$$\int_{0}^{\frac{\pi}{2}}\sqrt{sin\theta}d\theta\times\int_{0}^{\frac{\pi}{2}}\frac{1}{\sqrt{sin\theta}}d\theta=\pi$$
$\int _{0}^{\frac{\pi }{2}}\sqrt{\sin(\theta )}\mathrm{d}\theta \int _{0}^{\frac{\pi }{2}}\frac{1}{\sqrt{\sin(\theta )}}\mathrm{d}\theta =\pi $
