Question

The value $\int x{\sin }^{-1}xdx=\frac{x^2}{2}f\left(x\right)+\frac{1}{2}\left[\frac{x}{2}g\right(x)+\frac{1}{2}f(x\left)\right]-\frac{1}{2}f\left(x\right)+C$ then $f\left(x\right)$ and $g\left(x\right)$ are :

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

Answer 1

The correct option is B $f\left(x\right)={\sin }^{-1}x,g\left(x\right)=\sqrt{1-x^2}$

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

$=\frac{x^2}{2}{\sin }^{-1}x+\frac{1}{2}\int \frac{(1-x^2)-1}{\sqrt{1-x^2}}dx\frac{x^2}{2}{\sin }^{-1}x+\frac{1}{2}\int \sqrt{1-x^2}dx-\frac{1}{2}\int \frac{dx}{\sqrt{1-x^2}}$

$\frac{x^2}{2}{\sin }^{-1}+\frac{1}{2}[\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}x]-\frac{1}{2}{\sin }^{-1}+C$

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