Answer
$$\eqalign{
& \left( {\text{a}} \right)u\sin u \cr
& \left( {\text{b}} \right){\text{2}}\pi \cr} $$
Work Step by Step
$$\eqalign{
& \left( {\text{a}} \right){\text{ }}\sin u - u\cos u + C = \int {u\sin udu} \cr
& {\text{Differentiating }}\sin u - u\cos u + C \cr
& \frac{d}{{du}}\left[ {\sin u - u\cos u + C} \right] \cr
& = \frac{d}{{du}}\left[ {\sin u} \right] - u\frac{d}{{du}}\left[ {\cos u} \right] - \cos u\frac{d}{{du}}\left[ u \right] + \frac{d}{{du}}\left[ C \right] \cr
& = \cos u - u\left( { - \sin u} \right) - \cos u\left( 1 \right) + 0 \cr
& = u\sin u \cr
& \cr
& \left( {\text{b}} \right)\int_0^{{\pi ^2}} {\sin \sqrt x dx} \cr
& {\text{Let }}u = \sqrt x ,{\text{ }}du = \frac{1}{{2\sqrt x }}dx,{\text{ }}dx = 2\sqrt x du,{\text{ }}dx = 2udu \cr
& {\text{The new limits of integration are:}} \cr
& x = {\pi ^2} \to u = \pi \cr
& x = 0 \to u = 0 \cr
& {\text{Substituting}} \cr
& \int_0^{{\pi ^2}} {\sin \sqrt x dx} = \int_0^\pi {\sin u\left( {2u} \right)du} \cr
& = 2\int_0^\pi {u\sin udu} \cr
& {\text{Using the formula given in part }}\left( {\text{a}} \right) \cr
& 2\int_0^\pi {u\sin udu} = 2\left[ {\sin u - u\cos u} \right]_0^\pi \cr
& = 2\left[ {\sin \pi - \pi \cos \pi } \right] - 2\left[ {\sin 0 - 0\cos 0} \right] \cr
& = 2\left( {0 + \pi } \right) - 2\left( {0 - 0} \right) \cr
& = 2\pi \cr} $$
