Question

$\int \frac{x^2}{\sqrt{1-x^6}}dx=$

• A. $\frac{1}{3}{\sin }^{-1}\left(x^3\right)+c$
• B. $\frac{1}{3}{\cos }^{-1}\left(x^3\right)+c$
• C. $-\frac{1}{3}{\sin }^{-1}\left(x^3\right)+c$
• D. $\frac{1}{4}{\cos }^{-1}\left(x^3\right)+c$

Answer 1

Answer: (A)

$\frac{1}{3}{\sin }^{-1}\left(x^3\right)+c$

$\int \frac{x^2}{\sqrt{1-x^6}}dx$
$=\int \frac{x^2}{\sqrt{1-(x^3{)}^2}}dx$
Let $x^3=t$

$\Rightarrow 3x^{2}dx=dt$
Thus the above integral changes to
$=\frac{1}{3}\int \frac{dt}{\sqrt{1-(t{)}^2}}$
$=\frac{{\sin }^{-1}t}{3}+c$
$=\frac{{\sin }^{-1}\left(x^3\right)}{3}+c$