In order to calculate the integral $$ f(x) = \frac{2}{\pi}\int_0^{\pi/2}\delta\Big(x-\sqrt{1\pm\sqrt{1-\beta^2\sin^2t}}\Big)\mathrm{d}t $$ where $\beta\in(0,1]$. I am hunting for a better solution, which can handle the problem powerfully. Here below is my solution for case $-$.
Setting $u=\sin^2t$, then $\mathrm{d}t=\frac{1}{2\sqrt{u(1-u)}}\mathrm{d}u$. Therefore, $$ f(x) = \frac{2}{\pi}\int_0^{1}\delta\Big(x-\sqrt{1-\sqrt{1-\beta^2u}}\Big)\frac{1}{2\sqrt{u(1-u)}}\mathrm{d}u $$ By using of the integral property of delta function, we finally arrive at $$ f(x) = \frac{4x(1-x^2)}{\pi\sqrt(1-(1-x^2)^2)\sqrt(\beta^2-1+(1-x^2)^2)}. $$ Similarly, one can calculate the case for $+$.