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I need to show the following identity

$$ \int_a^b\frac{dx}{\sqrt{x-x^2}}=\int_{\tfrac{1-\sqrt{1-a}}{2}}^{\tfrac{1-\sqrt{1-b}}{2}}\frac{dx}{\sqrt{x-x^2}}+\int_{\tfrac{1+\sqrt{1-b}}{2}}^{\tfrac{1+\sqrt{1-a}}{2}}\frac{dx}{\sqrt{x-x^2}} $$

the sum of the integrals on the right side

$\arcsin \left(-\sqrt{-b+1}\right)-\arcsin \left(-\sqrt{-a+1}\right)+\arcsin \left(\sqrt{-a+1}\right)-\arcsin \left(\sqrt{-b+1}\right)$

and

$$\int_a^b\dfrac{dx}{\sqrt{x-x^2}}=\arcsin \left(2b-1\right)-\arcsin \left(2a-1\right). $$

Is there any trigonometric identity to show that they are the same? How does this show?

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The integrals in the right-hand side sum to $$ 2\arcsin\sqrt{1-a}-2\arcsin\sqrt{1-b} $$ If you look at https://math.stackexchange.com/a/3211273/62967, you'll see that $$ \arcsin(2x-1)=2\arcsin\sqrt{x}-\frac{\pi}{2} $$ For $x=1-a$, we have $2x-1=2-2a-1=1-2a$.

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