Question

Let $\int \frac{x^{1/2}}{\sqrt{1-x^3}}dx=\frac{2}{3}gof\left(x\right)+C$ (C being the constant of integration) then

• A. $f\left(x\right)=\sqrt{3}$
• B. $f\left(x\right)=x^{3/2}$
• C. $f\left(x\right)=x^{2/3}$
• D. $g\left(x\right)=si{n}^{-1}x$

Answer 1

Answer: (B), (D)

$f\left(x\right)=x^{3/2}$
$g\left(x\right)=si{n}^{-1}x$

Given : $\int \frac{x^{1/2}}{\sqrt{1-x^3}}dx$

Let $I=\int \frac{x^{1/2}}{\sqrt{1-x^3}}dx$

Substitute $x^{3/2}=t$

$\int \frac{x^{1/2}}{\sqrt{1-x^3}}dx=\frac{2}{3}\int \frac{dt}{\sqrt{1-t^2}}$

$=\frac{2}{3}{\sin }^{-1}t+c$

$I=\frac{2}{3}{\sin }^{-1}{x}^{3/2}+c$

Hence,

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