If $[x]$ is the greatest integer $\leq x$,then the value of the integral $\int_{-0.9}^{0.9} \left( [x^2] + \log \left( \frac{2-x}{2+x} \right) \right) dx$ is

Let $I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2-\cos 2 x}\left(\frac{3}{\pi}+\log \left(\frac{4+\sin x}{4-\sin x}\right)\right) d x$. Given that $\int \frac{d x}{1+k x^2}=\frac{1}{\sqrt{k}} \tan ^{-1}(\sqrt{k} x)+c, \tan ^{-1}(0)=0$ and $\tan ^{-1}(\sqrt{3})=\frac{\pi}{3}$. Then $3 I^2=$

$\int_0^{\frac{\pi}{2}} \frac{\sin^2 x}{\sin x + \cos x} dx =$

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

$\int_0^{2 \pi} (\sin x + |\sin x|) \, dx =$

$\int_{-1}^{3} \left[ \tan^{-1} \left( \frac{x}{x^{2}+1} \right) + \tan^{-1} \left( \frac{x^{2}+1}{x} \right) \right] dx =$