## Thomas' Calculus 13th Edition

$2 \ln 2 -\dfrac{3}{4}$
As we are given that $\int^{\ln 2}_{0} 4e^{-\theta} \sinh \theta d\theta$ Since, $\sinh \theta=\dfrac{e^{\theta} -e^{-\theta}}{2}$ The given integral $\int^{\ln 2}_{0} 4(e^{-\theta}) \sinh \theta d\theta$ can be re-written as:$\int^{\ln 2}_{0} 4e^{-\theta} [\dfrac{e^{\theta} -e^{-\theta}}{2}] d\theta= 2[ \theta +\dfrac{1}{2} e^{-2\theta}]^{\ln 2}_{0}$ or, $=2 [\ln 2+(\dfrac{1}{8})]-1$ Thus, $=2 \ln 2 -\dfrac{3}{4}$