<s>[INST] Evaluate: $\int \frac{1}{\sqrt{x}\sqrt{1-x}}\:dx$ Evaluate: $\int \frac{1}{\sqrt{x}\sqrt{1-x}}\:dx$
I solved this question using u-substitution.
But I got two answers:
1) When $x=u^2 \implies \int \frac{1}{\sqrt{x}\sqrt{1-x}}\:dx =  
$2arcsinx $\left(\sqrt{x}\right)$ +c and 
2) When $\sqrt{1-x}=u$ ===> $\int \frac{1}{\sqrt{x}\sqrt{1-x}}\:dx$ = -2arcsin $\sqrt{1-x}$       +c
So why did I get two answers? Are they equivalent? 
 [/INST] another way is :
$$\int \frac{dx}{\sqrt{x}\sqrt{1-x}}\\=\int \frac{dx}{\sqrt{x-x^2}}\\=\int \frac{dx}{\sqrt{\frac{1}{4}-\frac{1}{4}+x-x^2}}\\=\int \frac{dx}{\sqrt{\frac{1}{4}-(x-\frac{1}{2})^2}}\\=\int \frac{2dx}{\sqrt{1-(2x-1)^2}}\\$$now use this substittuion $u=2x-1$ so $du=2dx$ $$\int \frac{du}{\sqrt{1-u^2}}=arcsin(u)=arcsin(2x-1)$$ but  answers are equal $$(arcsin(2x-1))'=\frac{2}{\sqrt{1-(2x-1)^2}}=\frac{2}{\sqrt{4x-4x^2}}=\frac{1}{\sqrt{x-x^2}}$$
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