Answer
$$-\sin \left( 1 \right)$$
Work Step by Step
$$\eqalign{
& \int_0^{\sqrt {\ln \pi } } {2x{e^{{x^2}}}\cos \left( {{e^{{x^2}}}} \right)} dx \cr
& {\text{Use substitution}}{\text{. Let }}u = {e^{{x^2}}},{\text{ so that }}\frac{{du}}{{dx}} = 2x{e^{{x^2}}} \cr
& {\text{The new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}\theta = \sqrt {\ln \pi },{\text{ then }}u = {e^{{{\left( {\sqrt {\ln \pi } } \right)}^2}}} = \pi \cr
& \,\,\,\,\,\,{\text{If }}\theta = 0,{\text{ then }}u = {e^{{{\left( 0 \right)}^2}}} = 1 \cr
& {\text{Then}} \cr
& \int_0^{\sqrt {\ln \pi } } {2x{e^{{x^2}}}\cos \left( {{e^{{x^2}}}} \right)} dx = \int_1^\pi {\cos u} du \cr
& {\text{Integrate}} \cr
& \int_0^\pi {\cos u} du = \left( {\sin u} \right)_1^\pi \cr
& {\text{Use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr
& = \sin \left( \pi \right) - \sin \left( 1 \right) \cr
& {\text{Simplifying}} \cr
& = 0 - \sin \left( 1 \right) \cr
& = -\sin \left( 1 \right) \cr} $$
