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Question:

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

Doubt:

I took take $x=a\cos^2\theta+b\sin^2\theta$ and solved. The final answer that I got was: $$2\sin^{-1}\sqrt{\left(\frac{x-a}{b-a}\right)}+c$$

It matched the answer at the back of my book :)

I was thinking of another way in which I opened the brackets and got a quadratic equation. I then converted it into perfect square to get this form: $$\int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\left(\frac{x}{a}\right)+c$$ where $x$ was $x-(\frac{a+b}{2})$ and $a$ was $\frac{(a-b)}{2}$.

With this, I got the final answer which was different from the previous one.

Kindly tell what was wrong with the other method?

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    Why don't you just apply partial fraction decomposition plus a case division for $a=b$? – 2017-01-09

3 Answers 3

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Basically, we have, $$2 \arcsin \sqrt {\frac {x-a}{b-a}} = \arcsin ( 2 \sqrt {\frac {x-a}{b-a}} \sqrt {1- \frac {x-a}{b-a}}) =\arcsin (2\sqrt {\frac {(x-a)(b-x)}{(b-a)^2}})$$ (using $\arcsin A +\arcsin B =\arcsin (A\sqrt {1-B^2} +B\sqrt {1-A^2})$)

Can you take it from here?

On a side note, $a $ is $\frac {a-b}{2} $ and not $(\frac {a-b}{2})^2$.

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    Yeah.. I wrote $a^2$ instead of $a$. Thanks. – 2017-01-09
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Two comments which should help you figure out why your two primitives/anti-derivatives are actually equal up to a constant function, i.e. that $+c$ you add at the end :

  • Instead of keeping the symbol $x$ as integration variable when you do your completion of the square, name it $y$. Your integral at the end will be a function of $y$, which is then easily reverted to a function of $x$ by performing the substitution backwards (if $y=x-(a+b)/2$, then $x=y+(a+b)/2$).
  • Have a look at a list of trigonometric substitutions and find the ones involving $\sin^{-1}$! That should have been your first reflex!

Hope that helps,

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Although this is not directly the answer to the question posed by the OP, I still wanted to point out this alternative solution:

if $a=b$ one has $$\int\frac{1}{(x-a)(a-x)}\,dx=\int\frac{-1}{(x-a)^2}\,dx=\frac{1}{x-a}+c$$ if $a\neq b$ one has $$\frac{1}{(x-a)(b-x)}=\frac{1}{a-b}\left(\frac{1}{x-a}-\frac{1}{x-b}\right),$$ and hence

$$\int\frac{1}{(x-a)(a-x)}\,dx=\frac{1}{a-b}\Big(\log{(x-a)}-\log{(x-b)}+c\Big)=\frac{1}{a-b}\log\left(\frac{x-a}{x-b}\right)+c$$