If $\int {\frac{{\cos 4x + 1}}{{\cot x - \tan x}}} dx = k\,\cos 4x + c$,then:

If $\int \frac{dx}{(x^2+9) \sqrt{x^2+16}} = \frac{1}{3 \sqrt{7}} \operatorname{Tan}^{-1} \left( K \frac{x}{\sqrt{16+x^2}} \right) + c$, then $K=$

$\int \frac{d x}{\left(2 a x+x^2\right)^{\frac{3}{2}}} = $

If $\int {\frac{{\csc^2 x}}{{{{\left( {\csc x + \cot x} \right)}^{\frac{9}{2}}}}}\,dx} = {\left( {\csc x - \cot x} \right)^{\frac{7}{2}}}\left( {\frac{1}{\alpha } + \frac{{{{\left( {\csc x - \cot x} \right)}^2}}}{{11}}} \right) + C$ (where $C$ is the constant of integration and $\alpha \in N$),then $\alpha$ is:

The value of $\int \sqrt{1+\sec x} \, dx$ is

If $\int \frac{dx}{(x \tan x + 1)^2} = f(x) + c$,then $\lim_{x \rightarrow \frac{\pi}{2}} f(x) = $