Answer
$$\frac{23}{4880}$$
Work Step by Step
(a)
$\int_{0}^{\frac{\pi}{6}} \cos ^{3}(x) \sin ^{4}(x) d x=\frac{23}{4480} \approx 0.0051339$
(b) Use the given: $\int_{0}^{\pi / 6} \cos ^{3} x \sin ^{4} x d x=\int_{0}^{\pi / 6}\left(-\sin ^{6} x+\sin ^{4} x\right) \cos x d x$
Use substitution: $d u=\cos x d x$ and $u=\sin x$
$\int_{0}^{1 / 2}\left(-u^{6}+u^{4} x\right) d x=\left.\left(-\frac{u^{7}}{7}+\frac{u^{5}}{5}\right)\right|_{0} ^{1 / 2}=-\frac{1}{896}+\frac{1}{160}=\frac{23}{4880}$
