Answer
$2 \ln 2 -\dfrac{3}{4}$
Work Step by Step
Given: $\int^{\ln 2}_{0} 4e^{-\theta} \sinh \theta d\theta$
Since, $\sinh \theta=\dfrac{e^{\theta} -e^{-\theta}}{2}$
The given integral can be re-written as: $\int^{\ln 2}_{0} 4e^{-\theta} [\dfrac{e^{\theta} -e^{-\theta}}{2}] d\theta$
or, $= 2[ \theta +\dfrac{1}{2} e^{-2\theta}]^{\ln 2}_{0}$
or, $=2 [\ln 2+\dfrac{1}{8}]-1$
or, $=2 \ln 2 -\dfrac{3}{4}$
