Answer
$e+e^{-1}-2$
Work Step by Step
Given: $\int^{\pi/2}_{0} 2\sinh(\sin \theta) \cos \theta d\theta$
Plug in: $\sin \theta=t$ and $dt= \cos \theta d\theta$
Thus, the limits of integration will also get changed.
Then, $\int^{\pi/2}_{0} 2\sinh(\sin \theta) \cos \theta d\theta=\int^{1}_{0} 2 \sinh t dt$
or, $=[2 \cosh h t]^{1}_{0}$
or, $=2 \cosh (1) -2$
Since, $\cosh \theta=\dfrac{e^{\theta} + e^{-\theta}}{2}$
Thus, we have, $= 2[\dfrac{e^{1} + e^{-1}}{2}]-2$
or, $=e+e^{-1}-2$
