Oh k !

$\begin{aligned} Q & = \frac {\displaystyle \int_0^\frac \pi 2 \cos^{\sqrt 2+1}x \ dx}{\displaystyle \int_0^\frac \pi 2 \cos^{\sqrt 2-1}x \ dx} \\ & = \frac {\displaystyle 2 \int_0^\frac \pi 2 \sin^0 x \cos^{\sqrt 2+1}x \ dx}{\displaystyle 2 \int_0^\frac \pi 2 \sin^0 x \cos^{\sqrt 2-1}x \ dx} \\ & = \frac {B \left(\frac 12, 1+\frac 1{\sqrt 2} \right)}{B \left(\frac 12, \frac 1{\sqrt 2} \right)} & \small \color{#3D99F6} \text{where }B(m,n) \text{ is the beta function.} \\ & = \frac {\Gamma \left(\frac 12 \right)\Gamma \left(1+\frac 1{\sqrt 2} \right)}{\Gamma \left(\frac 32+\frac 1{\sqrt 2} \right)} \times \frac {\Gamma \left(\frac 12+\frac 1{\sqrt 2} \right)}{\Gamma \left(\frac 12 \right)\Gamma \left(\frac 1{\sqrt 2} \right)} & \small \color{#3D99F6} \text{where }\Gamma (z) \text{ is the gamma function.} \\ & = \frac {\frac 1{\sqrt 2}\Gamma \left(\frac 1{\sqrt 2} \right)}{\left(\frac 12+\frac 1{\sqrt 2} \right)\Gamma \left(\frac 12+\frac 1{\sqrt 2} \right)} \times \frac {\Gamma \left(\frac 12+\frac 1{\sqrt 2} \right)}{\Gamma \left(\frac 1{\sqrt 2} \right)} & \small \color{#3D99F6} \text{Note that } \Gamma (1+z) = z \Gamma (z) \\ & = \frac {\frac 1{\sqrt 2}}{\frac 12+\frac 1{\sqrt 2}} = \frac {\sqrt 2}{1+\sqrt 2} \\ & = \frac {\sqrt 2(\sqrt 2-1)}{(\sqrt 2+1)(\sqrt 2-1)} = 2 - \sqrt 2 \end{aligned}$

$\implies k = \boxed{2}$

References:

• Beta function
• Gamma function

Solution via process of elimination: Both integrands are strictly positive over the domain of integration, so their quotient is nonzero. Hence, $k\neq 1$ . The integrand of the numerator is less than or equal to the integrand of the denominator over the domain of integration, since $\cos^2(x)\leq 1$ . So, because $3-\sqrt{3}>1$ and $4-\sqrt{4}>1$ , $k\neq 3$ and $k\neq 4$ .