Question

Evaluate: $\int \sqrt{\frac{x}{4-x^{3}}} dx $

• A. $\frac{2}{3} \sin^{-1} \left(\frac{x^{3/2}}{2}\right) + c $
• B. $\frac{2}{3} \sin^{-1} \left(x^{3/2}\right) + c $
• C. $ 2 \sin^{-1} \left(\frac{x^{3/2}}{2}\right) + c $
• D. $\frac{1}{3} \sin^{-1} \left(\frac{x^{3/2}}{2}\right) + c $

Answer 1

Answer: (A)

$I = \int \sqrt{\frac{x}{4-x^{3} }} dx = \int \frac{\sqrt{x} dx}{\sqrt{4-x^{3}}} $
Here integral of $ \sqrt{x} = \frac{2}{3} $ and
Put $x^{3/2} 4 -x^{3} = 4 - \left(x^{3/2}\right)^{2} x^{3/2} = t $
$\Rightarrow \sqrt{x} dx = \frac{2}{3} dt $
So $I = \frac{2}{3} \int \frac{dt}{\sqrt{4-t^{2}} } $
$= \frac{2}{3} \sin^{-1}\left(\frac{x^{3/2}}{2}\right)+c$