If $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{96 x^2 \cos ^2 x}{\left(1+e^x\right)} \mathrm{d} x=\pi\left(\alpha \pi^2+\beta\right), \alpha, \beta \in \mathbb{Z}$, then $(\alpha+\beta)^2$ equals
Solution
<p>$\int_\limits{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{96 x^2 \cos ^2 x}{\left(1+e^x\right)} d x \quad$ (Apply King Property)</p>
<p>$$\begin{aligned}
& \int_0^{\frac{\pi}{2}} 96 x^2 \cos ^2 x=48 \int_0^{\frac{\pi}{2}} x^2(1+\cos 2 x) d x \\
& 48\left[\left(\frac{x^3}{3}\right)_0^{\pi / 2}+\int_0^{\frac{\pi}{2}} x_1^2 \cos _{I I} 2 x d x\right] \\
& \Rightarrow \text { On solving } \pi\left(2 \pi^2-12\right) \\
& \Rightarrow \alpha=2, \beta=-12 \\
& \Rightarrow(\alpha+\beta)^2=100
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Definite Integration · Topic: Properties of Definite Integrals
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