<s>[INST] 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..
 [/INST] 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 $y=\sqrt{x}$. 
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