# Which angle to pick for trigonometric substitution?

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

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!

Note that $\arcsin t$ and $-\arccos t$ differ by a constant. So both procedures, the "standard" one and the one that you suggest, are correct. To give a simpler example, $\displaystyle\int 2x\,dx=x^2+C$ and $\displaystyle\int 2x\,dx=x^2+17\pi+C$ are both correct.
As to why the common preference for $\arcsin$, it may be a simple matter of avoiding minus signs if possible. Perhaps the fact that $\arcsin 0=0$ is an added convenience factor.
• Let $G(x)=F(x)+17$. Then $G(b)-G(a)=F(b)-F(a)$, the $17$'s cancel. Thus any antiderivative will do, they all give the same answer. – André Nicolas Mar 10 '13 at 18:20
Remember the trigonometric identity $$\arcsin\theta+\arccos\theta=\frac\pi2,$$ so that $$-\arccos\theta=\arcsin\theta+\text{constant}.$$