If $f(x) = \frac{e^x}{1 + e^x}$,$I_1 = \int_{f(-a)}^{f(a)} x g\{x(1 - x)\} dx$,and $I_2 = \int_{f(-a)}^{f(a)} g\{x(1 - x)\} dx$,then the value of $\frac{I_2}{I_1}$ is

$\int_{\frac{-3\pi}{2}}^{\frac{-\pi}{2}} \left[ (x+\pi)^3 + \cos^2(x+3\pi) \right] dx = $

If $\int_{-a}^a f(x) dx = \int_0^a f(x) dx + \int_0^a g(x) dx$, then $g(x) =$

The value of $\int_{0}^{\pi /2} \frac{e^{x^2}}{e^{x^2} + e^{(\pi /2 - x)^2}} dx$ is

If the integral $\int_{0}^{10} \frac{[\sin 2 \pi x ]}{ e ^{ x -[ x ]}} dx =\alpha e ^{-1}+\beta e ^{-\frac{1}{2}}+\gamma$,where $\alpha, \beta, \gamma$ are integers and $[ x ]$ denotes the greatest integer less than or equal to $x$,then the value of $\alpha+\beta+\gamma$ is equal to ........ .

$\int_0^{\pi /2} {x\cot x\,dx} $ equals