Answer
$$ - 4$$
Work Step by Step
$$\eqalign{
& \int_{{\pi ^2}}^{4{\pi ^2}} {\frac{1}{{\sqrt x }}\sin \sqrt x } dx \cr
& u = \sqrt x ,{\text{ then }}du = \frac{1}{{2\sqrt x }}dx,{\text{ }}2du = \frac{1}{{\sqrt x }}dx \cr
& {\text{With this substitution}}{\text{,}} \cr
& {\text{if }}x = 4{\pi ^2},{\text{ }}u = \left( {\sqrt {4{\pi ^2}} } \right) = 2\pi \cr
& {\text{if }}x = {\pi ^2},{\text{ }}u = \sqrt {{\pi ^2}} = \pi \cr
& {\text{so}} \cr
& \int_{{\pi ^2}}^{4{\pi ^2}} {\frac{1}{{\sqrt x }}\sin \sqrt x } dx = \int_\pi ^{2\pi } {\sin u} \left( {2du} \right) \cr
& = 2\int_\pi ^{2\pi } {\sin u} du \cr
& {\text{find the antiderivative }} \cr
& = 2\left( { - \cos u} \right)_\pi ^{2\pi } \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& = - 2\left( {\cos 2\pi - \cos \pi } \right) \cr
& = - 2\left( {1 - \left( { - 1} \right)} \right) \cr
& = - 4 \cr} $$
