A) \[\frac{\pi }{2}(\alpha -\beta )\]
B) \[\frac{\pi }{2}(\beta -\alpha )\]
C) \[\pi (\alpha -\beta )\]
D) \[\pi (\beta -\alpha )\]
Correct Answer: B
Solution :
Let \[I=\int_{\alpha }^{\beta }{\sqrt{\frac{x-\alpha }{\beta -x}}}\,dx\] Put \[x=\alpha \,{{\cos }^{2}}\theta +\beta {{\sin }^{2}}\theta ,\] \[\therefore \] \[x-\alpha =(\beta -\alpha )\,{{\sin }^{2}}\,\theta ,\,\beta -x=(\beta -\alpha )\,{{\cos }^{2}}\theta \] and \[dx=2(\beta -\alpha )\,\sin \theta \,\cos \theta \,d\theta \] \[\therefore \] \[I=\int_{0}^{\pi /2}{\sqrt{\frac{(\beta -\alpha )\,{{\sin }^{2}}\theta }{(\beta -\alpha )\,{{\cos }^{2}}\theta }}}\] \[\times 2(\beta -\alpha )\,\sin \theta \,\cos \theta \,d\theta \] \[=(\beta -\alpha )\int_{0}^{\pi /2}{2{{\sin }^{2}}\,\theta \,d\theta }\] \[=(\beta -\alpha )\int_{0}^{\pi /2}{(1-cos\,2\theta )\,d\theta }\] \[=(\beta -\alpha )\left[ \theta -\frac{\sin \,2\theta }{2} \right]_{0}^{\pi /2}=(\beta -\alpha ).\frac{\pi }{2}\]
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