Integrals: Definite Integrals & Properties

Let $$I = \int_{0}^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx$$. Using the property $$\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx$$, we replace $$x$$ with $$\frac{\pi}{2} - x$$.

Since $$\sin(\frac{\pi}{2} - x) = \cos x$$ and $$\cos(\frac{\pi}{2} - x) = \sin x$$, we have:\n\n$$I = \int_{0}^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} \, dx$$

$$2I = \int_{0}^{\pi/2} \left( \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} \right) \, dx = \int_{0}^{\pi/2} 1 \, dx$$

$$2I = [x]_{0}^{\pi/2} = \frac{\pi}{2} - 0 = \frac{\pi}{2} \implies I = \frac{\pi}{4}$$.