Originally Posted by
simplependulum $\displaystyle \frac{1}{\cos(x) \sqrt{\sin(2x)}}$
$\displaystyle = \frac{1 + \cos(2x) - \cos(2x) }{\cos(x) \sqrt{\sin(2x)}}$
$\displaystyle = \frac{2\cos(x)}{\sqrt{\sin(2x)}} - \frac{\cos(2x)\sec(x)}{\sqrt{\sin(2x)}} $
its integral $\displaystyle = $
$\displaystyle \int \frac{2\cos(x)}{\sqrt{\sin(2x)}}~dx - \int \frac{\cos(2x)\sec(x)}{\sqrt{\sin(2x)}} ~dx $
$\displaystyle = \int \frac{2\cos(x)}{\sqrt{\sin(2x)}}~dx - \sqrt{\sin(2x)}\sec(x) + \int \sqrt{\sin(2x)} \sec(x) \tan(x) ~dx $
$\displaystyle = \int \frac{ 2\cos^2(x) + 2\sin^2(x)}{ \cos(x) \sqrt{\sin(2x)} } ~dx - \sqrt{\sin(2x)}\sec(x) +C $
therefore
$\displaystyle I = 2I - \sqrt{\sin(2x)}\sec(x) + C ~ , ~ I = \sqrt{\sin(2x)}\sec(x) + C' $
It is just a trial ... but luckily it is a successful trial