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Assam CEE Symmetry Trick — practice questions

Explore 1 free Symmetry Trick questions in Assam CEE Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Symmetry Trick questions and exercises with answers

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$\displaystyle\int_{-\pi/2}^{\pi/2} \dfrac{\sin^2 x}{1 + e^x}\, dx = ?$ (Glasser-type)Read the answer and worked solution →

Worked practice questions

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

  1. $\pi/4$
  2. $\pi/2$
  3. $0$
  4. $\pi$
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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 $I = \pi/4$.**

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