JEE Main & Advanced Sample Paper JEE Main Sample Paper-19

  • 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

    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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