$\int_0^{\pi /2} \log(\sin x) \, dx = $

On the real line $R$,we define two functions $f$ and $g$ as follows:
$f(x) = \min \{x - [x], 1 - x + [x]\}$
$g(x) = \max \{x - [x], 1 - x + [x]\}$
where $[x]$ denotes the greatest integer function. The positive integer $n$ for which $\int_0^n (g(x) - f(x)) \, dx = 100$ is:

$\int_{2 - \log 3}^{3 + \log 3} \frac{\log (4 + x)}{\log (4 + x) + \log (9 - x)} \, dx = $

Let $f: [0, \frac{\pi}{2}] \rightarrow [0, 1]$ be the function defined by $f(x) = \sin^2 x$ and let $g: [0, \frac{\pi}{2}] \rightarrow [0, \infty)$ be the function defined by $g(x) = \sqrt{\frac{\pi x}{2} - x^2}$.
(There are two questions based on this paragraph. The questions given below are those two.)
$(1)$ The value of $2 \int_0^{\frac{\pi}{2}} f(x) g(x) dx - \int_0^{\frac{\pi}{2}} g(x) dx$ is
$(2)$ The value of $\frac{16}{\pi^3} \int_0^{\frac{\pi}{2}} f(x) g(x) dx$ is

The value of $\int_{0}^{1} \tan ^{-1}\left(\frac{2 x-1}{1+x-x^{2}}\right) d x$ is

Statement $-1$: The value of the integral $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1 + \sqrt{\tan x}} = \frac{\pi}{6}$.
Statement $-2$: $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx$.