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$\displaystyle\int_{-\pi/2}^{\pi/2} \dfrac{\sin^2 x}{1 + e^x}\, dx = ?$ (Glasser-type)
A$\pi/4$
B$\pi/2$
C$0$
D$\pi$
Answer & Solution
Correct answer: A. $\pi/4$
Let $I = \int_{-\pi/2}^{\pi/2} \sin^2 x/(1 + e^x)\, dx$. Sub $x \to -x$: $I = \int_{-\pi/2}^{\pi/2} \sin^2 x/(1 + e^{-x})\, dx = \int \sin^2 x \cdot e^x/(1 + e^x)\, dx$. Add: $2I = \int_{-\pi/2}^{\pi/2} \sin^2 x\, dx = \pi/2$ (even integrand, area = $2 \cdot \pi/4 = \pi/2$). So $I = \pi/4$.
Related questions
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