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

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

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

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(A) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1+e^x} dx$ ... $(i)$Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we have $a+b = -\frac{\pi}{2} + \frac{\pi}{2} = 0$,so $f(x) 	o f(-x)$.$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{(-x)^2 \cos(-x)}{1+e^{-x}} dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1+e^{-x}} dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x \cdot e^x}{e^x+1} dx$ ... (ii)Adding $(i)$ and (ii):$2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x (1+e^x)}{1+e^x} dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x dx$Since $x^2 \cos x$ is an even function,$2I = 2 \int_{0}^{\frac{\pi}{2}} x^2 \cos x dx$.$I = \int_{0}^{\frac{\pi}{2}} x^2 \cos x dx$.Using integration by parts $\int u dv = uv - \int v du$ with $u=x^2, dv=\cos x dx$:$I = [x^2 \sin x]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} 2x \sin x dx$$I = (\frac{\pi^2}{4} \cdot 1 - 0) - 2 \int_{0}^{\frac{\pi}{2}} x \sin x dx$$I = \frac{\pi^2}{4} - 2 [x(-\cos x) - \int 1(-\cos x) dx]_{0}^{\frac{\pi}{2}}$$I = \frac{\pi^2}{4} - 2 [-x \cos x + \sin x]_{0}^{\frac{\pi}{2}}$$I = \frac{\pi^2}{4} - 2 [(-0 + 1) - (0 + 0)] = \frac{\pi^2}{4} - 2$.

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

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