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Question: The value of \(\int_{- \pi/2}^{\pi/2}\frac{\sin^{2}xdx}{1 + e^{x}}\) is-...

The value of ∫−π/2π/2sin⁡2xdx1+ex\int_{- \pi/2}^{\pi/2}\frac{\sin^{2}xdx}{1 + e^{x}} is-

A

π8\frac{\pi}{8}

B

π4\frac{\pi}{4}

C

π6\frac{\pi}{6}

D

π3\frac{\pi}{3}

Answer

π4\frac{\pi}{4}

Explanation

Solution

I = ∫−π/2π/2sin⁡2x1+exdx\int_{- \pi/2}^{\pi/2}{\frac{\sin^{2}x}{1 + e^{x}}dx} = ∫−π/2π/2sin⁡2(−x)1+e−xdx\int_{- \pi/2}^{\pi/2}{\frac{\sin^{2}( - x)}{1 + e^{- x}}dx}

Ž I + I = ∫−π/2π/2sin⁡2xex+11+exdx\int_{- \pi/2}^{\pi/2}{\sin^{2}x\frac{e^{x} + 1}{1 + e^{x}}dx}

Ž I = 12∫−π/2π/2sin⁡2xdx\frac{1}{2}\int_{- \pi/2}^{\pi/2}{\sin^{2}xdx}

or I = ∫0π/2sin⁡2xdx\int_{0}^{\pi/2}{\sin^{2}xdx}