If int sqrt{frac{x}{a^3-x^3}} d x=g(x)+c,then | vedclass

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(C) Let $I = \int \sqrt{\frac{x}{a^3-x^3}} dx$. To solve this,we substitute $x^{3/2} = t$. Then,$\frac{3}{2} x^{1/2} dx = dt$,which implies $x^{1/2} d

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$\frac{2}{3} \sin ^{-1}\left(\sqrt{\frac{x^3}{a^3}}\right)$

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(C) Let $I = \int \sqrt{\frac{x}{a^3-x^3}} dx$. To solve this,we substitute $x^{3/2} = t$. Then,$\frac{3}{2} x^{1/2} dx = dt$,which implies $x^{1/2} dx = \frac{2}{3} dt$. Substituting these into the integral: $I = \int \frac{x^{1/2} dx}{\sqrt{a^3 - (x^{3/2})^2}} = \int \frac{\frac{2}{3} dt}{\sqrt{(a^{3/2})^2 - t^2}}$. Using the standard integral formula $\int \frac{1}{\sqrt{A^2 - t^2}} dt = \sin^{-1}(\frac{t}{A}) + c$: $I = \frac{2}{3} \sin^{-1}(\frac{t}{a^{3/2}}) + c$. Replacing $t$ with $x^{3/2}$: $I = \frac{2}{3} \sin^{-1}(\sqrt{\frac{x^3}{a^3}}) + c$. Thus,$g(x) = \frac{2}{3} \sin^{-1}(\sqrt{\frac{x^3}{a^3}})$.

If $\int \sqrt{\frac{x}{a^3-x^3}} d x=g(x)+c$,then $g(x)$ is equal to :

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