Answer
$\int_{0}^{1}\sin~(3\pi~t)~dt = \frac{2}{3\pi}$
Work Step by Step
$\int_{0}^{1}\sin~(3\pi~t)~dt$
Let $u = 3\pi~t$
$\frac{du}{dt} = 3\pi$
$dt = \frac{du}{3\pi}$
When $t = 0$, then $u = 0$
When $t = 1$, then $u = 3\pi$
$\int_{0}^{3\pi} \frac{1}{3\pi}~sin~u~du$
$=\frac{1}{3\pi}(-cos~u)~\vert_{0}^{3\pi}$
$=\frac{1}{3\pi}[(-cos~3\pi)-(-cos~0)]$
$=\frac{1}{3\pi}[-(-1)-(-1)]$
$=\frac{1}{3\pi}(1+1)$
$=\frac{2}{3\pi}$
