Answer: Choice A
[tex]\frac{4}{\pi}\left[x*\sin^{-1}(x)+\sqrt{1-x^2}\right]-x+C\\\\[/tex]
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Proof:
Apply the derivative to choice A. The goal is to prove the derivative is equivalent to the given integrand.
[tex]y = \frac{4}{\pi}\left[x*\sin^{-1}(x)+\sqrt{1-x^2}\right]-x+C\\\\\frac{dy}{dx} = \frac{4}{\pi}\left[\sin^{-1}(x)+\frac{x}{\sqrt{1-x^2}}-\frac{x}{\sqrt{1-x^2}}\right]-1\\\\\frac{dy}{dx} = \frac{4}{\pi}\sin^{-1}(x)-1\\\\[/tex]
It's far from obvious, but we can apply a bit of algebra like so
[tex]\frac{dy}{dx} = \frac{4}{\pi}\sin^{-1}(x)-1\\\\\frac{dy}{dx} = \frac{4}{\pi}\sin^{-1}(x)-\frac{\pi}{\pi}\\\\\frac{dy}{dx} = \frac{4\sin^{-1}(x)-\pi}{\pi}\\\\[/tex]
Next, we divide every term by 2. This is so we can replace the pi terms with pi/2
[tex]\frac{dy}{dx} = \frac{4\sin^{-1}(x)-\pi}{\pi}\\\\\frac{dy}{dx} = \frac{2\sin^{-1}(x)-\frac{\pi}{2}}{\frac{\pi}{2}}\\\\[/tex]
We do this to take advantage of the trig identity [tex]\sin^{-1}(x)+\cos^{-1}(x) = \frac{\pi}{2}[/tex]
So we'll be replacing every instance of "pi/2" with the left hand side of that equation just mentioned.
Therefore,
[tex]\frac{dy}{dx} = \frac{2\sin^{-1}(x)-\frac{\pi}{2}}{\frac{\pi}{2}}\\\\\frac{dy}{dx} = \frac{2\sin^{-1}(x)-\left(\sin^{-1}(x)+\cos^{-1}(x)\right)}{\sin^{-1}(x)+\cos^{-1}(x)}\\\\\frac{dy}{dx} = \frac{2\sin^{-1}(x)-\sin^{-1}(x)-\cos^{-1}(x)}{\sin^{-1}(x)+\cos^{-1}(x)}\\\\\frac{dy}{dx} = \frac{\sin^{-1}(x)-\cos^{-1}(x)}{\sin^{-1}(x)+\cos^{-1}(x)}\\\\[/tex]
This shows that differentiating the expression in choice A leads to the given integrand.
If we reverse all of the steps mentioned, then we can show that:
[tex]\displaystyle\frac{dy}{dx} = \frac{\sin^{-1}(x)-\cos^{-1}(x)}{\sin^{-1}(x)+\cos^{-1}(x)}\\\\\\ \int\frac{dy}{dx}dx = \int\frac{\sin^{-1}(x)-\cos^{-1}(x)}{\sin^{-1}(x)+\cos^{-1}(x)}dx\\\\\\y = \frac{4}{\pi}\left[x*\sin^{-1}(x)+\sqrt{1-x^2}\right]-x+C\\\\[/tex]
since integrals and derivatives are tied together through inverse processes (one undoes the other more or less).
