If the value of the integral int_{-frac{pi}{2 | vedclass

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(A) Let $I = \int_{-\pi/2}^{\pi/2} \left( \frac{x^2 \cos x}{1+\pi^x} + \frac{1+\sin^2 x}{1+e^{\sin x^{323}}} \right) dx$. Using the property $\int_{-a

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$3$

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(A) Let $I = \int_{-\pi/2}^{\pi/2} \left( \frac{x^2 \cos x}{1+\pi^x} + \frac{1+\sin^2 x}{1+e^{\sin x^{323}}} \right) dx$. Using the property $\int_{-a}^{a} f(x) dx = \int_{0}^{a} (f(x) + f(-x)) dx$,we observe the terms. For the first part,$\frac{x^2 \cos x}{1+\pi^x} + \frac{(-x)^2 \cos(-x)}{1+\pi^{-x}} = \frac{x^2 \cos x}{1+\pi^x} + \frac{x^2 \cos x \cdot \pi^x}{\pi^x+1} = x^2 \cos x$. For the second part,since $\sin x^{323}$ is an odd function,let $g(x) = \frac{1+\sin^2 x}{1+e^{\sin x^{323}}}$. Then $g(x) + g(-x) = \frac{1+\sin^2 x}{1+e^{\sin x^{323}}} + \frac{1+\sin^2 x}{1+e^{-\sin x^{323}}} = 1+\sin^2 x$. Thus,$2I = \int_{-\pi/2}^{\pi/2} (x^2 \cos x + 1 + \sin^2 x) dx = 2 \int_{0}^{\pi/2} (x^2 \cos x + 1 + \sin^2 x) dx$. $I = \int_{0}^{\pi/2} x^2 \cos x dx + \int_{0}^{\pi/2} 1 dx + \int_{0}^{\pi/2} \sin^2 x dx$. Evaluating $\int_{0}^{\pi/2} x^2 \cos x dx = [x^2 \sin x]_0^{\pi/2} - \int_{0}^{\pi/2} 2x \sin x dx = \frac{\pi^2}{4} - 2[-x \cos x + \sin x]_0^{\pi/2} = \frac{\pi^2}{4} - 2$. Evaluating $\int_{0}^{\pi/2} 1 dx = \frac{\pi}{2}$. Evaluating $\int_{0}^{\pi/2} \sin^2 x dx = \int_{0}^{\pi/2} \frac{1-\cos 2x}{2} dx = \frac{\pi}{4}$. Summing these: $I = \frac{\pi^2}{4} - 2 + \frac{\pi}{2} + \frac{\pi}{4} = \frac{\pi^2}{4} + \frac{3\pi}{4} - 2 = \frac{\pi}{4}(\pi+3) - 2$. Comparing with $\frac{\pi}{4}(\pi+a)-2$,we get $a=3$.

If the value of the integral $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{x^2 \cos x}{1+\pi^x}+\frac{1+\sin ^2 x}{1+e^{\sin x^{323}}}\right) d x=\frac{\pi}{4}(\pi+a)-2$,then the value of $a$ is

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