If int_{0}^{a} sqrt{frac{a - x}{x}} dx = frac | vedclass

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(D) Let $I = \int_{0}^{a} \sqrt{\frac{a - x}{x}} dx$. Substitute $x = a \sin^{2} 	heta$,then $dx = 2a \sin 	heta \cos 	heta d 	heta$. When $x = 0$

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$\pi a$

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(D) Let $I = \int_{0}^{a} \sqrt{\frac{a - x}{x}} dx$. Substitute $x = a \sin^{2} 	heta$,then $dx = 2a \sin 	heta \cos 	heta d 	heta$. When $x = 0$,$	heta = 0$. When $x = a$,$	heta = \frac{\pi}{2}$. $I = \int_{0}^{\pi/2} \sqrt{\frac{a - a \sin^{2} 	heta}{a \sin^{2} 	heta}} (2a \sin 	heta \cos 	heta) d 	heta$ $I = \int_{0}^{\pi/2} \frac{\cos 	heta}{\sin 	heta} (2a \sin 	heta \cos 	heta) d 	heta$ $I = 2a \int_{0}^{\pi/2} \cos^{2} 	heta d 	heta$ Using the identity $\cos^{2} 	heta = \frac{1 + \cos 2 	heta}{2}$,we get: $I = 2a \int_{0}^{\pi/2} \frac{1 + \cos 2 	heta}{2} d 	heta = a \int_{0}^{\pi/2} (1 + \cos 2 	heta) d 	heta$ $I = a [	heta + \frac{\sin 2 	heta}{2}]_{0}^{\pi/2} = a [(\frac{\pi}{2} + 0) - (0 + 0)] = \frac{\pi a}{2}$. Given $\int_{0}^{a} \sqrt{\frac{a - x}{x}} dx = \frac{K}{2}$,so $\frac{\pi a}{2} = \frac{K}{2}$. Therefore,$K = \pi a$.

If $\int_{0}^{a} \sqrt{\frac{a - x}{x}} dx = \frac{K}{2}$,then $K = . . . . . .$.

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