If $\int e^x\left(\frac{x\sin^{-1}x}{\sqrt{1-x^2}}+\frac{\sin^{-1}x}{(1-x^2)^{3/2}}+\frac{x}{1-x^2}\right)dx=g(x)+C$, where $C$ is the constant of integration, then $g\left(\frac12\right)$ equals :

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] Let $h(x)=\frac{x\sin^{-1}x}{\sqrt{1-x^2}}$. Then $h'(x)=\frac{\sin^{-1}x}{(1-x^2)^{3/2}}+\frac{x}{1-x^2}$, so the integrand is $e^x\,(h+h')$ and $g(x)=e^x\frac{x\sin^{-1}x}{\sqrt{1-x^2}}$. $g\left(\frac12\right)=\sqrt e\cdot\frac{\frac12\cdot\frac\pi6}{\sqrt3/2}=\frac\pi6\sqrt{\frac e3}$. Answer (2).

Q. If $\int e^x\left(\frac{x\sin^{-1}x}{\sqrt{1-x^2}}+\frac{\sin^{-1}x}{(1-x^2)^{3/2}}+\frac{x}{1-x^2}\right)dx=g(x)+C$, where $C$ is the constant of integration, then $g\left(\frac12\right)$ equals :

Explanation

Reviewed by Dr. Abubacker

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