Answer
See below for detailed work.
Work Step by Step
$$A=\int \sqrt{1-x^2}dx$$
We set $u= \sqrt{1-x^2}$ and $dv=dx$
That makes $$du=\frac{(1-x^2)'}{2\sqrt{1-x^2}}dx=-\frac{2x}{2\sqrt{1-x^2}}dx=-\frac{x}{\sqrt{1-x^2}}dx$$ and $$v=x$$
Applying integration by parts $\int udv=uv-\int vdu$, we have
$$A=x\sqrt{1-x^2}-\int x\Big(-\frac{x}{\sqrt{1-x^2}}\Big)dx$$ $$A=x\sqrt{1-x^2}+\int\frac{x^2}{\sqrt{1-x^2}}dx$$ $$A=x\sqrt{1-x^2}+\int\frac{x^2+1-1}{\sqrt{1-x^2}}dx$$ $$A=x\sqrt{1-x^2}+\Big(\int\frac{x^2-1}{\sqrt{1-x^2}}dx+\int\frac{1}{\sqrt{1-x^2}}dx\Big)$$ $$A=x\sqrt{1-x^2}+\Big(-\int\sqrt{1-x^2}dx+\int\frac{1}{\sqrt{1-x^2}}dx\Big)$$
We can now replace $\int\sqrt{1-x^2}$ with $A$, since it is the given interval.
$$A=x\sqrt{1-x^2}+\Big(-A+\int\frac{1}{\sqrt{1-x^2}}dx$$ $$2A=x\sqrt{1-x^2}+\int\frac{1}{\sqrt{1-x^2}}dx$$ $$A=\frac{1}{2}x\sqrt{1-x^2}+\frac{1}{2}\int\frac{1}{\sqrt{1-x^2}}dx$$
So the formula has been proved.
