Answer
$$0$$
Work Step by Step
$$\eqalign{
& \int_0^{\sqrt \pi } {5x\cos \left( {{x^2}} \right)} dx \cr
& u = {x^2},{\text{ then }}du = 2xdx,{\text{ }}\frac{{du}}{2} = xdx \cr
& {\text{With this substitution}}{\text{,}} \cr
& {\text{if }}x = \sqrt \pi ,{\text{ }}u = {\left( {\sqrt \pi } \right)^2} = \pi \cr
& {\text{if }}x = 0,{\text{ }}u = {\left( 0 \right)^2} = 0 \cr
& {\text{so}} \cr
& \int_0^{\sqrt \pi } {5x\cos \left( {{x^2}} \right)} dx = \int_0^\pi {\cos u} \left( {\frac{1}{2}du} \right) \cr
& = \frac{1}{2}\int_0^\pi {\cos udu} \cr
& {\text{find the antiderivative }} \cr
& = \frac{1}{2}\left( {\sin u} \right)_0^\pi \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& = \frac{1}{2}\left( {\sin \left( \pi \right) - \sin \left( 0 \right)} \right) \cr
& = \frac{1}{2}\left( 0 \right) \cr
& = 0 \cr} $$
