Here is an approach.
Theorem.
Let $s$ be a real number such that $-1<s<1$. Then
\begin{equation}{\Large\int_{0}^{\!\Large \frac{\pi}{2}}} \frac{\cos \left(\! s \arctan \left(-\frac{x}{\ln \cos x}\right)\right)}{(x^2+\ln^2\! \cos x)^{\Large\frac{s}{2}}}\, \mathrm{d}x = \frac{\pi}{2}\frac{1}{\ln^{\Large s}\!2} \tag1\end{equation}
Proof. First assume that $0<s<1.$ Then we may write
$$
\begin{align}
\int_{0}^{\!\Large \frac{\pi}{2}} \frac{\cos \left(\! s \arctan \left(-\frac{x}{\ln \cos x}\right)\right)}{(x^2+\ln^2\! \cos x)^{s/2}} \mathrm{d}x
& = \frac{1}{\Gamma(s)}\int_{0}^{\!\Large \frac{\pi}{2}}\!\!\int_{0}^{+\infty} u^{s-1} \cos (u x) \:e^{u \ln \cos x}\:\mathrm{d}u \:\mathrm{d}x \tag2\\\\
& = \frac{1}{\Gamma(s)}\int_{0}^{+\infty}u^{s-1}\!\!\int_{0}^{\!\Large \frac{\pi}{2}} \cos^u\! x \cos (u x)\:\mathrm{d}x \:\mathrm{d}u \tag3\\\\
& = \frac{1}{\Gamma(s)}\int_{0}^{+\infty}u^{s-1} \frac{\pi}{2^{u+1}}\:\mathrm{d}u \tag4\\\\
& = \frac{\pi}{2} \frac{1}{\Gamma(s)}\int_{0}^{+\infty}u^{s-1} e^{-u\ln 2}\:\mathrm{d}u \tag5\\\\
& = \frac{\pi}{2}\frac{1}{\Gamma(s)}\frac{\Gamma(s)}{\ln^{s}\!2} \tag6\\\\
& = \frac{\pi}{2}\frac{1}{\ln^{s}\!2} \tag7
\end{align}
$$
where we have used Fubini's theorem and the classic results (here and there)
$$
\begin{align}
& \int_{0}^{+\infty} u^{s-1} \cos (a u) \:e^{-b u}\:\mathrm{d}u = \Gamma (s)\frac{\cos \left(\! s \arctan \left(\frac{a}{b}\right)\right)}{(a^2+b^2)^{s/2}}, \, \left(\Re(s)>0, b>0, a>0 \right) \tag8 \\
& \int_{0}^{\!\Large \frac{\pi}{2}} \cos^u\! x \cos (u x)\:\mathrm{d}x = \frac{\pi}{2^{u+1}}, \quad u>-1. \tag9
\end{align}
$$
We may extend identity $(7)$ by analytic continuation to obtain $(1)$.
Example 1. We have
$$
\int_{0}^{\pi/2}\displaystyle
\sqrt{\sqrt{x^2 + \ln^2\!\cos x}-\ln\! \cos x}\,\,\mathrm{d}x = \frac{\pi}{2}\sqrt{2\ln 2} \tag{10}
$$
and
$$
\int_{0}^{\pi/2}\displaystyle
\frac{1}{\sqrt{x^2 + \ln^2\!\cos x}} \sqrt{\sqrt{x^2 + \ln^2\!\cos x}-\ln\! \cos x}\,\,\mathrm{d}x = \frac{\pi}{\sqrt{2\ln 2}} \tag{11}
$$
Proof. Let $0<x<\frac{\pi}{2}$ and set $t:=\arctan \left(-\frac{x}{\ln \cos x}\right)$. Observe that $0 < t < \frac{\pi}{2}$ and
$$
\cos t=\cos \left(\!\arctan \left(-\frac{x}{\ln \cos x}\right)\right)= -\frac{\ln \cos x}{\sqrt{x^2 + \ln^2\!\cos x}},
$$
$$
\cos \left(\frac{t}{2}\right) = \sqrt{
\frac{1}{2}+ \frac{1}{2} \cos t}
$$
then put successively $\displaystyle s=-\frac{1}{2}$, $\displaystyle s=\frac{1}{2}$ in $(1)$ to obtain $(10)$ and $(11)$.
