IIT JEE , AIEEE Forum - Mathematics , Integral Calculus - Indefinite -- Another one

ramkumar_november

I = $\int\sqrt{\frac{x-a}{b-x}}\;dx$

$PUT \;\;x=\;acos^2\theta\;+bsin^2\theta$

differentiating we get.........

$dx\;=\;(b-a)sin2\theta\,d\theta$

$x-a\,=\,(b-a)sin^2\theta$

$b-x\,=\,(b-a)cos^2\theta$

so $\sqrt{\frac{x-a}{b-x}}\;=\;tan\theta$

substituting in the integral. we get......

I = $\int(b-a)sin2\theta\,tan\theta\,d\theta$

I = $2(b-a)\int\,sin^2\theta\,d\theta$

I = $2(b-a)\bigg(\frac{\theta}{2}\,-\,\frac{cos2\theta}{4}\bigg)$

$where\;\;\theta\;=\;sin^{-1}\sqrt{\frac{x-a}{b-a}}$

clear with my proof???