Answer
$e-e^{-1}$
Work Step by Step
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, $=e-e^{-1}$