Question

If $I=\int \frac{dx}{\sqrt{(x-\alpha )(\beta -x)}},(\beta <\alpha )$ then value of I is

• A. ${\sin }^{-1}\left(\frac{2x-\alpha -\beta }{\beta -\alpha }\right)+C$
• B. ${\sin }^{-1}\left(\frac{x+\alpha +\beta }{\beta -\alpha }\right)$
• C. ${\sin }^{-1}\left(\frac{2x+\beta -\alpha }{\beta +\alpha }\right)+C$
• D. none of these

Answer 1

The correct option is A ${\sin }^{-1}\left(\frac{2x-\alpha -\beta }{\beta -\alpha }\right)+C$

Let $I=\int \frac{dx}{\sqrt{(x-\alpha )(\beta -x)}}$

Putting $t=\frac{1}{2}(x-\alpha +x-\beta )\Rightarrow dt=dx$

$t=x-\frac{1}{2}(\alpha +\beta )$

So $(x-\alpha )(\beta -x)=(t+\frac{1}{2}(\alpha +\beta )-\alpha )(\beta -\frac{1}{2}(\alpha +\beta )-t)$

$(x-\alpha )(\beta -x)=\frac{1}{4}{(\beta -\alpha )}^2-t^2$

$I=\int \frac{dt}{\sqrt{\frac{1}{4}{(\beta -\alpha )}^2-t^2}}$

We know that, $\int \frac{1}{\sqrt{a^2-x^2}}dx={\sin }^{-1}\left(\frac{x}{a}\right)+c$

$I=-{\sin }^{-1}\left(\frac{2t}{\beta -\alpha }\right)+c={\sin }^{-1}\left(\frac{2x-\alpha -\beta }{\beta -\alpha }\right)+c$