<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] $$\int\dfrac{\cos xdx}{\cos^{3/2}2x}=\int\dfrac{\cos xdx}{(\cos^2x-\sin^2x)^{3/2}}=$$
$$\dfrac{\cos xdx}{(1-\tan^2x)^{3/2}\cos^3x}=\int\dfrac{\sec^2xdx}{(1-\tan^2x)^{3/2}}$$
$$u=\tan x,du=\sec^2xdx$$
$$\int\frac{du}{(1-u^2)^{3/2}}$$
$$u=\sin\theta,du=\cos\theta d\theta$$
$$\int\dfrac{\cos\theta d\theta}{\cos^3\theta}=\tan\theta+C=\frac{u}{\sqrt{1-u^2}}+C=\frac{\tan x}{\sqrt{1-\tan^2x}}+C=$$
$$\dfrac{\sin x}{\sqrt{\cos^2x-\sin^2x}}+C=\dfrac{\sin x}{\sqrt{\cos2x}}+C$$
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