$$\int_{0}^{\frac{\pi}{2}}\sqrt{sin\theta}d\theta\times\int_{0}^{\frac{\pi}{2}}\sqrt{sin\theta}d\theta=\pi$$
$\left(\int _{0}^{\frac{\pi }{2}}\sqrt{\sin(\theta )}\mathrm{d}\theta \right)^{2}-\pi =0$
$\exists n_{1}\in \mathrm{Z}\text{ : }\left(\theta \geq 2\pi n_{1}\text{ and }\theta \leq 2\pi n_{1}+\pi \right)$
