$\displaystyle\int_{-\pi/2}^{\pi/2} \dfrac{\sin^2 x}{1 + e^x}\, dx = ?…

Question

$\displaystyle\int_{-\pi/2}^{\pi/2} \dfrac{\sin^2 x}{1 + e^x}\, dx = ?$ (Glasser-type)

Accepted Answer

A. $\pi/4$

Let $I = \int_{-\pi/2}^{\pi/2} \sin^2 x/(1 + e^x)\, dx$.
Sub $x 	o -x$: $I = \int_{-\pi/2}^{\pi/2} \sin^2 x/(1 + e^{-x})\, dx = \int \sin^2 x \cdot e^x/(1 + e^x)\, dx$.
**Add:** $2I = \int_{-\pi/2}^{\pi/2} \sin^2 x\, dx = \pi/2$ (even integrand, area = $2 \cdot \pi/4 = \pi/2$).
**So $I = \pi/4$.**

Answer

$\pi/2$

Answer

$0$

Answer

$\pi$

$\displaystyle\int_{-\pi/2}^{\pi/2} \dfrac{\sin^2 x}{1 + e^x}\, dx = ?$ (Glasser-type)

Keep practising Unit 6B: Definite Integration & Riemann Sums