Find x a^3-x^3 d x. - Vidyadip

Question

Find $\int \frac{\sqrt{x}}{\sqrt{a^3-x^3}} d x$.

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Answer

Given $I=\int \frac{\sqrt{x}}{\sqrt{a^3-x^3}} d x=\int \frac{\sqrt{x}}{\sqrt{\left(a^{3 / 2}\right)^2-\left(x^{3 / 2}\right)^2}} d x$
$\text { Let, } x^{3 / 2}=a^{3 / 2} t$
$\Rightarrow \frac{3}{2} x^{1 / 2} d x=a^{3 / 2} d t$
$\Rightarrow \sqrt{x} d x=\frac{2}{3} a^{3 / 2} d t$
$\therefore I=\int \frac{\frac{2}{3} a^{3 / 2}}{\sqrt{\left(a^{3 / 2}\right)^2-\left(a^{3 / 2} t\right)^2}} d t$
$=\frac{2}{3} a^{3 / 2} \int \frac{d t}{a^{3 / 2} \sqrt{1-t^2}}$
$=\frac{2}{3} \int \frac{d t}{\sqrt{1-t^2}}=\frac{2}{3} \sin ^{-1}\left(\frac{t}{1}\right)+C$
$\left[\because \int \frac{d x}{a^2-x^2}=\sin ^{-1}\left(\frac{x}{a}\right)+c\right]$
$=\frac{2}{3} \sin ^{-1}\left(\frac{x^{3 / 2}}{a^{3 / 2}}\right)+C \left[\text { put } t=\frac{x^{3 / 2}}{a^{3 / 2}}\right]$
$=\frac{2}{3} \sin ^{-1}\left(\sqrt{\frac{x^3}{a^3}}\right)+C$

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