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If $\int \sqrt{\frac{x}{a^3-x^3}} d x=g(x)+c$, then $g(x)$ is equal to :
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Verified Answer
The correct answer is:
$\frac{2}{3} \sin ^{-1}\left(\sqrt{\frac{x^3}{a^3}}\right)$
Let $I=\int \sqrt{\frac{x}{a^3-x^3}} d x$
$=\int \sqrt{\frac{a^3}{a^3-x^3}} \cdot \sqrt{\frac{a^3}{x^3}} \cdot \frac{x^3}{a^3} d x$
$=\frac{2}{3} \sin ^{-1}\left(\sqrt{\frac{x^3}{a^3}}\right)+c$
But $\quad I=g(x)+c$
$\therefore \quad g(x)=\frac{2}{3} \sin ^{-1}\left(\sqrt{\frac{x^3}{a^3}}\right)$
$=\int \sqrt{\frac{a^3}{a^3-x^3}} \cdot \sqrt{\frac{a^3}{x^3}} \cdot \frac{x^3}{a^3} d x$
$=\frac{2}{3} \sin ^{-1}\left(\sqrt{\frac{x^3}{a^3}}\right)+c$
But $\quad I=g(x)+c$
$\therefore \quad g(x)=\frac{2}{3} \sin ^{-1}\left(\sqrt{\frac{x^3}{a^3}}\right)$
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