The value of the integral ∫ - π / 2 π / 2 sin 2 x 1 + e x d x is, π 6, π 4, π 2, π 2 2

I = ∫ 0 π / 2 sin 2 x 1 + e x + sin 2 x 1 + e - x d x = ∫ 0 π / 2 sin 2 x d x = 1 2 × π 2 = π 4

The value of the integral ∫ - π / 2 π / 2 sin 2 x 1 + e x d x is

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The value of the integral

$\int_{- π / 2}^{π / 2} \frac{\sin^{2} x}{1 + e^{x}} d x$ is

MathematicsDefinite IntegrationKVPYKVPY 2020 (SB/SX)

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Solution:

2175 Upvotes Verified Answer

The correct answer is:

\frac{π}{4}

I = \int_{0}^{π / 2} (\frac{\sin^{2} x}{1 + e^{x}} + \frac{\sin^{2} x}{1 + e^{- x}}) d x

= \int_{0}^{π / 2} \sin^{2} x d x = \frac{1}{2} \times \frac{π}{2} = \frac{π}{4}

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