If int_{-frac{pi}{2}}^{frac{pi}{2}} frac{96 x | vedclass

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(C) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{96 x^2 \cos^2 x}{1+e^x} dx$. Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we

Vedclass · Accepted Answer

$100$

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(C) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{96 x^2 \cos^2 x}{1+e^x} dx$. Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we have $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{96 x^2 \cos^2 x}{1+e^{-x}} dx$. Adding the two expressions for $I$: $2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 96 x^2 \cos^2 x \left( \frac{1}{1+e^x} + \frac{1}{1+e^{-x}} \right) dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 96 x^2 \cos^2 x dx$. Since $x^2 \cos^2 x$ is an even function,$I = \int_0^{\frac{\pi}{2}} 96 x^2 \cos^2 x dx = 48 \int_0^{\frac{\pi}{2}} x^2 (1 + \cos 2x) dx$. $I = 48 \left[ \frac{x^3}{3} \right]_0^{\frac{\pi}{2}} + 48 \int_0^{\frac{\pi}{2}} x^2 \cos 2x dx$. $I = 48 \left( \frac{\pi^3}{24} \right) + 48 \left[ x^2 \frac{\sin 2x}{2} - \int 2x \frac{\sin 2x}{2} dx \right]_0^{\frac{\pi}{2}}$. $I = 2\pi^3 + 48 \left[ 0 - \int_0^{\frac{\pi}{2}} x \sin 2x dx \right] = 2\pi^3 - 48 \left[ x \left( -\frac{\cos 2x}{2} \right) - \int -\frac{\cos 2x}{2} dx \right]_0^{\frac{\pi}{2}}$. $I = 2\pi^3 - 48 \left[ -\frac{x \cos 2x}{2} + \frac{\sin 2x}{4} \right]_0^{\frac{\pi}{2}} = 2\pi^3 - 48 \left[ -\frac{\pi}{2} \cdot \frac{(-1)}{2} - 0 \right] = 2\pi^3 - 12\pi = \pi(2\pi^2 - 12)$. Comparing with $\pi(\alpha \pi^2 + \beta)$,we get $\alpha = 2, \beta = -12$. Thus,$(\alpha + \beta)^2 = (2 - 12)^2 = (-10)^2 = 100$.

If $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{96 x^2 \cos^2 x}{1+e^x} dx = \pi(\alpha \pi^2 + \beta)$,where $\alpha, \beta \in \mathbb{Z}$,then $(\alpha + \beta)^2$ equals:

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