• # question_answer $\int{\frac{dx}{(x-\beta )\sqrt{(x-\alpha )(\beta -x)}}}$is A) $\frac{2}{\alpha -\beta }\sqrt{\frac{x-\alpha }{\beta -x}}+c$B) $\frac{2}{\alpha -\beta }\sqrt{(x-\alpha )(\beta -x)}+c$C) $\frac{\alpha -\beta }{2}(x-\alpha )\sqrt{\beta -x}$D)  none of these

$I=\int{\frac{dx}{(x-\beta )\sqrt{(x-\alpha )(\beta -x)}}}$ Put$x=\alpha {{\sin }^{2}}\theta +\beta {{\cos }^{2}}\theta$ [see the standard substitutions]                 $dx=2(\alpha -\beta )\sin \theta \cos \theta d\theta$ Also,      $(x-\alpha )=(\beta -\alpha ){{\cos }^{2}}\theta$                 $(x-\beta )=(\alpha -\beta ){{\sin }^{2}}\theta$ $\therefore$$I=\int{\frac{2(\alpha -\beta )\sin \theta \cos \theta d\theta }{(\alpha -\beta ){{\sin }^{2}}\theta (\beta -\alpha )\sin \theta \cos \theta }}$        $=\frac{2}{\beta -\alpha }\int{\frac{d\theta }{{{\sin }^{2}}\theta }=\frac{2}{\beta -\alpha }\int{\cos \text{e}{{\text{c}}^{2}}}\theta d\theta }$       $=\frac{2}{\beta -\alpha }(-\cot \theta )+C=\frac{2}{\alpha -\beta }\cot \theta +C$ Now,     $x=\alpha {{\sin }^{2}}\theta +\beta {{\cos }^{2}}\theta$ $\Rightarrow$               $x\cos \text{e}{{\text{c}}^{2}}\theta =\alpha +\beta {{\cos }^{2}}\theta$ $\Rightarrow$               $x(1+{{\cot }^{2}}\theta )=\alpha +\beta {{\cot }^{2}}\theta$ $\therefore$$\cot \theta =\sqrt{\frac{x-\alpha }{\beta -x}};\,\,\therefore \,\,I=\frac{2}{\alpha -\beta }\sqrt{\frac{x-\alpha }{\beta -x}}+C$