Answer
$$2$$
Work Step by Step
$$\eqalign{
& \int_0^{{\pi ^2}/4} {\sin \sqrt x } dx \cr
& {\text{Let }}t = \sqrt x ,{\text{ }}{t^2} = x,{\text{ }}2tdt = dx \cr
& {\text{Use the substitution}} \cr
& \int {\sin \sqrt x } dx = \int {\sin t} \left( {2t} \right)dt \cr
& = 2\int {t\sin t} dt \cr
& {\text{Integrate by parts,}} \cr
& {\text{Let }}u = t,{\text{ }}du = dt \cr
& dv = \sin tdt,{\text{ }}v = - \cos t \cr
& 2\int {t\sin t} dt = 2\left( { - t\cos t + \int {\cos t} dt} \right) \cr
& 2\int {t\sin t} dt = - 2t\cos t + 2\int {\cos t} dt + C \cr
& 2\int {t\sin t} dt = - 2t\cos t + 2\sin t + C \cr
& {\text{Back - substitute }}t = \sqrt x \cr
& \int {\sin \sqrt x } dx = - 2\sqrt x \cos \sqrt x + 2\sin \sqrt x + C \cr
& {\text{Therefore,}} \cr
& \int_0^{{\pi ^2}/4} {\sin \sqrt x } dx = \left[ { - 2\sqrt x \cos \sqrt x + 2\sin \sqrt x } \right]_0^{{\pi ^2}/4} \cr
& = \left[ { - 2\sqrt {\frac{{{\pi ^2}}}{4}} \cos \sqrt {\frac{{{\pi ^2}}}{4}} + 2\sin \sqrt {\frac{{{\pi ^2}}}{4}} } \right] - \left[ {0\cos \sqrt 0 + 2\sin \sqrt 0 } \right] \cr
& = \left[ { - 2\left( {\frac{\pi }{2}} \right)\cos \left( {\frac{\pi }{2}} \right) + 2\sin \left( {\frac{\pi }{2}} \right)} \right] - \left[ 0 \right] \cr
& = - 2\left( {\frac{\pi }{2}} \right)\left( 0 \right) + 2\sin \left( {\frac{\pi }{2}} \right) \cr
& = 2 \cr} $$
