The value of int_{-pi / 2}^{pi / 2} frac{cos | vedclass

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(A) Let $I = \int_{-\pi / 2}^{\pi / 2} \frac{\cos ^{2} x}{1+3^{x}} dx$  --- $(1)$Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,

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$\frac{\pi}{4}$

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(A) Let $I = \int_{-\pi / 2}^{\pi / 2} \frac{\cos ^{2} x}{1+3^{x}} dx$  --- $(1)$Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we get:$I = \int_{-\pi / 2}^{\pi / 2} \frac{\cos ^{2}(-x)}{1+3^{-x}} dx = \int_{-\pi / 2}^{\pi / 2} \frac{\cos ^{2} x}{1+\frac{1}{3^{x}}} dx = \int_{-\pi / 2}^{\pi / 2} \frac{3^{x} \cos ^{2} x}{3^{x}+1} dx$ --- $(2)$Adding $(1)$ and $(2)$:$2I = \int_{-\pi / 2}^{\pi / 2} \frac{\cos ^{2} x + 3^{x} \cos ^{2} x}{1+3^{x}} dx$$2I = \int_{-\pi / 2}^{\pi / 2} \frac{\cos ^{2} x(1+3^{x})}{1+3^{x}} dx = \int_{-\pi / 2}^{\pi / 2} \cos ^{2} x dx$Since $\cos^{2} x$ is an even function:$2I = 2 \int_{0}^{\pi / 2} \cos ^{2} x dx$$I = \int_{0}^{\pi / 2} \frac{1+\cos 2x}{2} dx = \frac{1}{2} [x + \frac{\sin 2x}{2}]_{0}^{\pi / 2}$$I = \frac{1}{2} [(\frac{\pi}{2} + 0) - (0 + 0)] = \frac{\pi}{4}$

The value of $\int_{-\pi / 2}^{\pi / 2} \frac{\cos ^{2} x}{1+3^{x}} d x$ is

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