Integrate $\int\frac{\cos x dx}{\cos^{3/2}2x}=\frac{\sin x}{\sqrt{\cos 2x}}+C$ Wolfram gives this nice result:
$$\int\frac{\cos x dx}{\cos^{3/2}2x}=\frac{\sin x}{\sqrt{\cos 2x}}+\text{constant}$$ 
I have tried writing $\cos 2x = \cos^2x - \sin^2x $ and doing Weierstrass substitution $\tan (x/2) = t$ but its getting very complicated. Any help/hints ?

Alternative method (akin to finding an integrating factor by inspection):
$\cos x = \cos (2x - x) = \cos 2x \cos x + \sin 2x \sin x$.
Then the integral becomes:
$$\begin{align}
\int \dfrac{\cos 2x \cos x + \sin 2x \sin x}{\cos^{\frac 3 2} 2x} dx
& = \int \dfrac{\cos x}{\sqrt {\cos 2x}} + \dfrac{\sin x \sin 2x}{\cos^\frac 3 2 2x}dx\\
& = \int \dfrac{d (\sin x)}{\sqrt {\cos 2x}} - \dfrac 1 2 \int  \dfrac{\sin x \, d(\cos 2x)}{\cos^\frac 3 2 2x}\\
& = \int \dfrac{du}{\sqrt v} - \dfrac 1 2 \int\dfrac{u\, dv}{v^{3/2}} \qquad [u = \sin x,\ v = \cos x]\\
& = \int d\left( \dfrac{u}{\sqrt v} \right)\\
& = \dfrac{u}{\sqrt v}\\
& = \dfrac{\sin x}{\sqrt {\cos 2x}}
\end{align}$$
