The value of int_{-pi}^{pi} frac{cos^2 x}{1 + | vedclass

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(C) Let $I = \int_{-\pi}^{\pi} \frac{\cos^2 x}{1 + a^x} dx$. Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we replace $x$ with

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$\frac{\pi}{2}$

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(C) Let $I = \int_{-\pi}^{\pi} \frac{\cos^2 x}{1 + a^x} dx$. Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we replace $x$ with $-\pi + \pi - x = -x$: $I = \int_{-\pi}^{\pi} \frac{\cos^2(-x)}{1 + a^{-x}} dx = \int_{-\pi}^{\pi} \frac{\cos^2 x}{1 + \frac{1}{a^x}} dx = \int_{-\pi}^{\pi} \frac{a^x \cos^2 x}{a^x + 1} dx$. Adding the two expressions for $I$: $2I = \int_{-\pi}^{\pi} \frac{\cos^2 x}{1 + a^x} dx + \int_{-\pi}^{\pi} \frac{a^x \cos^2 x}{1 + a^x} dx = \int_{-\pi}^{\pi} \frac{(1 + a^x) \cos^2 x}{1 + a^x} dx = \int_{-\pi}^{\pi} \cos^2 x dx$. Since $\cos^2 x$ is an even function,$2I = 2 \int_{0}^{\pi} \cos^2 x dx = \int_{0}^{\pi} (1 + \cos 2x) dx$. $2I = [x + \frac{\sin 2x}{2}]_{0}^{\pi} = (\pi + 0) - (0 + 0) = \pi$. Therefore,$I = \frac{\pi}{2}$.

The value of $\int_{-\pi}^{\pi} \frac{\cos^2 x}{1 + a^x} dx, a > 0$ is

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