Finding another way of doing this integral $\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}$ Problem : 
Integrate : $\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}$
I have the solution : We can substitute $\sqrt{x}= \cos^2t$ and proceeding further, 
I got the  the answer which is $-2\sqrt{1-x}+\cos^{-1}\sqrt{x}+\sqrt{x-x^2}+C$
Can we do this problem some other way as well.. Thanks..

Write $y = \sqrt{(1-\sqrt{x})/(1+\sqrt{x})}$.  Then $y^2 = (1 - \sqrt{x})/(1+\sqrt{x}) = -1 + 2/(1+\sqrt{x}),$
so 
$2/(1+y^2) = 1 + \sqrt{x},$
and so
$$x = \Bigl(\dfrac{2}{1+ y^2} - 1\Bigr)^2.$$ 
Thus $$y\, dx = y \, d\Bigl(\dfrac{2}{1+y^2} - 1\Bigr)^2
= -8 y^2 \Bigl(\dfrac{2}{1+y^2} - 1\Bigr)\dfrac{dy}{(1+y^2)^2}.$$
The indefinite integral of any rational function of $y$ can be
expressed in terms of elementary functions of $y$, and hence
our original integral, $\int y dx,$ can be expressed in terms of
elementary functions of $y$.
Working out the details will be painful, though, compared to the
trig-based substitutions.
