The value of the definite integral $\int_{0}^{\frac{\pi}{2}} \sin x \sin 2x \sin 3x \, dx$ is equal to:

For a real number $x$,let $[x]$ denote the largest integer less than or equal to $x$ and $\{x\}=x-[x]$. The smallest possible integer value of $n$ for which $\int_1^n [x]\{x\} dx$ exceeds $2013$ is

Let $a$ and $b$ be real constants such that the function $f$ defined by $f(x) = \begin{cases} x^2+3x+a, & x \leq 1 \\ bx+2, & x > 1 \end{cases}$ is differentiable on $\mathbb{R}$. Then,the value of $\int_{-2}^2 f(x) dx$ equals

$ \int_{0}^{\frac{\pi}{2}} \frac{dx}{1+\cos x} = $

$\int_0^\pi \frac{\cos x}{\sqrt{1-\sin ^2 x}} d x=$

By the definition of the definite integral,the value of $\lim _{n \rightarrow \infty}\left(\frac{1^4}{1^5+n^5}+\frac{2^4}{2^5+n^5}+\frac{3^4}{3^5+n^5}+\ldots+\frac{n^4}{n^5+n^5}\right)$ is