Integration Using Inverse Trig Functions 
$$\int \frac{1}{(36-4x^2)^{0.5}} dx$$

You can immediately see that,
$$6^2 = 36$$
And,
$$(2x)^2 = 4x^2$$
Using the integration technique as follows:
$$\int \frac{1}{(a^2-x^2)^{0.5}} dx = \arcsin(\frac{x}{a})+C$$
We get,
$$\int \frac{1}{(36-4x^2)^{0.5}} dx = \int \frac{1}{(6^2-(2x)^2)^{0.5}} dx$$
Where $a = 6$ and $x = 2x$. Substituting this into the above general rule, we get
$$\int \frac{1}{(6^2-(2x)^2)^{0.5}} dx = \arcsin(\frac{2x}{6})+C$$

$$=\arcsin(\frac{x}{3})+C$$

However, using this same procedure when we split the denominator gives us a different result:
$$\int \frac{1}{(36-4x^2)^{0.5}} dx = \int \frac{1}{((4)(9-x^2))^{0.5}} dx$$
$$=\int \frac{1}{(4)^{0.5}(9-x^2)^{0.5}} dx$$
$$=\int \frac{1}{2(9-x^2)^{0.5}} dx$$
$$=\frac{1}{2}\int \frac{1}{(9-x^2)^{0.5}} dx$$
Here, $a = 3$ and $x = x$ to give:

$$ = \frac{1}{2}\arcsin(\frac{x}{3})+C$$

Why are both results different? Through differentiation I know that the second result is correct - why does a direct substitution into the general formula yield an incorrect answer?

Here's the issue. The general form is $$\int\frac{1}{\sqrt{a^2-x^2}}\,dx=\arcsin\left(\frac{x}{a}\right)+C,$$
but your integral is 
$$\int\frac{1}{\sqrt{6^2-(2x)^2}}\,dx.$$
The $2x$ is different than $x$. The difference can be eliminated by making the substitution $2x=u$, so $dx=du/2$, and we get
$$\frac{1}{2}\int\frac{1}{\sqrt{6^2-u^2}}\,du=\frac{1}{2}\arcsin\left(\frac{u}{6}\right)+C=\frac{1}{2}\arcsin\left(\frac{x}{3}\right)+C.$$
