If e^x(x ^-1x√1-x^2+ ^-1x(1-x^2)^3/2+x1-x^2)dx=g(x)+C, where C is the constant of integrat…

Question

If $\int {e}^{x}(\frac{x{\sin }^{-1}x}{\sqrt{1-{x}^{2}}}+\frac{{\sin }^{-1}x}{{(1-{x}^{2})}^{3/2}}+\frac{x}{1-{x}^{2}})dx=g(x)+C$, where C is the constant of integration, then $g(\frac{1}{2})$ equals :

Accepted Answer

$\begin{gathered} \because \frac{d}{dx}(\frac{x{\sin }^{-1}x}{\sqrt{1-{x}^{2}}})=\frac{{\sin }^{-1}x}{{(1-{x}^{2})}^{3/2}}+\frac{x}{1-{x}^{2}} \ \Rightarrow \int {e}^{x}(\frac{x{\sin }^{-1}x}{\sqrt{1-{x}^{2}}}+\frac{{\sin }^{-1}x}{{(1-{x}^{2})}^{3/2}}+\frac{x}{1-{x}^{2}})dx \ ={e}^{x}\cdot \frac{x{\sin }^{-1}x}{\sqrt{1-{x}^{2}}}+c=g(x)+C \ g(x)=\frac{{xe}^{x}{\sin }^{-1}x}{\sqrt{1-{x}^{2}}} \ g(1/2)=\frac{{e}^{1/2}}{2}\cdot \frac{\frac{\pi }{6}	imes 2}{\sqrt{3}}=\frac{\pi }{6}\sqrt{\frac{e}{3}} \end{gathered}$

Comment : In this question we will not get a unique function $g(x)$, but in order to match the answer we will have to assume $g(x)=\frac{{xe}^{x}{\sin }^{-1}x}{\sqrt{1-{x}^{2}}}$.

Solutions · 1