Answer
$a)\int ^{5}_{0}\left( \sin x\right) ^{2}\cos xdx=\dfrac {\left( \sin x\right) ^{3}}{3}+c$$
$$b)\int ^{5}_{0}\left( \sin x\right) ^{2}\cos xdx=\dfrac {\left( \sin 5\right) ^{3}-\left( \sin 0\right) ^{3}}{3}\approx -0.29$
Work Step by Step
$u=\sin x\Rightarrow du=\cos xdx$
$$a)\int ^{5}_{0}\left( \sin x\right) ^{2}\cos xdx=\int ^{5}_{0}u^{2}du=\dfrac {u^{3}}{3}+c=\dfrac {\left( \sin x\right) ^{3}}{3}+c$$
$$b)\int ^{5}_{0}\left( \sin x\right) ^{2}\cos xdx=\dfrac {\left( \sin x\right) ^{3}}{3}]^{5}_{0}=\dfrac {\left( \sin 5\right) ^{3}-\left( \sin 0\right) ^{3}}{3}\approx -0.29$$
