## Calculus: Early Transcendentals 8th Edition

$\displaystyle \frac{2}{15}$
Leave one sine and use $\sin^{2}t+\cos^{2}t=1.$ $\displaystyle \int_{0}^{\pi/2}\sin^{3}\theta\cos^{2}\theta d\theta=\int_{0}^{\pi/2}\left(1-\cos^{2}\theta\right)\cos^{2}\theta\sin\theta d\theta=$ $\left[\begin{array}{ll} u=\cos\theta & \\ du=-\sin\theta d\theta & \\ \theta=0\Rightarrow u=1, & \theta=\pi/2\Rightarrow u=0 \end{array}\right]$ $=\displaystyle \int_{1}^{0}\left(1-u^{2}\right)u^{2}(-du)\quad$ $=\displaystyle \int_{0}^{1}\left(u^{2}-u^{4}\right)du$ $=\displaystyle \left[\frac{1}{3}u^{3}-\frac{1}{5}u^{5}\right]_{0}^{1}$ $=\displaystyle \left(\frac{1}{3}-\frac{1}{5}\right)-0$ $=\displaystyle \frac{2}{15}$