Answer
$\displaystyle \frac{3}{32}+\ln 2$
Work Step by Step
Evaluate the integral as follows:
$\displaystyle \cosh\theta=\frac{e^{\theta}+e^{-\theta}}{2}$
$\displaystyle \int_{-\ln 4}^{-\ln 2}2e^{\theta}\cosh\theta d\theta=\int_{-\ln 4}^{-\ln 2}2e^{\theta}(\frac{e^{\theta}+e^{-\theta}}{2})d\theta$
$=\displaystyle \int_{-\ln 4}^{-\ln 2}(e^{2\theta}+1)d\theta\qquad\left[\begin{array}{ll}
t=e^{2\theta} & dt=2e^{2\theta}d\theta\\
\theta=-\ln 4 & \rightarrow t=\frac{1}{e^{2\ln 4}}=\frac{1}{4^{2}}\\
\theta=-\ln 2 & \rightarrow t=\frac{1}{e^{2\ln 2}}=\frac{1}{2^{2}}
\end{array}\right]$
$=\displaystyle \int_{1/16}^{1/4}\frac{1}{2}dt+\int_{-\ln 4}^{-\ln 2}d\theta$
$=\displaystyle \frac{1}{2}[t]_{1/16}^{1/4}+[\theta]_{-\ln 4}^{-\ln 2}$
$=\displaystyle \frac{1}{4}(\frac{1}{4}-\frac{1}{16})+(-\ln 2+\ln 4)$
$=\displaystyle \frac{1}{4}(\frac{4-1}{16})+(-\ln 2+2\ln 2)$
$=\displaystyle \frac{3}{32}+\ln 2$
