Let $I={\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{{\displaystyle \frac{3}{2}}}x}{{\sin }^{\frac{3}{2}}x+{\cos }^{\frac{3}{2}}x}dx ...\left(1\right)$

$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{{\displaystyle \frac{3}{2}}}\left(\frac{\pi }{2}-x\right)}{{\sin }^{\frac{3}{2}}\left(\frac{\pi }{2}-x\right)+{\cos }^{\frac{3}{2}}\left(\frac{\pi }{2}-x\right)}dx$ (Using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$)

$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{{\displaystyle \frac{3}{2}}}x}{{\cos }^{\frac{3}{2}}x+{\sin }^{\frac{3}{2}}x}dx ...\left(2\right)$

Adding $\left(1\right) \&\hspace{0.17em}\left(2\right)$,

$\Rightarrow 2I={\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{\frac{3}{2}}x+{\sin }^{\frac{3}{2}}x}{{\cos }^{\frac{3}{2}}x+{\sin }^{\frac{3}{2}}x}dx$

$\Rightarrow I=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}dx=\frac{1}{2}{\left(x\right)}_0^{\frac{\pi }{2}}=\frac{\pi }{4}$