Evaluate int_{alpha}^{beta} sqrt{frac{x-alpha | vedclass

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(C) Let $I = \int_{\alpha}^{\beta} \sqrt{\frac{x-\alpha}{\beta-x}} dx$.Substitute $x = \alpha \cos^2 t + \beta \sin^2 t$.Then $dx = 2(\beta - \alpha)

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$\frac{\pi}{2}(\beta - \alpha)$

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(C) Let $I = \int_{\alpha}^{\beta} \sqrt{\frac{x-\alpha}{\beta-x}} dx$.Substitute $x = \alpha \cos^2 t + \beta \sin^2 t$.Then $dx = 2(\beta - \alpha) \sin t \cos t dt$.When $x = \alpha$,$t = 0$. When $x = \beta$,$t = \frac{\pi}{2}$.$x - \alpha = (\beta - \alpha) \sin^2 t$ and $\beta - x = (\beta - \alpha) \cos^2 t$.So,$\sqrt{\frac{x-\alpha}{\beta-x}} = \sqrt{\frac{(\beta-\alpha)\sin^2 t}{(\beta-\alpha)\cos^2 t}} = 	an t$.$I = \int_{0}^{\pi/2} 	an t \cdot 2(\beta - \alpha) \sin t \cos t dt = 2(\beta - \alpha) \int_{0}^{\pi/2} \sin^2 t dt$.Using the identity $\sin^2 t = \frac{1 - \cos 2t}{2}$,we get:$I = 2(\beta - \alpha) \int_{0}^{\pi/2} \frac{1 - \cos 2t}{2} dt = (\beta - \alpha) [t - \frac{\sin 2t}{2}]_{0}^{\pi/2}$.$I = (\beta - \alpha) [\frac{\pi}{2} - 0] = \frac{\pi}{2}(\beta - \alpha)$.

Evaluate $\int_{\alpha}^{\beta} \sqrt{\frac{x-\alpha}{\beta-x}} dx$

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