Let I=∫0^π/2(2sin ^3x+5cos ^3x)/(sin ^3x+cos ^3x) dx. Find the value of (4I)/(π).

Use the property $\int_0^{a}g(x)\,dx=\int_0^{a}g(a-x)\,dx$ with $a=\frac\pi2$. Under $x\to\frac\pi2-x$ sines and cosines swap, so $I=\int_0^{\pi/2}\frac{2\cos^3x+5\sin^3x}{\sin^3x+\cos^3x}dx$. Adding the two forms of $I$ gives $2I=\int_0^{\pi/2}\frac{7(\sin^3x+\cos^3x)}{\sin^3x+\cos^3x}dx=7\cdot\frac\pi2$. Hence $I=\frac{7\pi}{4}$, and $\frac{4I}{\pi}=7$. Numerical quadrature of the original integrand returns $5.4978=\frac{7\pi}{4}$, agreeing with the symmetry argument. The denominator never vanishes on $[0,\frac\pi2]$ because both cubes are non-negative and cannot be zero at the same point. Hence the symmetry trick removes the awkward denominator entirely, and the requested quantity $\frac{4I}{\pi}$ equals the integer $7$, which is just the sum $2+5$ of the two coefficients.