<s>[INST] 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 ?
 [/INST] $\cos2x=1-2\sin^2 x$, let $t=\sin x$ so $dt=\cos xdx$ in numerator.
SO:
$$I=\int\frac{dt}{(1-2t^2)^{3/2}}$$
Then use the substitution $u^2=(1-2t^2)$ so $udu=-2tdt$, so $\displaystyle dt=-\frac{udu}{2t}=\frac{-udu}{\sqrt2\sqrt{1-u^2}}$:
$$I=-\int\frac{udu}{\sqrt2\sqrt{1-u^2}u^3}=\frac{-1}{\sqrt2}\int\frac{du}{u^2\sqrt{1-u^2}}$$
Then use $v=1/u$ so $\displaystyle du=-\frac1{v^2}dv$ So:
$$I=\frac1{\sqrt2}\int\frac{vdv}{\sqrt{v^2-1}}=\frac1{\sqrt2}\sqrt{v^2-1}+C$$
Now:
$$\sqrt{v^2-1}=\frac{\sqrt{1-u^2}}u=\frac{\sqrt2t}{\sqrt{1-2t^2}}=\frac{\sqrt2\sin\theta}{\sqrt{1-2\sin^2\theta}}=\frac{\sqrt2\sin\theta}{\sqrt{\cos2\theta}}$$
So:
$$\large I=\frac{\sin\theta}{\sqrt{\cos2\theta}}+C$$
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