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

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(D) Let $I = \int_{-\pi}^{\pi} \frac{\cos^2 x}{1+\alpha^x} \, dx$  --- $(i)$Using the property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$

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$\frac{\pi}{2}$

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(D) Let $I = \int_{-\pi}^{\pi} \frac{\cos^2 x}{1+\alpha^x} \, dx$  --- $(i)$Using the property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$,we get:$I = \int_{-\pi}^{\pi} \frac{\cos^2(-x)}{1+\alpha^{-x}} \, dx = \int_{-\pi}^{\pi} \frac{\cos^2 x}{1+\frac{1}{\alpha^x}} \, dx = \int_{-\pi}^{\pi} \frac{\alpha^x \cos^2 x}{\alpha^x + 1} \, dx$ --- $(ii)$Adding $(i)$ and $(ii)$:$2I = \int_{-\pi}^{\pi} \frac{\cos^2 x + \alpha^x \cos^2 x}{1+\alpha^x} \, dx = \int_{-\pi}^{\pi} \frac{\cos^2 x(1+\alpha^x)}{1+\alpha^x} \, dx = \int_{-\pi}^{\pi} \cos^2 x \, dx$Since $\cos^2 x$ is an even function,$2I = 2 \int_{0}^{\pi} \cos^2 x \, dx$$I = \int_{0}^{\pi} \cos^2 x \, dx = \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 $\int_{-\pi}^{\pi} \frac{\cos^2 x}{1+\alpha^x} \, dx$ for $\alpha > 0$ is

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