Question

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

• A. $f\left(x\right)=\sqrt{x}$, $g\left(x\right)={\sin }^{-1}x$
• B. $f\left(x\right)=x^{\frac{3}{2}}$, $g\left(x\right)={\sin }^{-1}x$
• C. $f\left(x\right)=x^{\frac{2}{3}}$, $g\left(x\right)={\cos }^{-1}x$
• D. $f\left(x\right)=\sqrt{x}$, $g\left(x\right)={\cos }^{-1}x$

Answer 1

Answer: (B)

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

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

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

$=\int \frac{\sqrt{x}}{\sqrt{(1-{\left(x^{\frac{3}{2}}\right)}^2)}}dx$

$=\frac{2}{3}\int \frac{d\left(x^{\frac{3}{2}}\right)}{\sqrt{(1-{\left(x^{\frac{3}{2}}\right)}^2)}}$

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

Comparing this with the given problem we get,

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

So, option (B) is correct.