \cos 3 x \cdot 3 d x\\
&=\int_{0}^{\pi / 6}\left(1-\sin ^{2} 3 x\right)^{2} \cos 3 x \cdot 3 d x\\
&=\int_{0}^{\pi / 6}\left(1-2 \sin ^{2} 3 x+\sin ^{4} 3 x\right) \cos 3 x \cdot 3 d x\\
&=\int_{0}^{\pi / 6} \cos 3 x \cdot 3 d x-2 \int_{0}^{\pi / 6} \sin ^{2} 3 x \cos 3 x \cdot 3 d x+\int_{0}^{\pi / 6} \sin ^{4} 3 x \cdot \cos 3 x \cdot 3 d x\\
&=\left[\sin 3 x-2 \frac{\sin ^{3} 3x}{3}+\frac{\sin ^{3}3 x}{5}\right]_{0}^{\pi / 6}\\
&=\left[\sin 3\pi / 2-2 \frac{\sin ^{3} \pi / 2}{3}+\frac{\sin ^{3} \pi / 2}{5}\right] -\left[\sin 0-2 \frac{\sin ^{3} 0}{3}+\frac{\sin ^{3} 0}{5}\right] \\
&=\left(1-\frac{2}{3}+\frac{1}{5}\right)-(0)\\
&=\frac{8}{15}
\end{aligned}
