Polar area and do I need absolute value signs?

Question

Imagine a curve such as $$r=\sqrt{\sin x}$$

The area over $(0,\pi)$ would be calculated as follows:

$$\int_0^{\pi}\frac{1}{2}\sqrt{\sin x}^2dx=\int_0^{\pi}\frac{1}{2}|{\sin x}|dx, not \int_0^{\pi}\frac{1}{2}{\sin x}dx$$ correct?