Answer
$\sin \pi -\sin 1$
or,
$-\sin 1$
Work Step by Step
Solve $\int_{0}^{\sqrt{\ln \pi}} 2xe^{x^2} \cos (e^{x^2}) dx$
Let us $p=e^{x^2}$ and $dp=2pe^{x^2}dx$
This implies $\int_{0}^{\sqrt{\ln \pi}} 2xe^{x^2} \cos (e^{x^2}) dx=\int_{0}^{\sqrt{\ln \pi}} \cos p dp$
$=[sin p]_{0}^{\sqrt{\ln \pi}} +c$
$=[sin e^{x^2}]_{0}^{\sqrt{\ln \pi}} +c$
Also, $=[\sin e^{(\sqrt{\ln \pi})^2}- \sin e^0] +c$
or, $=\sin \pi -\sin 1$
or, $=-\sin 1$