int_{-pi/2}^{pi/2} frac{sin^2 x}{1 + 2^x} dx | vedclass

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(C) Let $I = \int_{-\pi/2}^{\pi/2} \frac{\sin^2 x}{1 + 2^x} dx$  $(i)$ Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we get: $I

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$\frac{\pi}{4}$

Vedclass · Answer

(C) Let $I = \int_{-\pi/2}^{\pi/2} \frac{\sin^2 x}{1 + 2^x} dx$  $(i)$ Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we get: $I = \int_{-\pi/2}^{\pi/2} \frac{\sin^2(-x)}{1 + 2^{-x}} dx = \int_{-\pi/2}^{\pi/2} \frac{\sin^2 x}{1 + \frac{1}{2^x}} dx = \int_{-\pi/2}^{\pi/2} \frac{2^x \sin^2 x}{2^x + 1} dx$  $(ii)$ Adding $(i)$ and $(ii)$: $2I = \int_{-\pi/2}^{\pi/2} \frac{\sin^2 x + 2^x \sin^2 x}{1 + 2^x} dx = \int_{-\pi/2}^{\pi/2} \sin^2 x dx$ Since $\sin^2 x$ is an even function,$2I = 2 \int_{0}^{\pi/2} \sin^2 x dx$ $I = \int_{0}^{\pi/2} \sin^2 x dx = \int_{0}^{\pi/2} \frac{1 - \cos 2x}{2} dx = \left[ \frac{x}{2} - \frac{\sin 2x}{4} \right]_{0}^{\pi/2} = \frac{\pi}{4} - 0 = \frac{\pi}{4}$

$\int_{-\pi/2}^{\pi/2} \frac{\sin^2 x}{1 + 2^x} dx = \dots$

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