Calculate integral $\int_0^{\pi/2} \frac{\cos^3x}{\sin^2x + \cos^3x}dx$. 
Calculate integral $$\int_0^{\pi/2} \frac{\cos^3x}{\sin^2x + \cos^3x}dx.$$

My direction: Since this integral can't calculate normally, I tried to use the property following:$$\int_{a}^{b}f(x)dx = \int_{a}^{b}f(a+b-x)dx.$$
Then, I have
$$I=\int_0^{\pi/2} \frac{\cos^3x}{\sin^2x + \cos^3x}dx = \int_0^{\pi/2} \frac{\sin^3x}{\cos^2x + \sin^3x}dx.$$
Therefore
$$2I = \int_0^{\pi/2} \left(\frac{\cos^3x}{\sin^2x + \cos^3x} + \frac{\sin^3x}{\cos^2x + \sin^3x}\right) dx.$$
I stucked here.

This integral does have a closed form despite being very messy. We have \begin{align}I&=\int_0^{\pi/2} \frac{\cos^3x}{\sin^2x + \cos^3x}\,dx\\&=\frac14\int_{-\pi}^\pi\frac{\cos^3(u/2)}{\sin^2(u/2)+\cos^3(u/2)}\,du\\&=\frac14\oint_{|z|=1}\frac{\frac18(z^{1/2}+z^{-1/2})^3}{-\frac14(z^{1/2}-z^{-1/2})^2+\frac18(z^{1/2}+z^{-1/2})^3}\frac{dz}{iz}\\&=-\frac i4\oint_{|z|=1}\frac{(z+1)^3}{z((z+1)^3-2z^{1/2}(z-1)^2)}\,dz.\end{align} Note that $(z+1)^3-2z^{1/2}(z-1)^2=(w^2+1)-2w(w^2-1)^2=\prod\limits_{i=1}^6(w-z_i)$ where $z_1=\overline{z_2}$, $z_3=\overline{z_4}$ and $z_5=\overline{z_6}$. Hence $$I=-\frac i4\cdot2\pi i\left(\operatorname{Res}(f,0)+\sum_{i=1}^6\operatorname{Res}(f,z_i^2)\right)=\frac\pi2\left(1+\sum_{i=1}^6\operatorname{Res}(f,z_i^2)\right).$$ An interesting property about this polynomial is that it is symmetric about its coefficients \begin{align}(w^2+1)-2w(w^2-1)^2&=w^6-2w^5+3w^4+4w^3+3w^2-2w+1\\&=w^3(w^3-2w^2+3w+4+3w^{-1}-2w^{-2}+w^{-3}).\end{align} This means we can make the substitution $w=v+1/v$ to obtain a cubic equation which is solvable in radicals, as in @Quanto's answer.
