Answer
$$1$$
Work Step by Step
$$\eqalign{
& \int_{\ln \left( {\pi /6} \right)}^{\ln \left( {\pi /2} \right)} {2{e^v}\cos {e^v}} dv \cr
& = 2\int_{\ln \left( {\pi /6} \right)}^{\ln \left( {\pi /2} \right)} {{e^v}\cos {e^v}} dv \cr
& {\text{Use substitution}}{\text{. Let }}u = {e^v},{\text{ so that }}du = {e^v}dv \cr
& {\text{The new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}\theta = \ln \left( {\pi /2} \right),{\text{ then }}u = {e^{\ln \left( {\pi /2} \right)}} = \frac{\pi }{2} \cr
& \,\,\,\,\,\,{\text{If }}\theta = \ln \left( {\pi /6} \right),{\text{ then }}u = {e^{\ln \left( {\pi /2} \right)}} = \frac{\pi }{6} \cr
& {\text{Then}} \cr
& 2\int_{\ln \left( {\pi /6} \right)}^{\ln \left( {\pi /2} \right)} {{e^v}\cos {e^v}} dv = 2\int_{\pi /6}^{\pi /2} {\cos u} du \cr
& {\text{Integrate}} \cr
& \int_{\pi /6}^{\pi /2} {\cos u} du = 2\left( {\sin u} \right)_{\pi /6}^{\pi /2} \cr
& {\text{Use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr
& = 2\sin \left( {\frac{\pi }{2}} \right) - 2\sin \left( {\frac{\pi }{6}} \right) \cr
& {\text{Simplifying}} \cr
& = 2 - 2\left( {\frac{1}{2}} \right) \cr
& = 1 \cr} $$
