Alternate approaches to solve this Integral

Question

Evaluate $$I=\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\:dx$$

I have used parts taking first function as Integrand and second function as $1$ we get

$$I=x\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}-\int \frac{d}{dx}\left(\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\right) \times x dx$$

now $$\frac{d}{dx}\left(\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\right)=\frac{1}{2\left(\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\right)} \times \left(\frac{\frac{1+\sqrt{x}}{-2\sqrt{x}}-\frac{1-\sqrt{x}}{2\sqrt{x}}}{(1+\sqrt{x})^2}\right)=\frac{-1}{2\sqrt{x}\sqrt{1-x}(1+\sqrt{x})}$$ so

$$I=x\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+\int\frac{xdx}{2\sqrt{x}\sqrt{1-x}(1+\sqrt{x})} $$ $\implies$

$$I=x\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+\frac{1}{2}\int \frac{\sqrt{x}dx}{\sqrt{1-x}(1+\sqrt{x})}$$ $\implies$

$$I=x\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+\frac{1}{2}\int\frac{\sqrt{x}(1-\sqrt{x})dx}{(1-x)^{\frac{3}{2}}}$$ $\implies$

$$I=x\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+\frac{1}{2}\int\frac{\sqrt{x}dx}{(1-x)^{\frac{3}{2}}}-\frac{1}{2}\int\frac{xdx}{(1-x)^{\frac{3}{2}}}$$

Now $$\int\frac{xdx}{(1-x)^{\frac{3}{2}}}=\int\frac{(x-1+1)dx}{(1-x)^{\frac{3}{2}}}$$ and if we split is straight forward to compute.

Also $$\int\frac{\sqrt{x}dx}{(1-x)^{\frac{3}{2}}}$$ can be evaluated using substitution $x=sin^2y$

I need any other better approaches to evaluate this integral.

Answer 1

$$ \sqrt{\dfrac{1-\sqrt{x}}{1+\sqrt{x}}}=\sqrt{\dfrac{1-\sqrt{x}}{1+\sqrt{x}}\times \dfrac{1-\sqrt{x}}{1-\sqrt{x}}}=\dfrac{1-\sqrt{x}}{\sqrt{1-x}} $$ thus we get $$ \int \dfrac{1}{\sqrt{1-x}}dx -\int \sqrt{\dfrac{x}{1-x}}dx $$ So I got to your final integrals quickly.

Answer 2

First rationalize: $$\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}$$

$$=\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\times\sqrt{\frac{1-\sqrt{x}}{1-\sqrt{x}}}$$

$$=\frac{\sqrt{1-\sqrt{x}}}{\sqrt{1+\sqrt{x}}}\times\frac{\sqrt{1-\sqrt{x}}}{\sqrt{1-\sqrt{x}}}$$

$$=\frac{1-\sqrt{x}}{\sqrt{1-x}}$$

$$=\frac{1}{\sqrt{1-x}}-\frac{\sqrt{x}}{\sqrt{1-x}}$$

Then to do the integral, for second half let $x=t^2$ so $dx=2tdt$:

$$\int\frac{1}{\sqrt{1-x}}-\frac{\sqrt{x}}{\sqrt{1-x}}dx$$

$$=-2\sqrt{1-x}-\int\frac{2t^2}{\sqrt{1-t^2}}dt$$

$$=-2\sqrt{1-x}+\int2\frac{1-t^2}{\sqrt{1-t^2}}-\frac{2}{\sqrt{1-t^2}}dt$$

$$=-2\sqrt{1-x}+\int2\sqrt{1-t^2}dt-2\arcsin t$$

$$=-2\sqrt{1-x}+t\sqrt{1-t^2}+\int\frac{1}{\sqrt{1-t^2}}dt-2\arcsin \sqrt{x}$$

$$=-2\sqrt{1-x}+\sqrt{x}\sqrt{1-x}+\arcsin t-2\arcsin \sqrt{x}+c;c\in\mathbb{R}$$

$$=-2\sqrt{1-x}+\sqrt{x}\sqrt{1-x}-\arcsin \sqrt{x}+c$$

Answer 3

$\sqrt x=\cos2t~$ seems like the most natural substitution, since there are well-known trigonometric formulas for $1\pm\cos2t,~$ and it is fairly obvious that $x\in\Big[0,1\Big]$.