Question

The value of ${\int }_0^{\pi /2}\frac{dx}{1+{\tan }^{3}x}$ is

Answer 1

Let $I={\int }_0^{\pi /2}\frac{dx}{1+{\tan }^{3}x}\cdots \cdots \left(1\right)$

as we know that ${\int }_b^{a}f\left(x\right)dx={\int }_b^{a}f(a+b-x)dx$

$⟹I={\int }_0^{\pi /2}\frac{dx}{1+{\tan }^3(\frac{\pi }{2}-x)}={\int }_0^{\pi /2}\frac{dx}{1+{\cot }^{3}x}={\int }_0^{\pi /2}\frac{dx}{1+\frac{1}{{\tan }^{3}x}}={\int }_0^{\pi /2}\frac{{\tan }^{3}xdx}{1+{\tan }^{3}x}...\left(2\right)$

adding $\left(1\right)$ and $\left(2\right)$

$2I={\int }_0^{\pi /2}dx=[x{]}_0^{\pi /2}=\frac{\pi }{2}$

$⟹I=\frac{\pi }{4}$