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

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(C) Let $I = \int \sqrt{\frac{x}{a^3-x^3}} \, dx$. To solve this, we rewrite the integrand: $I = \int \sqrt{\frac{x}{a^3(1 - (x/a)^{3})}} \, dx = \int

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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 rewrite the integrand: $I = \int \sqrt{\frac{x}{a^3(1 - (x/a)^{3})}} \, dx = \int \frac{\sqrt{x}}{\sqrt{a^3} \sqrt{1 - (x^{3/2}/a^{3/2})^2}} \, dx$. Let $u = \left(\frac{x}{a}\right)^{3/2} = \sqrt{\frac{x^3}{a^3}}$. Then $du = \frac{3}{2} \sqrt{\frac{x}{a^3}} \, dx$, which implies $\sqrt{\frac{x}{a^3}} \, dx = \frac{2}{3} \, du$. Substituting these into the integral: $I = \int \frac{1}{\sqrt{1 - u^2}} \cdot \frac{2}{3} \, du = \frac{2}{3} \sin^{-1}(u) + c$. Substituting back $u = \sqrt{\frac{x^3}{a^3}}$, we get $I = \frac{2}{3} \sin^{-1}\left(\sqrt{\frac{x^3}{a^3}}\right) + c$. Comparing this with $g(x) + c$, we find $g(x) = \frac{2}{3} \sin^{-1}\left(\sqrt{\frac{x^3}{a^3}}\right)$.

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

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