The value of the integral $\int_{0}^{\pi / 2} \frac{1}{1+(\tan x)^{-101}} d x$ is equal to

Let $f:[-2, 3] \to [0, \infty)$ be a continuous function such that $f(1-x) = f(x)$ for all $x \in [-2, 3]$. If $R_1$ is the numerical value of the area of the region bounded by $y = f(x)$,$x = -2$,$x = 3$ and the $x$-axis,and $R_2 = \int_{-2}^3 x f(x) dx$,then:

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

$\int_0^{400 \pi} \sqrt{1-\cos 2 x} \, dx =$ (in $\sqrt{2}$)

$\int_{\pi / 5}^{3 \pi / 10} \frac{d x}{\sec ^2 x+\left(\tan ^{2022} x-1\right)\left(\sec ^2 x-1\right)}=$

If $f(x)$ is a continuous periodic function with period $T,$ then the integral $I = \int_a^{a + T} {f(x)\,dx} $ is