\begin{align}
\int\frac{\sqrt{\cos 2x}}{\sin x}\ dx&=\int\frac{\sqrt{\cos^2x-\sin^2x}}{\sin x}\ dx\\
&\stackrel{\color{red}{[1]}}=\int\frac{\sqrt{t^4-6t^2+1}}{t^3+t}\ dt\\
&\stackrel{\color{red}{[2]}}=\frac12\int\frac{\sqrt{u^2-6u+1}}{u^2+u}\ du\\
&\stackrel{\color{red}{[3]}}=\int\frac{(y^2-6y+1)^2}{(y-1)(y-3)(y+1)(y^2+2t-7)}\ dy\\
&\stackrel{\color{red}{[4]}}=\int\left[\frac1{y-1}+\frac1{y-3}-\frac1{y+1}-\frac{16}{y^2+2y-7}\right]\ dt\\
&=\int\left[\frac1{y-1}+\frac1{y-3}-\frac1{y+1}-\frac{16}{(y+1)^2-8}\right]\ dt
\end{align}
The rest is yours.
Notes :
$\color{red}{[1]}\;\;\;$Use Weierstrass substitution, $\tan\left(\dfrac{x}{2}\right)=t$.
$\color{red}{[2]}\;\;\;$Use substitution $u=t^2$.
$\color{red}{[3]}\;\;\;$Use Euler substitution, $y-u=\sqrt{u^2-6u+1}\;\color{blue}{\Rightarrow}\;y=\dfrac{u^2-1}{2u-6}$.
$\color{red}{[4]}\;\;\;$Use partial fractions decomposition.
As said, at the price of the "monster" given by @Raffaele in comments, you can compute it.
What you could also do (but at the price of another infinite summation, is to use
$$\sqrt{\cos(x)}=t \implies x=\cos ^{-1}\left(t^2\right)\implies dx=-\frac{2 t}{\sqrt{1-t^4}}\,dt$$ to make
$$I=-2\int \frac{t^2}{\sqrt{1-t^4}}\, \cos ^{-1}\left(t^2\right)\,dt$$
Now, using
$$\frac{t^2}{\sqrt{1-t^4}}=\sum_{n=0}^\infty (-1)^n \binom{-\frac{1}{2}}{n}\, t^{4 n+2}$$ we face the problem of
$$J_n=\int t^{4 n+2}\, \cos ^{-1}\left(t^2\right)\,dt$$ which are
$$J_n=\frac{t^{4 n+3} \left(2 t^2 \,
_2F_1\left(\frac{1}{2},\frac{4n+5}{4};\frac{4n+9}{4};t^4\right)+(4 n+5) \cos
^{-1}\left(t^2\right)\right)}{(4 n+3) (4 n+5)}$$
But, these integrals express also in terms of elliptic integrals of the first kind. Their general form is
$$J_n= t \sqrt{1-t^4}P_{n}(t^4)+\frac {t^{4n+3}}{4n+3} \, \cos ^{-1}\left(t^2\right)+a_n\, F\left(\left.\sin ^{-1}(t)\right|-1\right)$$
Concerning the $a_n$'s, they form the sequence
$$\left\{\frac{2}{9},\frac{10}{147},\frac{30}{847},\frac{26}{1155},\frac{442}{27797},
\frac{1326}{110561},\frac{11050}{1168101},\frac{320450}{41575743},\cdots\right\}$$ which do not seem to be known in $OEIS$.
However, in a private discussion, @Raymond Manzoni did identify the sequence
$$a_n=\frac{2}{3 (4 n+3)}\prod_{k=1}^n \frac{4 k+1}{4 k+3}=\frac{2}{3 (4 n+3)}\frac{\Gamma \left(\frac{7}{4}\right) \Gamma \left(n+\frac{5}{4}\right)}{\Gamma
\left(\frac{5}{4}\right) \Gamma \left(n+\frac{7}{4}\right)}$$
I think that the above would be interesting from a computing point of view.
Edit
If we are concerned by
$$I=\int_0^{\frac \pi 2} x\sqrt{\cos (x)}\,dx=\frac{4}{15} \,
_3F_2\left(1,\frac{5}{4},\frac{5}{4};\frac{7}{4},\frac{9}{4};1\right)$$
Now, using
$$I=-2\sum_{n=0}^\infty (-1)^n \binom{-\frac{1}{2}}{n}\, J_n$$ Integrated between $0$ and $1$, the first two terms of the expressions for $J_n$ are $0$ and we are left with
$$J_n=a_n\, K(-1)=\frac{2\, \sqrt{\pi } \,\Gamma \left(n+\frac{5}{4}\right)}{(4 n+3)^2 \,\Gamma \left(n+\frac{3}{4}\right)}$$
$$I=-4 \sqrt{\pi }\sum_{n=0}^\infty (-1)^n \binom{-\frac{1}{2}}{n}\,\frac{ \Gamma
\left(n+\frac{5}{4}\right)}{(4 n+3)^2 \,\Gamma \left(n+\frac{3}{4}\right)}$$
$$I=-\frac{\sqrt{\pi }\, \Gamma \left(\frac{1}{4}\right)}{9 \,\Gamma
\left(\frac{3}{4}\right)}\,
_3F_2\left(\frac{1}{2},\frac{3}{4},\frac{5}{4};\frac{7}{4},\frac{7}{4};1\right
)$$
Update
All of the above is so complex that, may be, approximations could be used.
For example, using the $1,400$ years od approximation
$$\cos(x) \simeq\frac{\pi ^2-4x^2}{\pi ^2+x^2}\qquad \text{for}\qquad -\frac \pi 2 \leq x\leq\frac \pi 2 $$ in this range
$$\int x\sqrt{\cos (x)}\,dx \sim \frac{1}{2} \sqrt{\left(\pi ^2-4 x^2\right) \left(x^2+\pi ^2\right)}+\frac{5}{4}
\pi ^2 \sin ^{-1}\left(\frac{2 \sqrt{x^2+\pi ^2}}{\sqrt{5} \pi }\right)$$ which, integrated between $0$ and $\frac \pi 2$ would give
$$\frac{5 \pi ^3}{8}-\frac{\pi ^2}{4} \left(2+5 \sin
^{-1}\left(\frac{2}{\sqrt{5}}\right)\right)\approx 0.785221$$ while the exact value is $\approx 0.784608$.
Best Answer
Rewrite the integrand as
$$\begin{align*} & \int\frac{\sin(x)-x\cos(x)}{x\sqrt{x\sin(x)}} \, dx \\ &= \int \left(\frac{\sqrt{\sin x}}{x\sqrt x} - \frac{\cos x}{\sqrt{x \sin x}}\right) \, dx \\ &= \int \frac{\sin x-x\cos x}{x\sin x} \sqrt{\frac{\sin x}x} \, dx \end{align*}$$
Now substitute
$$u=\sqrt{\frac{\sin x}x} \implies du = -\frac12 \sqrt{\frac x{\sin x}} \frac{\sin x-x\cos x}{x^2}\,dx = -\frac12 \frac{\sin x-x\cos x}{x\sqrt{x\sin x}} \, dx$$