Real quick: If I have the function $$\int { \sqrt { { a }^{ 2 }-{ x }^{ 2 } } } dx$$
I can easily substitute by setting $x$ equal to $a\sin \theta$.
But why actually is that? If I draw a right triangle it can also be $x = a\cos \theta$, depending on where you choose the sides and the angle $\theta$ to be.
If it has nothing to do with it then I ask myself why people derive trig substitution by drawing a right triangle...
You can use either. Remember, though, that in the integration when you use $x = a \sin \theta$ you also need to substitute $dx = (\cos \theta) d \theta$. If you were to use $x = a \cos \theta$ then you would instead need to substitute $dx = -(\sin \theta) d \theta$. You will get the same final answer either way. The reason you probably see the former substitution of $x = a \sin \theta$ more commonly is that you avoid the negative sign in the substitution and the work might be slightly cleaner looking that way.