Answer
$ \frac{1}{5}(1-\frac{9\sqrt3}{32})$
Work Step by Step
Since $u=\cos x$, then $du=-\sin x dx$; moreover, when $x:0 \to \pi/6$, then $u:1 \to \sqrt 3/2$. Now, we have
$$
\int_{0}^{\pi / 6} \sin x \cos ^{4} x d x=-\int_{1}^{\sqrt 3/ 2} u ^{4} d u\\
=-\frac{1}{5} u^5|_1^{\sqrt 3/2}= \frac{1}{5}(1-\frac{9\sqrt3}{32}).$$
