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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 ?
$\cos2x=1-2\sin^2 x$, 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: $$\large I=\frac{\sin\theta}{\sqrt{\cos2\theta}}+C$$
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:
$$\begin{align} \int \dfrac{\cos 2x \cos x + \sin 2x \sin x}{\cos^{\frac 3 2} 2x} dx & = \int \dfrac{\cos x}{\sqrt {\cos 2x}} + \dfrac{\sin x \sin 2x}{\cos^\frac 3 2 2x}dx\\ & = \int \dfrac{d (\sin x)}{\sqrt {\cos 2x}} - \dfrac 1 2 \int \dfrac{\sin x \, d(\cos 2x)}{\cos^\frac 3 2 2x}\\ & = \int \dfrac{du}{\sqrt v} - \dfrac 1 2 \int\dfrac{u\, dv}{v^{3/2}} \qquad [u = \sin x,\ v = \cos x]\\ & = \int d\left( \dfrac{u}{\sqrt v} \right)\\ & = \dfrac{u}{\sqrt v}\\ & = \dfrac{\sin x}{\sqrt {\cos 2x}} \end{align}$$
$$\int\dfrac{\cos xdx}{\cos^{3/2}2x}=\int\dfrac{\cos xdx}{(\cos^2x-\sin^2x)^{3/2}}=$$ $$\dfrac{\cos xdx}{(1-\tan^2x)^{3/2}\cos^3x}=\int\dfrac{\sec^2xdx}{(1-\tan^2x)^{3/2}}$$ $$u=\tan x,du=\sec^2xdx$$ $$\int\frac{du}{(1-u^2)^{3/2}}$$ $$u=\sin\theta,du=\cos\theta d\theta$$ $$\int\dfrac{\cos\theta d\theta}{\cos^3\theta}=\tan\theta+C=\frac{u}{\sqrt{1-u^2}}+C=\frac{\tan x}{\sqrt{1-\tan^2x}}+C=$$ $$\dfrac{\sin x}{\sqrt{\cos^2x-\sin^2x}}+C=\dfrac{\sin x}{\sqrt{\cos2x}}+C$$
Integrate by parts $$I=\int\frac{\cos x}{\cos^{3/2}2x}dx= \int\frac{\sqrt {2\sin x}}{\cos^{1/4}2x} d\left(\frac{\sqrt {2\sin x}}{\cos^{1/4}2x} \right)=\frac{2\sin x}{\sqrt{\cos 2x}}-I$$
