Using the King's property, evaluate $\int_0^{\pi/2} \dfrac{dx}{1 + \sq…

Question

Using the King's property, evaluate $\int_0^{\pi/2} \dfrac{dx}{1 + \sqrt[3]{	an x}}$:

Accepted Answer

A. $\pi/4$

Let $I = \int_0^{\pi/2} dx/(1+\sqrt[3]{	an x}) = \int_0^{\pi/2} \sqrt[3]{\cos x}/(\sqrt[3]{\cos x}+\sqrt[3]{\sin x})\,dx$.
**Apply King:** $I = \int_0^{\pi/2} \sqrt[3]{\sin x}/(\sqrt[3]{\sin x}+\sqrt[3]{\cos x})\,dx$.
Adding gives $2I = \int_0^{\pi/2} 1\,dx = \pi/2$, so $I = \pi/4$.
Beautiful symmetry trick that works for ANY $f$, not just cube root.

Answer

$\pi/2$

Answer

$\pi$

Answer

$\pi/8$

Using the King's property, evaluate $\int_0^{\pi/2} \dfrac{dx}{1 + \sqrt[3]{\tan x}}$:

Keep practising Unit 6B: Definite Integration & Riemann Sums