TITLE: Calculate integral $\int_0^{\pi/2} \frac{\cos^3x}{\sin^2x + \cos^3x}dx$.
QUESTION [14 upvotes]: 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.

REPLY [2 votes]: A Physicist's Point of View.
The closed form solution to the integral is very impressive, of course.
Sometimes when needed a quick numerical estimate for such a hard-to-calculate integral it is useful to take a closer look at the integrand.
Consider the denominator of the integrand
$$\sin^2x + \cos^3x$$
What happens if we replace here $\cos^3x$ with $\cos^2x$?
Probably not much because $\cos x$ varies within $\left [0,1  \right ]$
But we get a considerable simplification
$$I_{approx}\approx\int_0^{\pi/2} \cos^3x dx.=\frac{2}{3}$$
Absolute error from the exact value is about $0.05$