Answer
$1$
Work Step by Step
Solve $\int_{\ln(\pi/6)}^{\ln(\pi/2)} 2e^{v} \cos e^v dv$
Let us $p=e^{v}$ and $dp=e^{v} dv$
This implies $\int_{\ln(\pi/6)}^{\ln(\pi/2)} 2e^{v} \cos e^v dv=2\int_{\ln(\pi/6)}^{\ln(\pi/2)} \cos p dp$
$=2[sin p]_{\ln(\pi/6)}^{\ln(\pi/2)} +c$
$=2[sin e^{v}]_{\ln(\pi/6)}^{\ln(\pi/2)} +c$
Also, $=2[sin e^{\ln(\pi/2)}-sin e^{\ln(\pi/6)}] +c$
or, $=2[sin (\pi/2)-sin (\pi/6)] +c$
$=2(1-0.5)$
$=1$
