Answer
$$\int_{0}^{\pi/2}sin^{7}\theta\,cos^{5}\theta\,d\theta=\frac{1}{120}$$
Work Step by Step
$$\int_{0}^{\pi/2}sin^{7}\theta\,cos^{5}\theta\,d\theta=\int_{0}^{\pi/2}sin^{7}\theta\,(1-sin^{2}\theta)^{2}cos\theta\,d\theta$$
$Let\,t=sin\theta,\,dt=cos\theta\,d\theta,\int_{0}^{\pi/2}=>\int_{0}^{1}$
$$\int_{0}^{\pi/2}sin^{7}\theta\,(1-sin^{2}\theta)^{2}cos\theta\,d\theta=\int_{0}^{1}t^{7}(1-t^{2})^{2}dt$$
$$=\int_{0}^{1}(t^{7}-2t^{9}+t^{11})dt$$
$$==\left [\frac{t^{8}}{8}-\frac{t^{10}}{5}+\frac{t^{12}}{12} \right ]_{0}^{1}$$
$$=\frac{1}{120}$$
