Let $p(x)$ be a function defined on $R$ such that $p'(x) = p'(1 - x)$ for all $x \in [0, 1]$,$p(0) = 1$,and $p(1) = 41$. Then $\int_{0}^{1} p(x) dx = $

Let $g_i: \left[\frac{\pi}{8}, \frac{3\pi}{8}\right] \rightarrow \mathbb{R}, i=1, 2$,and $f: \left[\frac{\pi}{8}, \frac{3\pi}{8}\right] \rightarrow \mathbb{R}$ be functions such that $g_1(x)=1, g_2(x)=|4x-\pi|$ and $f(x)=\sin^2 x$,for all $x \in \left[\frac{\pi}{8}, \frac{3\pi}{8}\right]$.
Define $S_i = \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} f(x) \cdot g_i(x) dx, i=1, 2$.
$(1)$ The value of $\frac{16S_1}{\pi}$ is.
$(2)$ The value of $\frac{48S_2}{\pi^2}$ is.

If $I_n = \int_0^{\frac{\pi}{4}} \tan^n \theta \, d\theta$,then $I_{12} + I_{10} =$

If ${I_n} = \int_0^{\pi /4} {{\tan ^n}\theta \,d\theta }$,then ${I_8} + {I_6}$ equals

If $\int_{ - a}^a {\sqrt {\frac{{a - x}}{{a + x}}} \,dx = k\pi ,} $ then $k = $

If $f(x) = \frac{2 - x \cos x}{2 + x \cos x}$ and $g(x) = \ln x$ for $x > 0$,then the value of the integral $\int_{-\pi/4}^{\pi/4} g(f(x)) dx$ is