University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{99}{10}-2 \ln 10$
Given: $\int^{\ln 10}_{0} 4\sinh^2(\dfrac{x}{2}) dx$ Since, $\sinh x=\dfrac{e^{x} - e^{-x}}{2}$ Re-write as:$\int^{\ln 10}_{0} 4 [\dfrac{e^{x/2} - e^{-x/2}}{2}]^2 dx$ Then, $\int^{\ln 10}_{0} 4 [\dfrac{e^{x/2} - e^{-x/2}}{2}]^2 dx= \int^{\ln 10}_{0} (e^{x} + e^{-x} -2) dx$ or, $=[ e^{x} - e^{-x}-2x]^{\ln 10}_{0}$ or, $=10 -\dfrac{1}{10}-2 \ln 10$ or, $=\dfrac{99}{10}-2 \ln 10$