$\int\limits_{0}^{5} \cos \left(\pi\left(x-\left[\frac{x}{2}\right]\right)\right) d x$,where $[t]$ denotes the greatest integer less than or equal to $t$,is equal to:

For a real number $x$,let $[x]$ denote the greatest integer less than or equal to $x$. The smallest positive integer $n$ for which the integral $\int_{1}^{n} [x][\sqrt{x}] \, dx$ exceeds $60$ is

Let $\lim _{c \rightarrow 0} \int_c^x \frac{b t \cos 4 t - a \sin 4 t}{t^2} d t = \frac{a \sin 4 x}{x} - 1$. Find the values of $a$ and $b$.

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

If $[2,6]$ is divided into four intervals of equal length,then the approximate value of $\int_2^6 \frac{1}{x^2-x} dx$ using Simpson's rule is

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin (x-[x]) \, dx=$