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Integrate $\int \frac{dx}{\sqrt{(x-a)(b-x)}}$ where $b>a$

My Attempt $$ \begin{align} &\int \frac{1}{\sqrt{(x-a)(b-x)}}dx=\int\frac{dx}{\sqrt{-x^2+(a+b)x-ab}}\\ &=\int\frac{dx}{\sqrt{-(x^2-2.\frac{a+b}{2}x+ab+\frac{(a+b)^2}{4}-\frac{(a+b)^2}{4})}}\\ &=\int\frac{dx}{\sqrt{\frac{a^2+b^2+2ab}{4}-ab-\big[x-\frac{a+b}{2}\big]^2}}=\int\frac{dx}{\sqrt{\frac{(a-b)^2}{4}-\big(x-\frac{a+b}{2}\big)^2}}\\ &=\sin^{-1}\frac{x-\frac{a+b}{2}}{\frac{b-a}{2}}+C\color{red}{=\sin^{-1}\frac{2x-(a+b)}{b-a}+C} \end{align} $$ My reference shows the solution $I=2\sin^{-1}\sqrt{\frac{x-a}{b-a}}$, yet why am I getting a different solution or both the same ?

Thanx@lab bhattacharjee for the hint.

$$ \begin{align} \sin^{-1}\frac{2x-(a+b)}{b-a}+C&=\frac{\pi}{2}-\cos^{-1}\frac{2x-(a+b)}{b-a}+C\\&=\frac{\pi}{2}-\pi+\cos^{-1}\frac{(a+b)-2x}{b-a}+C_1\\ &=\cos^{-1}\frac{(a+b)-2x}{b-a}+C_2 \end{align} $$ Let, $$ y=\cos^{-1}\frac{(a+b)-2x}{b-a}\implies\cos y=\frac{(a+b)-2x}{b-a}\\ 2\sin^2\frac{y}{2}=1-\cos y=1-\frac{(a+b)-2x}{b-a}=\frac{b-a-a-b+2x}{b-a}=\frac{2x-2a}{b-a}\\ \sin^2\frac{y}{2}=\frac{x-a}{b-a}\implies \sin\frac{y}{2}=\sqrt{\frac{x-a}{b-a}}\\ \frac{y}{2}=\sin^{-1}\sqrt{\frac{x-a}{b-a}}\implies\boxed{y=2\sin^{-1}\sqrt{\frac{x-a}{b-a}}} $$

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Hint:

From your answer if $\sin^{-1}\sqrt{\dfrac{x-a}{b-a}}=y$

$\sin y=\sqrt{\dfrac{x-a}{b-a}}$

$\implies\cos2y=1-2\sin^2y=\dfrac{a+b-2x}{b-a}$

$2y=\cos^{-1}\dfrac{a+b-2x}{b-a}=\dfrac\pi2-\sin^{-1}\dfrac{a+b-2x}{b-a}=\dfrac\pi2+\sin^{-1}\dfrac{2x-a-b}{b-a}$

So, you have reached to the right answer.

See also: Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $

Alternatively,

As $(x-a)(b-x)=-\{x^2-(a+b)x+ab\}=\dfrac{(a-b)^2}4-\left(x-\dfrac{a+b}2\right)^2$

choose $x-\dfrac{a+b}2=\dfrac{(a-b)}2\cos2t$

$$2x=a+b+(a-b)(\cos^2t-\sin^2t)=2(a\cos^2t+b\sin^2t)$$

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Use the Euler substition $$\sqrt{(x-a)(x-b)}=(x-a)t$$ Then is $$x=\frac{at^2+b}{1+t^2}$$ and then is $$dx=\frac{2(a-b)t}{(1+t^2)^2}dt$$

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