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

$\dfrac{1}{2}\ln (\dfrac{17}{8})$
Given: $\int^{\ln 2}_{0} \tanh 2x dx$ This can be re-written as:$\int^{\ln 2}_{0} \dfrac{\sinh 2x}{\cosh 2x} dx$ Plug in: $\cosh 2x=t$ and $dt=2 \sinh 2x dx$ Thus, the limits of integration will also get changed. Then, $\int^{\ln 2}_{0} \dfrac{\sinh 2x}{\cosh 2x} dx=\dfrac{1}{2}\int^{17/8}_{1} \dfrac{dt}{t} =\dfrac{1}{2}[\ln |t|]^{17/8}_{1}=\dfrac{1}{2}[\ln \dfrac{17}{8}-\ln(1)]$ or, $=\dfrac{1}{2}\ln (\dfrac{17}{8})$