Integrating $\sqrt{1-x^2}$ by substitution: Why is this wrong?

Question

When solving the following integral by substitution, I can get two different solutions (one is wrong) depending on the substitution. I'm not sure why the second option produces the incorrect solution. $$\int_0^{\frac12}\sqrt{1-x^2}\;dx$$

Option 1: Let $x=\sin(u)$.

Therefore $\;\sqrt{1-x^2}=\cos(u)\;$ and $\;dx=\cos(u)\,du\;$ which gives: $$\int_0^{\frac\pi6}\cos^2(u)\;du=\frac{\sqrt3}8+\frac{\pi}{12}$$

Option2: Let $x=\cos(u)$.

Therefore $\;\sqrt{1-x^2}=\sin(u)\;$ and $dx=-\sin(u)\,du$ which gives: $$\int_0^{\frac\pi3}-\sin^2(u)\;du=\frac{\sqrt3}8-\frac{\pi}6$$

$\mathbf{My\;question\;is}$: What is wrong with the second option?

Answer 1

The second solution's limit went wrong.

$x = cos(u),u = \arccos(x) \in[\frac{\pi}{3},\frac{\pi}{2}] $

$$\int_\frac\pi2^{\frac\pi3}-\sin^2(u) =\frac{\sqrt3}8+\frac{\pi}{12} $$.