int sin ^{-1} sqrt{frac{x}{a+x}} d x= | vedclass

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(D) Let $I = \int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x$. Substitute $x = a 	an^2 	heta$, then $dx = 2a 	an 	heta \sec^2 	heta d	heta$. Then $\sqr

Vedclass · Accepted Answer

$	an ^{-1}\left(\sqrt{\frac{x}{a}}ight)(a+x)-\sqrt{a x}+c$

Vedclass · Answer

(D) Let $I = \int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x$. Substitute $x = a 	an^2 	heta$, then $dx = 2a 	an 	heta \sec^2 	heta d	heta$. Then $\sqrt{\frac{x}{a+x}} = \sqrt{\frac{a 	an^2 	heta}{a(1+	an^2 	heta)}} = \sqrt{\sin^2 	heta} = \sin 	heta$. So, $I = \int 	heta \cdot 2a 	an 	heta \sec^2 	heta d	heta = 2a \int 	heta 	an 	heta \sec^2 	heta d	heta$. Using integration by parts, let $u = 	heta$ and $dv = 	an 	heta \sec^2 	heta d	heta$. Then $du = d	heta$ and $v = \frac{1}{2} 	an^2 	heta$. $I = 2a \left[ \frac{1}{2} 	heta 	an^2 	heta - \int \frac{1}{2} 	an^2 	heta d	heta ight] = a 	heta 	an^2 	heta - a \int (\sec^2 	heta - 1) d	heta$. $I = a 	heta 	an^2 	heta - a (	an 	heta - 	heta) + c = a 	heta (	an^2 	heta + 1) - a 	an 	heta + c$. Since $	an^2 	heta = \frac{x}{a}$, we have $	heta = 	an^{-1} \sqrt{\frac{x}{a}}$. $I = a 	an^{-1} \sqrt{\frac{x}{a}} (\frac{x}{a} + 1) - a \sqrt{\frac{x}{a}} + c = (x+a) 	an^{-1} \sqrt{\frac{x}{a}} - \sqrt{ax} + c$.

$\int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x=$

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