int sin ^{-1}left(sqrt{frac{x-a}{x}}right) d | vedclass

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Question

(A) Let $I = \int \sin ^{-1}\left(\sqrt{\frac{x-a}{x}}ight) d x$. Substitute $x = a \sec^2 	heta$,then $dx = 2a \sec^2 	heta 	an 	heta \, d	het

Vedclass · Accepted Answer

$x \cos ^{-1} \sqrt{\frac{a}{x}}-\sqrt{a x-a^2}+c$

Vedclass · Answer

(A) Let $I = \int \sin ^{-1}\left(\sqrt{\frac{x-a}{x}}ight) d x$. Substitute $x = a \sec^2 	heta$,then $dx = 2a \sec^2 	heta 	an 	heta \, d	heta$. Since $\sqrt{\frac{x-a}{x}} = \sqrt{\frac{a \sec^2 	heta - a}{a \sec^2 	heta}} = \sqrt{\frac{	an^2 	heta}{\sec^2 	heta}} = \sin 	heta$,the integral becomes: $I = \int 	heta \cdot (2a \sec^2 	heta 	an 	heta) \, d	heta = 2a \int 	heta \sec^2 	heta 	an 	heta \, d	heta$. Using integration by parts,let $u = 	heta$ and $dv = 2 \sec^2 	heta 	an 	heta \, d	heta = 	an^2 	heta \, d	heta$ is not correct,rather $dv = (2 \sec^2 	heta 	an 	heta) d	heta = d(	an^2 	heta)$. $I = 	heta 	an^2 	heta - \int 	an^2 	heta \, d	heta = 	heta 	an^2 	heta - \int (\sec^2 	heta - 1) \, d	heta = 	heta 	an^2 	heta - 	an 	heta + 	heta + C = 	heta(	an^2 	heta + 1) - 	an 	heta + C = 	heta \sec^2 	heta - 	an 	heta + C$. Since $x = a \sec^2 	heta$,$\sec^2 	heta = \frac{x}{a} \Rightarrow \cos^2 	heta = \frac{a}{x} \Rightarrow 	heta = \cos^{-1} \sqrt{\frac{a}{x}}$. Also $	an 	heta = \sqrt{\sec^2 	heta - 1} = \sqrt{\frac{x}{a} - 1} = \sqrt{\frac{x-a}{a}}$. Thus,$I = x \cos^{-1} \sqrt{\frac{a}{x}} - a \sqrt{\frac{x-a}{a}} + C = x \cos^{-1} \sqrt{\frac{a}{x}} - \sqrt{a(x-a)} + C = x \cos^{-1} \sqrt{\frac{a}{x}} - \sqrt{ax-a^2} + C$.

$\int \sin ^{-1}\left(\sqrt{\frac{x-a}{x}}\right) d x=$

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