$\int_0^{\pi /2} |\sin x - \cos x| \, dx = $

The value of the integral $\int_{-2}^{2} \frac{\sin^2 x}{[\frac{x}{\pi}] + \frac{1}{2}} \, dx$ (where $[x]$ denotes the greatest integer less than or equal to $x$) is

The function $F(x) = \int_0^x \log \left( \frac{1 - t}{1 + t} \right) \,dt$ is

Let $f:[-1,2] \rightarrow[0, \infty)$ be a continuous function such that $f(x)=f(1-x), \forall x \in[-1,2]$. If $R_1=\int_{-1}^2 x f(x) d x$ and $R_2$ is the area of the region bounded by $y=f(x), x=-1, x=2$ and the $X$-axis,then:

Let $I_n = \int_0^{\pi / 2} x^n \cos x \, dx$,where $n$ is a non-negative integer. Then,$\sum_{n=2}^{\infty} \left( \frac{I_n}{n!} + \frac{I_{n-2}}{(n-2)!} \right)$ equals

If $g(x)$ is the inverse of $f(x)$ and $f(x)$ has domain $x \in [1, 5]$,where $f(1) = 2$ and $f(5) = 10$,then the value of $\int_{1}^{5} f(x) dx + \int_{2}^{10} g(y) dy$ is equal to: