## Thomas' Calculus 13th Edition

$\dfrac{99}{10}-2 \ln 10$
As we are given that $\int^{\ln 10}_{0} 4\sinh^2(\dfrac{x}{2}) dx$ Use formula, $\sinh x=\dfrac{e^{x} - e^{-x}}{2}$ Thus, $\int^{\ln 10}_{0} 4\sinh^2(\dfrac{x}{2}) dx=\int^{\ln 10}_{0} 4 [\dfrac{e^{x/2} - e^{-x/2}}{2}]^2 dx$ and $\int^{\ln 10}_{0} 4 [\dfrac{e^{x/2} - e^{-x/2}}{2}]^2 dx= \int^{\ln 10}_{0} (e^{x} + e^{-x} -2) dx=[ e^{x} - e^{-x}-2x]^{\ln 10}_{0}$ or, $=(10 -\dfrac{1}{10}-2 \ln 10)$ Hence, $\int^{\ln 10}_{0} 4\sinh^2(\dfrac{x}{2}) dx=\dfrac{99}{10}-2 \ln 10$