Integrate $\int \sqrt{(\sec{x} +\tan{x})}\ \cdot \sec^2x\,dx$ Integrate: $$\int \sqrt{(\sec{x} +\tan{x})}\ \cdot \sec^2x\,dx$$  
My attempt :
I substituted $\sec{x} + \tan{x} $ as $t^2$
Then, $$ (\sec{x} \cdot \tan{x} + \sec^2x) dx =2tdt$$
$$\sec{x}(  \tan{x} + \sec{x}) dx =2tdt$$  
$$\sec{x}\cdot t^2 dx =2tdt$$
But there is  $\sec{x}$ left outside.

Hmmm...  It's possible double angle identities will work here.
$$\int\sqrt{\sec x+\tan x}\sec^2xdx=\int\sec^{5/2}x\sqrt{1+\sin x}dx=$$
$$\int\sec^{5/2}x\sqrt{\sin^2\frac x2+\cos^2\frac x2+2\sin\frac x2\cos\frac x2}dx=$$
$$\int(\cos^2\frac x2-\sin^2 \frac x2)^{5/2}(\sin\frac x2+\cos\frac x2)dx=$$
$$\int(\cos\frac x2-\sin\frac x2)^{5/2}(\cos\frac x2+\sin\frac x2)^{7/2}dx$$
Not sure where to go from here.  Maybe someone else can find a way to make this work.
