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Question

(A) Let $I = \int \sqrt{\frac{a-x}{x}} \, dx$. Substitute $x = a \sin^2 	heta$,then $dx = 2a \sin 	heta \cos 	heta \, d	heta$. $I = \int \sqrt{\fr

Vedclass · Accepted Answer

$a \left[ \sin^{-1} \sqrt{\frac{x}{a}} + \sqrt{\frac{x}{a}} \sqrt{\frac{a-x}{a}} \right] + c$

Vedclass · Answer

(A) Let $I = \int \sqrt{\frac{a-x}{x}} \, dx$. Substitute $x = a \sin^2 	heta$,then $dx = 2a \sin 	heta \cos 	heta \, d	heta$. $I = \int \sqrt{\frac{a - a \sin^2 	heta}{a \sin^2 	heta}} \cdot (2a \sin 	heta \cos 	heta) \, d	heta$ $I = \int \frac{\cos 	heta}{\sin 	heta} \cdot 2a \sin 	heta \cos 	heta \, d	heta = 2a \int \cos^2 	heta \, d	heta$ $I = a \int (1 + \cos 2	heta) \, d	heta = a(	heta + \frac{\sin 2	heta}{2}) + c = a(	heta + \sin 	heta \cos 	heta) + c$ Since $x = a \sin^2 	heta$,we have $\sin 	heta = \sqrt{\frac{x}{a}}$ and $	heta = \sin^{-1} \sqrt{\frac{x}{a}}$. Also,$\cos 	heta = \sqrt{1 - \sin^2 	heta} = \sqrt{1 - \frac{x}{a}} = \sqrt{\frac{a-x}{a}}$. Substituting these back,$I = a \left[ \sin^{-1} \sqrt{\frac{x}{a}} + \sqrt{\frac{x}{a}} \sqrt{\frac{a-x}{a}} ight] + c$.

$\int \sqrt{\frac{a-x}{x}} \, dx = $

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