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.

$\bf{My\; Solution::}$ Let $\displaystyle I = \int \sqrt{\sec x+\tan x}\cdot \sec^2 xdx = \int\frac{\sqrt{\sec^2 x-\tan^2 x}}{\sqrt{\sec x-\tan x}}\cdot \sec^2 xdx$
Now Integral $\displaystyle I = \int\frac{\sec^2 x}{\left(\sec x-\tan x \right)^{\frac{1}{2}}}dx$
Now Let $\left(\sec x-\tan x \right) = t^2\;,$ Then $\sec x\cdot \left(\tan x-\sec x\right)dx = 2tdt\Rightarrow \sec x\cdot t dx =-2tdt$
So $\sec xdx = -2dt$ So Integral $\displaystyle I = -2\int \frac{\sec x }{t}dt$
Now Using $\displaystyle \left(\sec x- \tan x\right) = t$ and $\displaystyle \left(\sec x+\tan x \right) = \frac{1}{t}$
Using $\left(\sec^2 x-\tan^2 x\right) = 1$
So $\displaystyle 2\sec x = t+\frac{1}{t} = \frac{t^2+1}{t}$
So Integral $\displaystyle I = -\int\frac{t^2+1}{t^2}dt = -t+\frac{1}{t}+\mathbb{C}$
So $\displaystyle I = \int \sqrt{\sec x+\tan x}\cdot \sec^2 xdx = -\left(\sec x-\tan x\right)+\frac{1}{\left(\sec x -\tan x\right)}+\mathbb{C}$
$\displaystyle  = -\left(\sec x-\tan x\right)+\left(\sec x+\tan x\right)+\mathbb{C} = 2\tan x+\mathbb{C}$
