$\int_{\pi}^{16\pi} |\sin x| dx = $

$\int_{0}^{\pi} [\cot x] dx = $

Evaluate the definite integral: $\int_0^\pi \frac{x \cos^2 x}{1+\sin x} dx$

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:

$\int_{-\pi}^\pi \frac{x \sin ^3 x}{4-\cos ^2 x} d x=$

$\int_0^{\pi /2} \frac{\sin^{3/2} x}{\cos^{3/2} x + \sin^{3/2} x} dx = $