Question

Evaluate $\int \frac{dx}{\sqrt{(x-\alpha )(\beta -x)}},\beta >\alpha$

Answer 1

$\int \frac{1}{(x-\alpha )(\beta -x)}dx$

$x=\alpha {\cos }^2\theta +\beta {\cos }^2\theta$

$x-\alpha =\alpha {\sin }^2\theta +\beta {\cos }^2\theta -\alpha$

$=(\beta -\alpha ){\cos }^2\theta$

$\beta -x=\beta -\alpha {\sin }^2\theta -\beta {\cos }^2\theta$

$=(\beta -\alpha ){\sin }^2\alpha$

$dx=[\alpha .2\sin \theta \cos \theta +\beta 2\cos \theta (-\sin \theta \left)\right]$

$=(\alpha -\beta )\sin \theta d\theta$

$\int \frac{1}{(x-\alpha )(\beta -x)}dx=\int \frac{(\alpha -\beta )\sin 2\theta d\theta }{\sqrt{(\beta -\alpha ){\cos }^2\theta (\beta -\alpha ){\sin }^2\theta }}$

$=\int \frac{(\alpha -\beta )\sin 2\theta d\theta }{(\beta -\alpha )\sin \theta \cos \theta }$

$=\int -2d\theta =-2\theta +c$

$x=\alpha {\sin }^2\theta +\beta {\cos }^2\theta$

$=\alpha (1-{\cos }^2\theta )+\beta {\cos }^2\theta$

${\cos }^2\theta =\frac{x-\alpha }{\beta -x}\Rightarrow \cos \theta =\sqrt{\frac{x-\alpha }{\beta -x}}$

$\theta =\frac{\pi }{2}-{\sin }^{-1}\sqrt{\frac{x-\alpha }{\beta -x}}$

$=2.{\sin }^{-1}\sqrt{\frac{x-\alpha }{\beta -x}}+c-\pi$