The value $9 \int_0^9 \left[ \sqrt{\frac{10x}{x+1}} \right] dx$,where $[t]$ denotes the greatest integer less than or equal to $t$,is . . . . . . .

If the value of the integral $\int_{0}^{1/2} \frac{x^{2}}{(1-x^{2})^{3/2}} dx$ is $\frac{k}{6}$,then $k$ is equal to

$\int_0^{\pi /6} {(2 + 3{x^2})\cos 3x\,dx = } $

$\int_{1 / 2}^{1 / \sqrt{2}} \frac{1}{\left(x+\sqrt{1-x^2}\right)\left(1-x^2\right)} d x=$

$\int_0^4 \frac{x+2}{\sqrt{4x-x^2}} dx =$

For a real number $x$,let $[x]$ denote the greatest integer less than or equal to $x$ and $\{x\} = x - [x]$. Let $n$ be a positive integer. Then,$\int_0^n \cos(2 \pi [x] \{x\}) dx$ is equal to