$\begin{eqnarray*} \int {\cos x\sqrt {\sin 2x} dx} &=& \int {\cos x\sqrt {2\sin x\cos x} dx} \\ &=& \sqrt 2 \int {\sqrt {\cos x\sin x} \cos xdx} \\ \begin{cases}\sin x = u\\ \cos xdx = du\end{cases} \\ &=& \sqrt 2 \int {{u^{1/2}}{{\left( {1 - {u^2}} \right)}^{1/4}}} du \end{eqnarray*} $
Do you know how to integrate differential binomials?
See this answer of mine. Since
$\frac{{m + 1}}{n} + p = \frac{3}{4} + \frac{1}{4} = 1$ is an integer, you should be able to integrate this in terms of elementary functions with the instructions provided in the answer I linked to. Letting $u^2=z$ gives
$ = \frac{{\sqrt 2 }}{2}\int {{{\left( {\frac{{1 - z}}{z}} \right)}^{1/4}}} dz$
Now let $\frac{{1 - z}}{z} = {m^4}$ whence $dz = \frac{{4{m^3}dm}}{{{{\left( {{m^4} + 1} \right)}^2}}}$
and get
$ = 2\sqrt 2 \int {\frac{{{m^4}}}{{{{\left( {{m^4} + 1} \right)}^2}}}dm} $ which is a treatable rational function.