If $\int_{0}^{\pi} \log (\sin x) dx = 8 k$,then $\int_{0}^{\pi / 4} \log (1 + \tan x) dx =$

Let $[t]$ denote the greatest integer $\leq t$. Then $\frac{2}{\pi} \int_{\pi/6}^{5\pi/6} (8[\operatorname{cosec} x] - 5[\cot x]) \, dx$ is equal to

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:

Let $f(x) = \frac{x}{(1+x^n)^{1/n}}$,$x \in R - \{-1\}$,$n \in N$,$n > 2$. If $f^n(x) = (f \circ f \circ f \dots \text{upto } n \text{ times})(x)$,then $\lim_{n \to \infty} \int_0^1 x^{n-2} (f^n(x)) dx$ is equal to $...............$.

The value of $\int_{-1}^1 \left(\sqrt{1+x+x^2} - \sqrt{1-x+x^2}\right) dx$ is

$\tan ^{-1}\left[\int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^x} d x\right]=$