Evaluate $\int^{441}_0\frac{\pi\sin \pi \sqrt x}{\sqrt x} dx$ Evaluate this definite integral:
$$\int^{441}_0\frac{\pi\sin \pi \sqrt x}{\sqrt x} dx$$
 A: This integral (even the indefinite one) can be easily solved by observing:
$$\frac{\mathrm d}{\mathrm dx}\pi\sqrt x = \frac{\pi}{2\sqrt x}$$
which implies that:
$$\frac{\mathrm d}{\mathrm dx}\cos\pi\sqrt x = -\frac{\pi \sin\pi\sqrt x}{2\sqrt x}$$
Finally, we obtain:
$$\int\frac{\pi\sin\pi\sqrt x}{\sqrt x}\,\mathrm dx = -2\cos\pi\sqrt x$$
whence the definite integral with bounds $0, n^2$ evaluates to $2(1-(-1)^n)$.
A: $$\int^{441}_0\frac{\pi\sin \pi \sqrt x}{\sqrt x} dx$$
$$\implies\pi\int^{441}_0\frac{\sin \pi \sqrt x}{\sqrt x} dx$$
$$\implies\pi\int^{21}_0\frac{2u \sin \pi u}{u} du$$
$$\implies2\pi\int^{21}_0 \sin{\pi u}\;du$$
$$\implies2\pi\int^{21\pi}_0 \frac{\sin{v}}{\pi}\;dv$$
$$\implies2\int^{21\pi}_0 \sin{v}\;dv$$
$$\implies -2 \left[\cos \pi \sqrt x\right]^{21}_{0}$$
$$\implies-2(\cos 21\pi - \cos 0)$$
$$\implies-2(-1 - 1)$$
$$\implies-2\times (-2)$$
$$\implies 4$$
An interesting thing is that for upper bounds that are squares of even numbers this evaluates to zero, while for squares of odd numbers (such as $21^2 = 441$) this evaluates to 4. 
