$\displaystyle\int_0^{\pi/2} \dfrac{dx}{1 + an^3 x} = ?$

Question

$\displaystyle\int_0^{\pi/2} \dfrac{dx}{1 + 	an^3 x} = ?$

Accepted Answer

A. $\pi/4$

Same King's trick as $\sin/(\sin+\cos)$: $1/(1 + 	an^3 x) = \cos^3 x/(\cos^3 x + \sin^3 x)$.
Sub $x 	o \pi/2 - x$ swaps $\sin↔\cos$: $I = \int \sin^3 x/(\sin^3 x + \cos^3 x)$.
**Add:** $2I = \int_0^{\pi/2} dx = \pi/2$.
So $I = \pi/4$.
(Indeed, $\int_0^{\pi/2} dx/(1 + 	an^n x) = \pi/4$ for ANY $n$ — striking invariance.)

Answer

$\pi/2$

Answer

$0$

Answer

$\ln 2$

$\displaystyle\int_0^{\pi/2} \dfrac{dx}{1 + \tan^3 x} = ?$

Keep practising Unit 6B: Definite Integration & Riemann Sums