Section 5.8 #102: Let $x>0$ and $b>0$. Show that
$$\displaystyle\int_{-b}^be^{xt} \mathrm{d}t = \dfrac{2\sinh(bx)}{x}.$$
Solution: Note that $t$ is the variable being integrated and so the variable "$x$" in the integral is treated as a constant. This means we should use a $u$-substitution with $u=xt$ so that $\mathrm{d}u = x \mathrm{d}t$ (note we didn't do $\mathrm{d}u=t \mathrm{d}x$ which is not "untrue", but it is not useful for our purposes because the integral we're computing has a "$\mathrm{d}t$" in it) and so $\dfrac{1}{x} \mathrm{d}u = \mathrm{d}t$. Consequently, if $t=-b$ then $u=-xb$ and if $t=b$ then $u=xb$. Now compute
$$\begin{array}{ll}
\displaystyle\int_{-b}^b e^{xt} \mathrm{d}t &= \dfrac{1}{x} \displaystyle\int_{-xb}^{xb} e^u \mathrm{d}u \\
&= \dfrac{1}{x} e^u \Bigg|_{-xb}^{xb} \\
&= \dfrac{e^{xb} - e^{-xb}}{x}.
\end{array}$$
But recall the definition of $\sinh$ which was $\sinh(z)=\dfrac{e^z-e^{-z}}{2}$. From this we see that $2\sinh(z)=e^z-e^{-z}$. Therefore we see, using $z=xb$, that
$$\displaystyle\int_{-b}^b e^{xt}\mathrm{d}t = \dfrac{e^{xb}-e^{-xb}}{x} = \dfrac{2\sinh(xb)}{x},$$