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                        |

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!!


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.

Remark: In a number of ways, the cosine function behaves more nicely than the sine. But for historical reasons, it seems to be condemned to be viewed forever as secondary.

  • $\begingroup$ right ill fix the minus sign right now $\endgroup$ – DanZimm Mar 10 '13 at 18:14
  • $\begingroup$ what about when integrating with limits, wouldn't this change the answer since they are not in fact the same? $\endgroup$ – DanZimm Mar 10 '13 at 18:16
  • $\begingroup$ 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. $\endgroup$ – André Nicolas Mar 10 '13 at 18:20
  • $\begingroup$ Oh jeeze that's a good point ok thank you very much!! $\endgroup$ – DanZimm Mar 10 '13 at 18:29

Remember the trigonometric identity $$ \arcsin\theta+\arccos\theta=\frac\pi2, $$ so that $$ -\arccos\theta=\arcsin\theta+\text{constant}. $$


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