Beautiful integral

All of the solutions given are in the form $\sqrt{\frac{sec(2x)+C}{2}}$ , so if we can find what the constant $C$ is we'll have our answer. To do this, we take the derivative of our answer and set it equal to the formula in our question, $cos(x) \sqrt{sec^{3}(2x)}$ .

$\frac{d}{dx} \sqrt{\frac{sec(2x)+C}{2}} = \frac{\sqrt{2}}{2} \frac{d}{dx} \sqrt{sec(2x)+C} = \frac{\sqrt{2}}{2} \frac{1}{2} 2tan(2x)sec(2x) \frac{1}{\sqrt{sec(2x)+C}}$

Simplification and replacing $tan(2x)$ with $sin(2x)sec(2x)$ :

$\frac{\sqrt{2}}{2} sin(2x)sec^{2}(2x) \frac{1}{\sqrt{sec(2x)+C}} = \frac{\sqrt{2}}{2} sin(2x) \frac{sec^{2}(2x)}{\sqrt{sec(2x)+C}}$

Replace $sin(2x)$ with $2sin(x)cos(x)$ and simplify:

$\frac{\sqrt{2}}{2} 2sin(x)cos(x) \frac{sec^{2}(2x)}{\sqrt{sec(2x)+C}} = \sqrt{2} sin(x)cos(x) \frac{sec^{2}(2x)}{\sqrt{sec(2x)+C}}$

Using the distributive property we have: $\sqrt{sec(2x)+C} = \sqrt{sec(2x)(1+C\times cos(2x))} = \sqrt{sec(2x)}\sqrt{1+C\times cos(2x)}$

Subbing in, simplifying:

$\sqrt{2} sin(x)cos(x) \frac{sec^{2}(2x)}{\sqrt{sec(2x)}\sqrt{1+C\times cos(2x)}} = \frac{\sqrt{2} sin(x)}{\sqrt{1+C\times cos(2x)}} cos(x) \sqrt{sec^{3}(2x)}$

Setting equal to our desired function:

$\frac{\sqrt{2} sin(x)}{\sqrt{1+C\times cos(2x)}} cos(x) \sqrt{sec^{3}(2x)} = cos(x) \sqrt{sec^{3}(2x)}$

From this we can see that $\frac{\sqrt{2} sin(x)}{\sqrt{1+C\times cos(2x)}}$ must equal $1$ for our equation to work. We can solve this equation for $C$ :

$\frac{\sqrt{2} sin(x)}{\sqrt{1+C\times cos(2x)}} = 1$

$\sqrt{2} sin(x) = \sqrt{1+C\times cos(2x)}$

$2sin^{2}(x) = 1+C\times cos(2x)$

$2sin^{2}(x) - 1= C\times cos(2x)$

Replace $cos(2x)$ with $1-2sin^{2}(x)$ :

$2sin^{2}(x) - 1= C(1-2sin^{2}(x))$

$\frac{2sin^{2}(x) - 1}{1-2sin^{2}(x)} = C$

$\frac{(-1)(1-2sin^{2}(x))}{1-2sin^{2}(x)} = C$

$C = -1$

Finally, plugging in C gives us our answer: $\boxed{\sqrt{\frac{sec(2x)-1}{2}}}$