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:

$\int_{-1/24}^{1/24} \sec x \log \left(\frac{1-x}{1+x}\right) dx =$

$\int_0^{\frac{\pi}{2}} \frac{x \tan x \sec^2 x}{\tan^4 x + 1} dx =$

If $m, l, r, s, n$ are integers such that $9 > m > l > s > n > r > 2$ and $\int_{-2 \pi}^{2 \pi} \sin ^m x \cos ^n x \, dx = 4 \int_0^\pi \sin ^m x \cos ^n x \, dx$, $\int_{-\pi}^\pi \sin ^r x \cos ^s x \, dx = 4 \int_0^{\pi / 2} \sin ^r x \cos ^s x \, dx$ and $\int_{-\pi / 2}^{\pi / 2} \sin ^l x \cos ^m x \, dx = 0$, then which of the following is true?

Let $f:R \to R$ and $g:R \to R$ be continuous functions,then the value of the integral $\int_{-\pi/2}^{\pi/2} [f(x) + f(-x)][g(x) - g(-x)] \, dx$ is:

$\int_{\frac{1}{2}}^2 \frac{1}{x} \operatorname{cosec}^{101}\left(x-\frac{1}{x}\right) d x=$