Let $f(x) = \begin{cases} -2, & -2 \leq x \leq 0 \\ x-2, & 0 < x \leq 2 \end{cases}$ and $h(x) = f(|x|) + |f(x)|$. Then $\int_{-2}^2 h(x) dx$ is equal to:

$ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{dx}{e^{\sin x}+1} $ is equal to

The value of the integral $\int \limits_1^3 \left((x-2)^4 \sin^3(x-2) + (x-2)^{2019} + 1\right) dx$ is

$\int_{\frac{\pi}{4}}^{\frac{5 \pi}{4}} (|\cos t| \sin t + |\sin t| \cos t) dt =$

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:

$\int\limits_0^\pi {\frac{{x\cos x}}{{{{\left( {1 + \sin x} \right)}^2}}}} dx$ is equal to :