By using the properties of definite integrals,evaluate the integral $\int_{0}^{\frac{\pi}{2}} \frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x} dx$.

Let $I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x} dx$ ..... $(1)$
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get:
$I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^{\frac{3}{2}}(\frac{\pi}{2}-x)}{\sin^{\frac{3}{2}}(\frac{\pi}{2}-x) + \cos^{\frac{3}{2}}(\frac{\pi}{2}-x)} dx$
Since $\sin(\frac{\pi}{2}-x) = \cos x$ and $\cos(\frac{\pi}{2}-x) = \sin x$,we have:
$I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^{\frac{3}{2}} x}{\cos^{\frac{3}{2}} x + \sin^{\frac{3}{2}} x} dx$ ..... $(2)$
Adding $(1)$ and $(2)$:
$2I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x} dx$
$2I = \int_{0}^{\frac{\pi}{2}} 1 dx$
$2I = [x]_{0}^{\frac{\pi}{2}}$
$2I = \frac{\pi}{2} - 0 = \frac{\pi}{2}$
$I = \frac{\pi}{4}$