If $\int_0^{\frac{\pi}{2}} \frac{\cot x}{\cot x + \csc x} dx = m(\pi + n)$,then $m \cdot n$ is equal to

$\int_0^{\frac{\pi}{2}} \frac{d x}{\cos x-\sqrt{3} \sin x}=$

The true solution set of the inequality $\sqrt{5x-6-x^2} + \left( \frac{\pi}{2} \int_{0}^{x} dz \right) > x \int_{0}^{\pi} \sin^2 x dx$ is:

Let for $x \in R$,$S_0(x) = x$,$S_k(x) = C_k x + k \int_0^x S_{k-1}(t) dt$,where $C_0 = 1$,$C_k = 1 - \int_0^1 S_{k-1}(x) dx$,$k = 1, 2, 3, \ldots$. Then $S_2(3) + 6C_3$ is equal to $...........$.

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

$\int_{ - \pi /2}^{\pi /2} {\sqrt {\frac{1}{2}(1 - \cos 2x)} } \,dx = $