Arcsin and Natural Log

Let us define a function $G(x) = g(x) + k = ln(x) + k$ (for $k > 0).$ If $G(x)$ is to share a common tangent with $f(x) = arcsin(x)$ , then we require equal slopes at a common point $x = c$ . This occurs for:

$f'(x) = G'(x) \Rightarrow \frac{1}{\sqrt{1-x^2}} = \frac{1}{x} \Rightarrow \frac{1}{1-x^2} = \frac{1}{x^2} \Rightarrow 2x^2 = 1 \Rightarrow x = \frac{1}{\sqrt{2}}.$

So our common point is $c = \frac{1}{\sqrt{2}}$ , and we're now left to compute the translation value of $k$ . A simple substitution gives:

$f(c) = G(c) \Rightarrow arcsin(\frac{1}{\sqrt{2}}) = ln(\frac{1}{\sqrt{2}}) + k \Rightarrow \frac{\pi}{4} = -\frac{ln(2)}{2} + k \Rightarrow k = \frac{\pi + ln(4)}{4}$

which leaves us with $G(x) = ln(x) + \frac{\pi + ln(4)}{4}$ . Finally, we can compute the area between $f$ and $G$ over $x \in (0, \frac{1}{\sqrt{2}}]$ via the definite integral:

$A = \displaystyle{\int_{0}^{\frac{1}{\sqrt{2}}} \arcsin(x) - (\ln(x) + \frac{\pi + ln(4)}{4})\,\mathrm{d}x}$ ;

or $x \cdot arcsin(x) + \sqrt{1 - x^2} - (xln(x) - x) - \frac{\pi+ln(4)}{4}x$ ;

or $[(\frac{1}{\sqrt{2}} \cdot arcsin(\frac{1}{\sqrt{2}}) + \sqrt{1 - (\frac{1}{\sqrt{2}})^2} - (\frac{1}{\sqrt{2}} ln(\frac{1}{\sqrt{2}}) - \frac{1}{\sqrt{2}}) - \frac{\pi + ln(4)}{4} \cdot (\frac{1}{\sqrt{2}})] - [0 + 1 - 0 + 0 - 0];$

or $\frac{1}{\sqrt{2}} \cdot \frac{\pi}{4} + \frac{1}{\sqrt{2}} - (\frac{1}{\sqrt{2}} \cdot -ln(\sqrt{2})) + \frac{1}{\sqrt{2}} - \frac{\pi+ln(4)}{4} \cdot (\frac{1}{\sqrt{2}}) - 1;$

or $\sqrt{2} - 1 + \frac{ln(2)}{2\sqrt{2}} - \frac{ln(2)}{2\sqrt{2}};$

or $A = \boxed{\sqrt{2} - 1}.$