$\int\limits_{ - a}^a {f(x)\,dx} = $

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:

Let $m, n, p, q$ be four positive integers. If $\int_0^{2 \pi} \sin^m x \cos^n x \, dx = 4 \int_0^{\pi/2} \sin^m x \cos^n x \, dx$, $\int_0^{2 \pi} \sin^p x \cos^n x \, dx = 0$, $\int_0^{\pi} \sin^p x \cos^q x \, dx = 0$, $a = m + n + p$ and $b = m + n + q$, then:

$\int_0^{\pi / 2} \frac{\sin ^3 x \cos x \, dx}{\sin ^4 x+\cos ^4 x} = ?$

The value of $\int_{e^2}^{e^4} \frac{1}{x} \left( \frac{e^{((\ln x)^2+1)^{-1}}}{e^{((\ln x)^2+1)^{-1}} + e^{((6-\ln x)^2+1)^{-1}}} \right) dx$ is

If $\int_{-1}^{4} f(x) dx = 4$ and $\int_{2}^{4} (3 - f(x)) dx = 7$,then the value of $\int_{2}^{-1} f(x) dx$ is