Answer
$$\int_0^\pi\sin^2{t}\cos{t}\hspace{0.5mm}dt=0$$
Work Step by Step
Let $u=\sin{t}$
then $du=\cos{t}\hspace{0.5mm}dt$
and therefore $\frac{du}{\cos{t}}=dt.$
Substituting into our integral we get
$\int \sin^2{t}\cos{t}\hspace{0.5mm}dt=\int u^2\cos{t}\times\frac{du}{\cos{t}}=\int u^2\hspace{0.5mm}du$
$=\frac{u^3}{3}+C=\frac{\sin^3{t}}{3}+C$
Plugging in our integration limits we get
$\frac{\sin^3{t}}{3}\bigg\vert_0^\pi=\frac{\sin^3{\pi}}{3}-\frac{\sin^3{0}}{3}=0-0=0$
