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

$e-e^{-1}$
Given: $\int^{\pi/4}_{-\pi/4} \cosh(\tan \theta)sec^2 \theta d\theta$ Plug in: $\tan \theta=t$ and $dt= \sec^2 \theta d\theta$ Thus, the limits of integration will also get changed. Then, $\int^{\pi/4}_{-\pi/4} \cosh(\tan \theta)sec^2 \theta d\theta=\int^{1}_{-1} \cosh t dt$ or, $=[\sin h t]^{1}_{-1}$ or, $=2 \sinh (1)$ Since, $\sinh \theta=\dfrac{e^{\theta} -e^{-\theta}}{2}$ Thus, we have, $= 2[\dfrac{e^{1} -e^{-1}}{2}]$ or, $=e-e^{-1}$