first timer on this stack exchange so I apologize if this is the wrong place to ask this question

I was wondering how one is supposed to properly pick an angle when using trig substitution to solve an integral.

Say I have 
$$
\int \frac{1}{\sqrt{a^2 - x^2}} \,\mathbb{d}x
$$
First I see that the triangle that goes along with this is

```

                                  |
                                / |
                              /   |
                            /  phi|
                          /       |
                        /         |
                      /           |
              a     /             |
                  /               |
                /                 |
              /                   |
            /                     |
          /                       |
        /                         |
      /                           | sqrt(a^2-x^2)
    /                             |
  /                               |
/    theta                        |
----------------------------------
               x

```

So my question is why do we pick `phi` in this picture rather than `theta` to base all of the formulas around.

In other words why don't we use $\cos\theta = \frac{x}{a}$ but use $\sin\phi = \frac{x}{a}$ for our substitutions?

If we use theta to base all of our formulas around then
$$
x = a\cos\theta \\
dx = -a \sin\theta\, \mathbb{d}\theta
$$
so the integral becomes
$$
\int \frac{1}{a\sqrt{1-\cos^2 \theta}} (-a \sin\theta) \; \mathbb{d}\theta \\
= -1 \int \frac{\sin \theta}{\sqrt{1-\cos^2 \theta}}\,\mathrm{d}\theta
$$
And since $\sin^2 \theta + \cos^2 \theta = 1 \implies \sin \theta = \sqrt{1-\cos^2 \theta}$ the integral becomes
$$
-1 \int \frac{\sin \theta}{\sin \theta} \,\mathbb{d} \theta \\
= -1 \int 1 \,\mathbb{d}\theta \\
= -\theta + C
= - \arccos \left(\frac{x}{a} \right) + C
$$
But this is wrong as according to everything I have looked at... so why do we choose the `phi` in that diagram and not the `theta`??

I apologize if this is a silly question

Thanks in advanced!!

EDIT:

I just wanted to add an example with limits and a real value for a, so let's do
$$
\int\_0^{\pi} \frac{1}{\sqrt{1-x^2}} \, \mathbb{d}x
$$
By the work above we know that this becomes
$$
- \arccos{x} |\_0^{\pi} \\
= - \arccos{\pi} - (- \arccos{0}) \\
= \arccos{0} - \arccos{\pi}
$$
Now if we use the sin this becomes
$$
\arcsin{x} |\_0^{\pi} \\
= \arcsin{\pi} - \arcsin{0}
$$
So I am still a bit confused unless those turn out to be the same value
EDIT \* 2:
Just realized that they are in fact the same :P thanks Andre for helping me out!