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
Correct Answer: A
Solution :
\[I=\int{\frac{dx}{(x-\beta )\sqrt{(x-\alpha )(\beta -x)}}}\] 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, \[x=\alpha {{\sin }^{2}}\theta +\beta {{\cos }^{2}}\theta \] \[\Rightarrow \] \[x\cos \text{e}{{\text{c}}^{2}}\theta =\alpha +\beta {{\cos }^{2}}\theta \] \[\Rightarrow \] \[x(1+{{\cot }^{2}}\theta )=\alpha +\beta {{\cot }^{2}}\theta \] \[\therefore \]\[\cot \theta =\sqrt{\frac{x-\alpha }{\beta -x}};\,\,\therefore \,\,I=\frac{2}{\alpha -\beta }\sqrt{\frac{x-\alpha }{\beta -x}}+C\]
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