int_2^5 sqrt{frac{5-x}{x-2}} , dx = | vedclass

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(C) Let $I = \int_2^5 \sqrt{\frac{5-x}{x-2}} \, dx$. Put $x = 2 \cos^2 	heta + 5 \sin^2 	heta = 2 + 3 \sin^2 	heta$. Then $dx = 6 \sin 	heta \cos

Vedclass · Accepted Answer

$\frac{3\pi}{2}$

Vedclass · Answer

(C) Let $I = \int_2^5 \sqrt{\frac{5-x}{x-2}} \, dx$. Put $x = 2 \cos^2 	heta + 5 \sin^2 	heta = 2 + 3 \sin^2 	heta$. Then $dx = 6 \sin 	heta \cos 	heta \, d	heta$. When $x=2, 	heta=0$. When $x=5, 	heta=\frac{\pi}{2}$. $I = \int_0^{\pi/2} \sqrt{\frac{5-(2+3\sin^2 	heta)}{(2+3\sin^2 	heta)-2}} \cdot 6 \sin 	heta \cos 	heta \, d	heta$ $I = \int_0^{\pi/2} \sqrt{\frac{3-3\sin^2 	heta}{3\sin^2 	heta}} \cdot 6 \sin 	heta \cos 	heta \, d	heta$ $I = \int_0^{\pi/2} \frac{\cos 	heta}{\sin 	heta} \cdot 6 \sin 	heta \cos 	heta \, d	heta = 6 \int_0^{\pi/2} \cos^2 	heta \, d	heta$ $I = 6 \int_0^{\pi/2} \frac{1+\cos 2	heta}{2} \, d	heta = 3 \left[ 	heta + \frac{\sin 2	heta}{2} ight]_0^{\pi/2} = 3 \left( \frac{\pi}{2} + 0 - 0 ight) = \frac{3\pi}{2}$.

$\int_2^5 \sqrt{\frac{5-x}{x-2}} \, dx =$

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