Integrate indefinite function :$\int {dx\over\sqrt x \sqrt {1-x}}$ How do you integrate indefinite $\int {dx\over \sqrt x \sqrt {1-x}}$ using substitution? I know that the answer is $2*\arcsin(\sqrt x)+c$ by substituting $u=\sqrt x$ and $2du={dx\over\sqrt x}$ but I don't understand how plugging it in will make the function $\int {2du\over\sqrt {1-u^2}}$  
 A: After appropriately substituting, you do get an integral of the form $$\int \frac {2\,du}{\sqrt{1 - u^2}}$$
After all, you've correctly computed $ u = \sqrt x \implies \color{blue}{\bf u^2 = x}\;\;\text{and}\;\;\color{red}{\bf 2 \,du = \frac{dx}{\sqrt x}}$. Now, just substitute, piece for piece (match up the colors, if needed), into $$\int \dfrac{\color{red}{\bf dx}}{\color{red}{\bf \sqrt x}\cdot \sqrt{1 - \color{blue}{\bf x}}}$$
This is the form that applies here: Now you can use the substitution $u = \sin \theta$.
Recall that when $a^2 - u^2$ occurs in an integrand, you can use the substitution $u = a\sin \theta$. 
A: Let $x=\sin^2 t$. Then $dx=2\sin t\cos t\,dt$ and $\sqrt{x}\sqrt{1-x}=\sin t\cos t$. Thus
$$\int \frac{1}{\sqrt{x}\sqrt{1-x}}\,dx=\int 2\,dt=2t+C.$$
Thus our integral is $2\arcsin(\sqrt{x})+C$.
Remark: The "normal" thing is to complete the square. However, $\sqrt{x(1-x)}$ is in many ways just as nice as $\sqrt{1-u^2}$, and one can take advantage of the special form to bypass the standard completing the square.  
