Question

Prove that:

${\int }_0^{\pi /2}\frac{1}{1+{\tan }^{3}x}dx=\frac{\pi }{4}$

Answer 1

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

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

$={\int }_0^{\pi /2}\frac{{\sin }^{3}x}{{\cos }^{3}x+{\sin }^{3}x}dx[\because {\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx]$

By adding equation (1) and (2)

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

$2I={\int }_0^{\pi /2}1dx$

$={\left[x\right]}_0^{\pi /2}=\frac{\pi }{2}$

$I=\frac{\pi }{4}$