Integral Calculus
Integration By Parts
MMTS_Full_Test_14
Grade 12
Question:
$\displaystyle\int\sqrt{\dfrac{x}{1-x^3}}dx=$
$\dfrac{2}{3}\sin^{-1}(x^{3/2})+c$
$\dfrac{2}{3}\cos^{-1}(x^{3/2})+c$
$\dfrac{1}{3}\sin^{-1}(x^{3/2})+c$
$\dfrac{1}{3}\cos^{-1}(x^{3/2})+c$
Step-by-Step Solution
Key Concept: Substitute $x^{3/2}=t$; $\frac{3}{2}x^{1/2}dx=dt$
$t=x^{3/2}$, $dt=\frac{3}{2}\sqrt{x}dx$. $\int\frac{\sqrt{x}dx}{\sqrt{1-x^3}}=\frac{2}{3}\int\frac{dt}{\sqrt{1-t^2}}=\frac{2}{3}\sin^{-1}(t)+c=\frac{2}{3}\sin^{-1}(x^{3/2})+c$.
Correct Answer: 1