Evaluate $\int^{441}_0\frac{\pi\sin \pi \sqrt x}{\sqrt x} dx$

Question

Evaluate this definite integral: $$\int^{441}_0\frac{\pi\sin \pi \sqrt x}{\sqrt x} dx$$

Answer 1

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 $2(1-(-1)^n)$.

Answer 2

$$\int^{441}_0\frac{\pi\sin \pi \sqrt x}{\sqrt x} dx$$ $$\implies\pi\int^{441}_0\frac{\sin \pi \sqrt x}{\sqrt x} dx$$ $$\implies\pi\int^{21}_0\frac{2u \sin \pi u}{u} du$$ $$\implies2\pi\int^{21}_0 \sin{\pi u}\;du$$ $$\implies2\pi\int^{21\pi}_0 \frac{\sin{v}}{\pi}\;dv$$ $$\implies2\int^{21\pi}_0 \sin{v}\;dv$$ $$\implies -2 \left[\cos \pi \sqrt x\right]^{21}_{0}$$ $$\implies-2(\cos 21\pi - \cos 0)$$ $$\implies-2(-1 - 1)$$ $$\implies-2\times (-2)$$ $$\implies 4$$

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.