$\int_{\log \frac{1}{2}}^{\log 2} \sin \left(\frac{e^{x}-1}{e^{x}+1}\right) dx=$

Let $g(x) = \int_{0}^{x} f(t) dt$,where $f$ is a continuous function in $[0, 3]$ such that $\frac{1}{3} \leq f(t) \leq 1$ for all $t \in [0, 1]$ and $0 \leq f(t) \leq \frac{1}{2}$ for all $t \in (1, 3]$. The largest possible interval in which $g(3)$ lies is:

The value of $\int_{-\pi/2}^{\pi/2} \frac{dx}{[x] + [\sin x] + 4}$,where $[t]$ denotes the greatest integer less than or equal to $t$,is

If $f(x)$ is a continuous periodic function with period $T,$ then the integral $I = \int_a^{a + T} {f(x)\,dx} $ is

If $f(a + b - x) = f(x)$,then $\int_{a}^{b} x \cdot f(a + b - x) \, dx = $

If $b = \int_{0}^{1} \frac{e^{t}}{t+1} dt$, then the value of $\int_{a-1}^{a} \frac{e^{-t}}{t-a-1} dt$ is