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..
 A: A relative of your substitution is $\sqrt{x}=\sin\theta$. Then $dx=2\sin\theta\cos\theta\,d\theta$. 
And $\dfrac{1-\sin\theta}{1+\sin\theta}=\dfrac{(1-\sin\theta)^2}{1-\sin^2\theta}$. Taking the square root, we get $\dfrac{1-\sin\theta}{\cos\theta}$. So our integral becomes 
$$\int 2\sin\theta(1-\sin\theta)\,d\theta,$$ 
which is quite straightforward. 
Remark: A sort of fun idea is to make a rationalizing non-trigonometric substitution. For instance, we can let $\sqrt{x}=\dfrac{1-t^2}{1+t^2}$, or, it turns out equivalently, let $\dfrac{1-\sqrt{x}}{1+\sqrt{x}}=t^2$. We can solve simply for $\sqrt{x}$ (and therefore $x$) in terms of $t$, and then find an expression for $dx$. (The result is closely related to the $\tan(\theta/2)$ substitution.) After some calculation, we find that we want
$$\int -\frac{8(1-t^2)t^2}{(1+t^2)^3}.$$
Now in principle it's all over, since we have a general procedure for integrating rational functions. But principle and practice are not precisely the same.
A: Regarding to @Ethan's comment, you can write the integrand as: $$\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}={\frac{1-\sqrt{x}}{\sqrt{1-x}}}=\frac{1}{\sqrt{1-x}}-\frac{\sqrt{x}}{\sqrt{1-x}}$$ Now by using the method called Differential binomial, you can defeat both fractions above. I personally prefer @Andre's answer cause mine may be tedious.
A: 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.
A: You could also try $\sqrt{x}=tan\theta$. 
$dx=2tan\theta sec^2\theta d\theta$
$\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}=\sqrt{\frac{1-tan\theta}{1+tan\theta}}$
Let $tan\theta=y$. Then $dy=sec^2\theta$. Substituting, we have
$\int{2y\sqrt{\frac{1-y}{1+y}}dy}=\int{\frac{2y(1-y)}{\sqrt{1-y^2}}dy}=\int{\frac{2y}{\sqrt{1-y^2}}dy}-\int{\frac{2y^2}{\sqrt{1-y^2}}dy}=I_{1}+I_{2}$
$I_{1}=-ln[\sqrt{1-y^2}]$; $I_{2}$: Let $y=sin\phi$. $dy=cos\phi d\phi$. $I_{2}=\int{2sin^2\phi d\phi}=\phi-\frac{sin2\phi}{2}=sin^{-1}y-y\sqrt{1-y^2}$ 
$I_{1}+I_{2}=y\sqrt{1-y^2} -ln[\sqrt{1-y^2}]-sin^{-1}y+C$. Now substiute $y=\sqrt{x}$. 
