Are there restrictions I have forgotten for Integration by Trigonometric Substitution or am I making some other mistake? I have been playing around with some integration problems I had previously solved correctly.  I attempted an approach that was a bit different on one in particular, and I am getting what seems to be a slightly incorrect answer.  The original problems is:
$$
\int \frac{x+2}{\sqrt{4-x^2}}\,dx \\[30pt]
$$
I previously solved the problem by splitting it up as:
$$
\int \frac{x}{\sqrt{4-x^2}}\,dx + 2\int \frac{1}{\sqrt{4-x^2}}\,dx \\[30pt]
$$
I ultimately found the correct answer to be:
$$
2\cdot\sin^{-1}\Big(\frac{x}{2}\Big)-\sqrt{4-x^2}+C \\[30pt]
$$
What I tried next was to solve the problem using trigonometric substitution without splitting it up.  My work is as follows:
$$
\int \frac {x+2}{\sqrt{4-x^2}}\,dx = \int \frac {x+2}{\sqrt{4(1-\frac{1}{4}x^2)}}\,dx = \frac{1}{2}\int\frac{x+2}{\sqrt{1-\frac{1}{4}x^2}}\,dx \\[30pt]
$$
Here I begin the trigonometric substitution with:
$$
\frac{1}{2}x=\sin(\theta)\implies x=2\cdot \sin(\theta); dx=2\cdot \cos(\theta)\,d\theta \\[30pt]
$$
Thus my work continues as follows:
$$
\frac{1}{2}\int\frac{x+2}{\sqrt{1-\frac{1}{4}x^2}}\,dx = \frac{1}{2}\int\frac{(2\cdot\sin(\theta)+2)}{\sqrt{1-\sin^2(\theta)}}\cdot(2\cdot \cos(\theta))\,d\theta = \\[30pt] \frac{1}{2}\int\frac{(2\cdot\sin(\theta)+2)}{\sqrt{\cos^2(\theta)}}\cdot(2\cdot \cos(\theta))\,d\theta = \frac{1}{2}\int\frac{(2\cdot\sin(\theta)+2)}{\cos(\theta)}\cdot(2\cdot \cos(\theta))\,d\theta = \\[30pt] \int (2\cdot\sin(\theta)+2)\,d\theta = 2\int (\sin(\theta)+1)\,d\theta = 2(\theta-\cos(\theta))+C = \\[30pt] 2\theta - 2\cos(\theta)+C \\[30pt]
$$
Now solving for $\theta$ I get $\theta = \sin^{-1}\big(\frac{x}{2}\big)$ and solving with a right triangle I find that $\cos(\theta) = \sqrt{4-x^2} \\[30pt]$.
Thus -- substituting back in for x -- the final answer would seem to be:
$$
2\cdot \sin^{-1}\Big( \frac{x}{2}\Big)-2\sqrt{4-x^2}+C \\[30pt]
$$
But:
$$
2\cdot \sin^{-1}\Big( \frac{x}{2}\Big)-2\sqrt{4-x^2}+C \neq 2\cdot\sin^{-1}\Big(\frac{x}{2}\Big)-\sqrt{4-x^2}+C \\[30pt]
$$
So what am I missing?  Is there a rule necessitating that problems like this be split up before using trigonometric substitution?  Have I made an error somewhere in my calculations/algebra?

In one of the last steps:
$\cos\theta=\cos(\arcsin(\frac{x}{2}))=\sqrt{1-\frac{x^2}{4}}$
(not $\sqrt{4-x^2}$)
