Evaluate $\int \frac{\sqrt{x}}{\sqrt{a^3 - x^3}} dx$.

$\frac{2}{3} \sin^{-1} \sqrt{\frac{x^3}{a^3}} + C$

The correct answer is Option (3) → $\frac{2}{3} \sin^{-1} \sqrt{\frac{x^3}{a^3}} + C$

Let $I = \int \frac{\sqrt{x}}{\sqrt{a^3 - x^3}} dx = \int \frac{\sqrt{x}}{\sqrt{(a^{3/2})^2 - (x^{3/2})^2}} dx$

Put $x^{3/2} = t \Rightarrow \frac{3}{2} x^{1/2} dx = dt$

$∴I = \frac{2}{3} \int \frac{dt}{\sqrt{(a^{3/2})^2 - t^2}} = \frac{2}{3} \sin^{-1} \frac{t}{a^{3/2}} + C \quad \left[ ∵ \int \frac{1}{\sqrt{a^2 - x^2}} dx = \sin^{-1} \left( \frac{x}{a} \right) + C \right]$

$= \frac{2}{3} \sin^{-1} \frac{x^{3/2}}{a^{3/2}} + C = \frac{2}{3} \sin^{-1} \sqrt{\frac{x^3}{a^3}} + C \quad [∵t = x^{3/2}]$