Let ${I_1} = \int_a^{\pi - a} {xf(\sin x)dx}$ and ${I_2} = \int_a^{\pi - a} {f(\sin x)dx}$,then ${I_2}$ is equal to

If $h(a) = h(b)$,the value of the integral $\int_a^b {[f(g(h(x)))]^{-1} f'(g(h(x))) \cdot g'(h(x)) \cdot h'(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 $-2$: $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx$.

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \cos^2 x(\sin x + \cos x) dx =$

$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin^{2} x \, dx =$

$\int_{0}^{1} \sin \left( 2 \tan^{-1} \sqrt{\frac{1+x}{1-x}} \right) \, dx = $