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!