Let $\int_0^1 f(x) \, dx = 1$,$\int_0^1 x f(x) \, dx = a$,and $\int_0^1 x^2 f(x) \, dx = a^2$. Then the value of $\int_0^1 (x - a)^2 f(x) \, dx$ is:

$\int_{\pi /6}^{\pi /4} \text{cosec} \, 2x \, dx = $

If $\int_0^{\frac{1}{2}} \frac{x^2}{\left(1-x^2\right)^{\frac{3}{2}}} \,d x=\frac{k}{6}$, then the value of $k$ is

$\int_0^1 \frac{e^{-x}}{1 + e^{-x}} \,dx = $

Let $f:[0,1] \rightarrow [0,1]$ be a continuous function such that $x^2+(f(x))^2 \leq 1$ for all $x \in [0,1]$ and $\int_0^1 f(x) dx = \frac{\pi}{4}$. Then,$\int_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}} \frac{f(x)}{1-x^2} dx$ equals

$I = \int_{\pi/4}^{\pi/3} \frac{8 \sin x - \sin 2x}{x} dx$. Then