$\int_0^\pi x f(\sin x) dx = $

$\int_0^\pi \frac{x \, dx}{1 + \sin x}$ is equal to

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

To find the numerical value of $\int_{-2}^{2} (px^2 + qx + s) \, dx$,it is necessary to know the values of which constants?

Let $f:[-1, 2] \rightarrow [0, \infty)$ be a continuous function such that $f(x) = f(1-x)$ for all $x \in [-1, 2]$. Let $R_1 = \int_{-1}^2 x f(x) dx$ and $R_2$ be the area of the region bounded by $y = f(x)$,$x = -1$,$x = 2$,and the $X$-axis. Then $R_2$ is:

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{2-\sin \theta}{2+\sin \theta}\right) d \theta$ is equal to