The value of the integral int_{-pi}^{pi} frac | 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 get: $I = \int_{-\p

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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 get: $I = \int_{-\pi}^{\pi} \frac{\cos^2(-\pi+\pi-x)}{1+a^{-\pi+\pi-x}} dx = \int_{-\pi}^{\pi} \frac{\cos^2 x}{1+a^{-x}} dx = \int_{-\pi}^{\pi} \frac{a^x \cos^2 x}{a^x+1} dx$. Adding the two expressions for $I$: $2I = \int_{-\pi}^{\pi} \frac{\cos^2 x}{1+a^x} dx + \int_{-\pi}^{\pi} \frac{a^x \cos^2 x}{1+a^x} dx = \int_{-\pi}^{\pi} \cos^2 x \left( \frac{1+a^x}{1+a^x} \right) dx = \int_{-\pi}^{\pi} \cos^2 x dx$. Since $\cos^2 x$ is an even function,$2I = 2 \int_{0}^{\pi} \cos^2 x dx$,so $I = \int_{0}^{\pi} \cos^2 x dx$. Using $\cos^2 x = \frac{1+\cos 2x}{2}$,we have $I = \int_{0}^{\pi} \frac{1+\cos 2x}{2} dx = \left[ \frac{x}{2} + \frac{\sin 2x}{4} \right]_{0}^{\pi} = \frac{\pi}{2} + 0 - (0+0) = \frac{\pi}{2}$.

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

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