$\int_{-\pi}^{\pi} \frac{2 x}{1+\cos ^{2} x} d x=$

Let $f:[-2, 3] \to [0, \infty)$ be a continuous function such that $f(1-x) = f(x)$ for all $x \in [-2, 3]$. If $R_1$ is the numerical value of the area of the region bounded by $y = f(x)$,$x = -2$,$x = 3$ and the $x$-axis,and $R_2 = \int_{-2}^3 x f(x) dx$,then:

$\int_0^{\pi /2} {x\cot x\,dx} $ equals

If $f$ and $g$ are continuous functions on $[0, a]$ satisfying $f(x) = f(a - x)$ and $g(x) + g(a - x) = 2$,then $\int_0^a f(x)g(x) dx = $

$\int_{-1}^1 \frac{\sin x-x^2}{3-|x|} d x=$

By using the properties of definite integrals,evaluate the integral $\int_{-5}^{5}|x+2| d x$.