Example 2.
$$
\frac{2}{\pi}\int_{0}^{\pi/2}\sqrt[\normalsize{8}]{x^2 + \ln^2\!\cos x}
\sqrt{ \frac{1}{2}+\frac{1}{2}\sqrt{
\frac{1}{2}+ \frac{1}{2} \sqrt{ \frac{\ln^{2}\!\cos x}{
x^2 + \ln^2\! \cos x}}}}\,\mathrm{d}x = \sqrt[\normalsize{4}]{\ln 2}
\tag{12} $$
and
$$
\frac{2}{\pi}\int_{0}^{\pi/2}\frac{1}{\sqrt[\normalsize{8}]{x^2 + \ln^2\!\cos x}}
\sqrt{ \frac{1}{2}+\frac{1}{2}\sqrt{
\frac{1}{2}+ \frac{1}{2} \sqrt{ \frac{\ln^{2}\!\cos x}{
x^2 + \ln^2\! \cos x}}}}\,\mathrm{d}x=\frac{1}{\sqrt[\normalsize{4}]{\ln 2}}
\tag{13} $$
Proof. Let $0<x<\frac{\pi}{2}$ and set $t:=\arctan \left(-\frac{x}{\ln \cos x}\right)$. Observe that $0 < t < \frac{\pi}{2}$ and
$$
\cos t=\cos \left(\!\arctan \left(-\frac{x}{\ln \cos x}\right)\right)=\sqrt{ \frac{\ln^{2}\!\cos x}{x^2 + \ln^2\! \cos x}},
$$
$$
\cos \left(\frac{t}{4}\right) = \sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{
\frac{1}{2}+ \frac{1}{2} \cos t}},
$$ then put successively $\displaystyle s=-\frac{1}{4}$, $\displaystyle s=\frac{1}{4}$ in $(1)$ to obtain $(12)$ and $(13)$.
Example n.
Set
$$
R_n^{+}:=\frac{2}{\pi}\int_{0}^{\pi/2}\sqrt[\Large {2^n}]{x^2 + \ln^2\!\cos x}
\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\cdots+\frac{1}{2}\sqrt{
\frac{1}{2}+ \frac{1}{2}\sqrt{ \frac{\ln^{2}\!\cos x}{
x^2 + \ln^2\! \cos x}}}}}\,\mathrm{d}x
$$
and
$$
R_n^{-}:=\frac{2}{\pi}\int_{0}^{\pi/2}\frac{1}{\sqrt[\Large {2^n}]{x^2 + \ln^2\!\cos x}}
\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\cdots+\frac{1}{2}\sqrt{
\frac{1}{2}+ \frac{1}{2}\sqrt{ \frac{\ln^{2}\!\cos x}{
x^2 + \ln^2\! \cos x}}}}}\,\mathrm{d}x.
$$
Then
$$
R_n^{+}= \sqrt[\Large {2^n}]{\ln 2}
\tag{14}
$$
and
$$
R_n^{-}= \frac{1}{\sqrt[\Large {2^n}]{\ln 2}}
\tag{15}.
$$
Proof. Let $0<x<\frac{\pi}{2}$ and set $t:=\arctan \left(-\frac{x}{\ln \cos x}\right)$. Observe that $0 < t < \frac{\pi}{2}$ and
$$
\cos t=\sqrt{ \frac{\ln^{2}\!\cos x}{x^2 + \ln^2\! \cos x}},
$$
$$
\cos \left(\frac{t}{2^n}\right) = \sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\cdots+\frac{1}{2}\sqrt{
\frac{1}{2}+ \frac{1}{2} \cos t}}}},
$$ then put successively $\displaystyle s=-\frac{1}{2^n}$, $\displaystyle s=\frac{1}{2^n}$, $n\geq 1,$ in $(1)$ to obtain $(14)$ and $(15)$.
