## University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{3}{32}+\ln 2$
Given: $\int^{-\ln 2}_{-\ln 4} 2e^{\theta} \cosh \theta d\theta$ Since, $\cosh \theta=\dfrac{e^{\theta}+e^{-\theta}}{2}$ The given integral can be re-written as: $\int^{-\ln 2}_{-\ln 4} 2e^{\theta} [\dfrac{e^{\theta}+e^{-\theta}}{2}] d\theta$ or, $=\int^{-\ln 2}_{-\ln 4}(e^{2\theta} +1) d\theta$ or, $=[\dfrac{1}{2}e^{2\theta}+\theta]^{-\ln 2}_{-\ln 4}$ or, $=[\dfrac{1}{8}-\ln 2]-[\dfrac{1}{32}- 2 \ln 2]$ or, $=\dfrac{3}{32}+\ln 2$