Question

Let $\int \frac{x^{\frac{1}{2}}}{\sqrt{1-x^3}}dx=\frac{2}{3}gof\left(x\right)+c$ then

• A. $f\left(x\right)=\sqrt{3}$
• B. $f\left(x\right)=x^{\frac{3}{2}}$ and $g\left(x\right)={\sin }^{-1}x$
• C. $f\left(x\right)=x^{\frac{2}{3}},g\left(x\right)={\sin }^{-1}x$
• D. None of these

Answer 1

Answer: (B)

$f\left(x\right)=x^{\frac{3}{2}}$ and $g\left(x\right)={\sin }^{-1}x$

$\int \frac{x^{\frac{1}{2}}}{\sqrt{1-x^3}}dx=\int \frac{x^{\frac{1}{2}}}{\sqrt{1-{\left(x^{\frac{3}{2}}\right)}^2}}dx$
$=\frac{2}{3}\int \frac{\frac{3}{2}{x}^{\frac{1}{2}}}{\sqrt{1-{\left(x^{\frac{3}{2}}\right)}^2}}dx$
$\therefore \int \frac{x^{\frac{1}{2}}}{\sqrt{1-x^3}}dx=\frac{2}{3}{\sin }^{-1}\left(x^{\frac{3}{2}}\right)+c=\frac{2}{3}gof\left(x\right)+c$
$\therefore f\left(x\right)=x^{\frac{3}{2}}$ and $g\left(x\right)={\sin }^{-1}x$
Hence choice (b) is correct answer.