Question

Evalute ${\int }_0^{\pi /2}\frac{{\sin }^{2}x-{\cos }^{2}x}{{\sin }^{3}x+{\cos }^{3}x}dx=$

Answer 1

Let $I={\int }_0^{\pi /2}\frac{{\sin }^{2}x-{\cos }^{2}x}{{\sin }^{3}x+{\cos }^{3}x}dx$ ...(1)
Using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$
$I={\int }_0^{\pi /2}\frac{{\cos }^{2}x-{\sin }^{2}x}{{\sin }^{3}x+{\cos }^{3}x}dx={\int }_0^{\pi /2}\frac{{\sin }^{2}x-{\cos }^{2}x}{{\sin }^{3}x+{\cos }^{3}x}dx=-1$ ...(2)
Adding (1) and (2)
$\therefore 2I=0\Rightarrow I=0$