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

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

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$(a+x) 	an^{-1} \sqrt{\frac{x}{a}} - \sqrt{ax}$

Vedclass · Answer

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

If $\int \sin^{-1}\left(\sqrt{\frac{x}{a+x}}\right) dx = A(x) + \text{constant}$, then $A(x) =$

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