The value of I = int_{-frac{pi}{2}}^{frac{pi} | vedclass

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

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$\frac{\pi^2}{4}-2$

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(A) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1+e^{-x}} \,dx \quad ...(i)$Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$, we get:$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{(-x)^2 \cos(-x)}{1+e^{-(-x)}} \,dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1+e^x} \,dx \quad ...(ii)$Adding $(i)$ and $(ii)$:$2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x \left( \frac{1}{1+e^{-x}} + \frac{1}{1+e^x} \right) dx$Since $\frac{1}{1+e^{-x}} + \frac{1}{1+e^x} = \frac{e^x}{e^x+1} + \frac{1}{1+e^x} = 1$, we have:$2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x \,dx$Since $f(x) = x^2 \cos x$ is an even function, $2I = 2 \int_{0}^{\frac{\pi}{2}} x^2 \cos x \,dx$, so $I = \int_{0}^{\frac{\pi}{2}} x^2 \cos x \,dx$Using integration by parts:$I = [x^2 \sin x]_0^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} 2x \sin x \,dx$$I = [x^2 \sin x]_0^{\frac{\pi}{2}} - 2([-x \cos x]_0^{\frac{\pi}{2}} + \int_{0}^{\frac{\pi}{2}} \cos x \,dx)$$I = [x^2 \sin x + 2x \cos x - 2 \sin x]_0^{\frac{\pi}{2}}$$I = ((\frac{\pi}{2})^2 \sin(\frac{\pi}{2}) + 2(\frac{\pi}{2}) \cos(\frac{\pi}{2}) - 2 \sin(\frac{\pi}{2})) - (0 + 0 - 0)$$I = \frac{\pi^2}{4}(1) + 0 - 2(1) = \frac{\pi^2}{4} - 2$

The value of $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1+e^{-x}} \,dx$ is equal to

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