If int frac{x^{frac{1}{2}}}{sqrt{a^3-x^3}} dx | vedclass

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Question

(C) Let $I = \int \frac{x^{\frac{1}{2}}}{\sqrt{a^3-x^3}} dx$. Substitute $t = x^{\frac{3}{2}}$,then $dt = \frac{3}{2} x^{\frac{1}{2}} dx$,which implie

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $I = \int \frac{x^{\frac{1}{2}}}{\sqrt{a^3-x^3}} dx$. Substitute $t = x^{\frac{3}{2}}$,then $dt = \frac{3}{2} x^{\frac{1}{2}} dx$,which implies $x^{\frac{1}{2}} dx = \frac{2}{3} dt$. Also,$t^2 = (x^{\frac{3}{2}})^2 = x^3$. Substituting these into the integral: $I = \int \frac{\frac{2}{3} dt}{\sqrt{a^3 - t^2}} = \frac{2}{3} \int \frac{dt}{\sqrt{(\sqrt{a^3})^2 - t^2}}$. Using the standard integral formula $\int \frac{du}{\sqrt{A^2 - u^2}} = \sin^{-1}(\frac{u}{A}) + c$: $I = \frac{2}{3} \sin^{-1}\left(\frac{t}{\sqrt{a^3}}\right) + c$. Substituting $t = x^{\frac{3}{2}}$ back: $I = \frac{2}{3} \sin^{-1}\left(\frac{x^{\frac{3}{2}}}{a^{\frac{3}{2}}}\right) + c = \frac{2}{3} \sin^{-1}\left(\left(\frac{x}{a}\right)^{\frac{3}{2}}\right) + c$. Thus,$P(x) = \frac{2}{3} \sin^{-1}\left(\left(\frac{x}{a}\right)^{\frac{3}{2}}\right)$.

If $\int \frac{x^{\frac{1}{2}}}{\sqrt{a^3-x^3}} dx = P(x) + c$,then $P(x) =$

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