$\int_{-a}^{a} \sqrt{\frac{a - x}{a + x}} dx =$

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.

$\int_{-\pi / 2}^{2 \pi} \sin ^{-1}(\sin x) d x=$

If $f(a+b-x)=f(x)$,then $\int_a^b x f(x) d x=$ . . . . . . .

By using the properties of definite integrals,evaluate the integral $\int_{0}^{\frac{\pi}{2}} \frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x} dx$.

The greater of $\int_0^{\pi /2} \frac{\sin x}{x} \, dx$ and $\frac{\pi}{2}$ is: