Single Variable Calculus, Chapter 5, Review Exercises, Section Review Exercises, Problem 18

Find the intergral $\displaystyle \int^1_0 \sin (3 \pi t) dt$, if it exists.
If we let $u = 3 \pi t$, then $du = 3 \pi dt$, so $\displaystyle dt = \frac{du}{3\pi}$. When $t = 0$, $u = 0$ and when $t = 1$, $u = 3\pi$. Therefore

$\begin{equation} \begin{aligned} \int^1_0 \sin (3 \pi t) dt &= \int^1_0 \sin u \frac{du}{3\pi}\\ \\ \int^1_0 \sin (3 \pi t) dt &= \frac{1}{3\pi} \int^1_0 \sin u du\\ \\ \int^1_0 \sin (3 \pi t) dt &= \left. \cdot - \cos u \right|^1_0\\ \\ \int^1_0 \sin (3 \pi t) dt &= \frac{1}{3\pi} \left[ - \cos ( 3 \pi ) + \cos (0) \right]\\ \\ \int^1_0 \sin (3 \pi t) dt &= \frac{1}{3\pi} (2)\\ \\ \int^1_0 \sin (3 \pi t) dt &= \frac{2}{3 \pi} \end{aligned} \end{equation}$