By using the properties of definite integrals,evaluate the integral $\int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}} d x$.

Let $I = \int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}} d x$ $(1)$
We use the property $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x$.
Applying this property to $(1)$,we get:
$I = \int_{0}^{a} \frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{a-(a-x)}} d x$
$I = \int_{0}^{a} \frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}} d x$ $(2)$
Adding $(1)$ and $(2)$,we obtain:
$2I = \int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}} d x + \int_{0}^{a} \frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}} d x$
$2I = \int_{0}^{a} \frac{\sqrt{x}+\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}} d x$
$2I = \int_{0}^{a} 1 d x$
$2I = [x]_{0}^{a}$
$2I = a - 0 = a$
$I = \frac{a}{2}$