$\int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^{x}} d x$ is equal to

Let $f(x)$ be a positive function,$I_1 = \int_{-\frac{1}{2}}^1 2x f(2x(1-2x)) dx$,and $I_2 = \int_{-1}^2 f(x(1-x)) dx$. Then the value of $\frac{I_2}{I_1}$ is equal to:

If $I_n = \int_{-\pi}^{\pi} \frac{\sin(nx)}{(1+\pi^x) \sin x} dx$,$n=0, 1, 2, \ldots$,then
$(A)$ $I_n = I_{n+2}$
$(B)$ $\sum_{m=1}^{10} I_{2m+1} = 10\pi$
$(C)$ $\sum_{m=1}^{10} I_{2m} = 0$
$(D)$ $I_n = I_{n+1}$

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

$ \int_0^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1-\sin x \cos x} d x = $

If $\int_0^\pi {x\,f({{\cos }^2}x + {{\tan }^4}x)\,dx} = k\int_0^{\pi /2} {f({{\cos }^2}x + {{\tan }^4}x)\,dx,}$ then the value of $k$ is