$ \int_{-3}^{3} \cot^{-1} x \, dx = $

$\int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x=$

If $g(x) = \int_0^x \cos^4 t \,dt$, then $g(x+\pi)$ equals

$\int \limits_{6}^{16} \frac{\log _{e} x^{2}}{\log _{e} x^{2}+\log _{e}\left(x^{2}-44 x+484\right)} d x$ is equal to:

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:

If $\int_{-\infty}^{\infty} f(x) dx = 1$,then $\int_{-\infty}^{\infty} f\left(x - \frac{1}{x}\right) dx$ is equal to