To solve the definite integral

$$ I = \int_{-a}^{a} \frac{dx}{\pi \sqrt{a^2-x^2}}$$

I used the substitution $x = a \sin \theta$ and tried to solve the integral without its interval definition, which yields

$$ I = \frac{\theta}{\pi} $$

Now according to what I have been taught, from this point on I can do either of two things:

  • write the result in terms of $x$ and evaluate with the initial limits; or
  • transform $x$'s limits into limits for $\theta$ and evaluate.

Selecting the first option listed, the result of the evaluation is 1, but selecting the second option I get as result $$ x = a \Rightarrow \theta = \frac{\pi}{2} + 2\pi n$$ $$x=-a\Rightarrow \theta = \frac{3 \pi}{2} + 2 \pi k$$ $$ I = \frac{\theta}{\pi} = \left(\frac{1}{2} + 2n\right) - \left(\frac{3}{2} + 2k \right) = 2(n-k)-1$$

(where $n$ and $k$ are integers.) Why would $n-k=1$ so this result would also be correct?

  • 2
    $\begingroup$ Substitution is only allowed for bijections. You should double-check your bounds on $\theta$ to trace $[-a,a]$ exactly once. You should therefor pick $\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ and see $n=0, k=-1$. $\endgroup$ – AlexR Sep 8 '13 at 21:42
  • $\begingroup$ Do you mean that $\theta$ has to cross its bounds in the same order $x$ does? In this case, since $x$ is first $-a$, $\theta$ can't first be $\pi/2$; did I get it right? $\endgroup$ – Severo Raz Sep 8 '13 at 21:54
  • $\begingroup$ Not necessarily, but if you allow more than a range of $[a, a+\pi]$ where $a = 2\pi k-\frac{pi}{2}$, you won't have a bijection anymore. The order (i.e. from $a$ to $-a$ or reversed) doesn't matter and will be annulled by the derivative (it will change signs). $\endgroup$ – AlexR Sep 8 '13 at 21:58
  • $\begingroup$ So since it takes $\pi$ for sine to go from $u$ to $-u$ once (where $|u|=1$), my $\theta$ range has to be $\pi$ long? I can get my mind around that, but I don't get why my range needs to be of the form $[a, a+\pi$] if order doesn't matter. (Perhaps it would help if I let you know I am considering a bijection only within $\theta$'s limits, I don't know if this is wrong.) $\endgroup$ – Severo Raz Sep 8 '13 at 22:15
  • $\begingroup$ I guess I now understand your bijection rule,. I read up on a PDF taken from one of Steward's Calculus books on trigonometric substitutions where it says that the intervals of the original variable of integration and that of the substitution must be able to be converted from one to the other and back. Is that it? $\endgroup$ – Severo Raz Sep 8 '13 at 23:44

Assuming $a\gt0$, here is the result of your substitution $$ \begin{align} I &=\int_{-a}^a\frac{\mathrm{d}x}{\pi\sqrt{a^2-x^2}}\\ &=\int_{-\pi/2}^{\pi/2}\frac{a\cos(\theta)\,\mathrm{d}\theta}{\pi a\cos(\theta)}\\ &=\int_{-\pi/2}^{\pi/2}\frac{\mathrm{d}\theta}{\pi}\\ \end{align} $$ You have to change the limits to match the change of variable. That is, as $x$ varies from $-a$ to $a$, $\theta$ varies from $-\pi/2$ to $\pi/2$. Note that $\cos(\theta)$ is positive over this range, so $\sqrt{a^2-x^2}=a\cos(\theta)$.

Changing direction of the parameterization It is possible to use the interval $[\pi/2,3\pi/2]$, as long as you match things up. We then use $x=-a\sin(\theta)$ and $\sqrt{a^2-x^2}=-a\cos(\theta)$ $$ \begin{align} I &=\int_{-a}^a\frac{\mathrm{d}x}{\pi\sqrt{a^2-x^2}}\\ &=\int_{\pi/2}^{3\pi/2}\frac{-a\cos(\theta)\,\mathrm{d}\theta}{-\pi a\cos(\theta)}\\ &=\int_{\pi/2}^{3\pi/2}\frac{\mathrm{d}\theta}{\pi}\\ \end{align} $$

  • $\begingroup$ I didn't understand why $\cos{\theta}$ being possitive is important, can you elaborate? $\endgroup$ – Severo Raz Sep 8 '13 at 21:57
  • $\begingroup$ $\cos(\theta) \neq 0$ a.e. is important. $\endgroup$ – AlexR Sep 8 '13 at 21:58
  • $\begingroup$ @WolterHellmund: $\cos(\theta)\gt0$ is important since $\sqrt{a^2-x^2}\gt0$. If $\cos{\theta}$ were negative, then $\sqrt{a^2-x^2}$ would be $-a\cos{\theta}$. $\endgroup$ – robjohn Sep 8 '13 at 22:01
  • $\begingroup$ So could this be another way of picking my $\theta$ interval? (Making sure there are no non-real possible evaluations throughout the integration steps.) $\endgroup$ – Severo Raz Sep 8 '13 at 22:19
  • $\begingroup$ @WolterHellmund: You can pick any interval for $\theta$ for which $a\sin(\theta)$ varies from $-a$ to $a$. $\endgroup$ – robjohn Sep 8 '13 at 22:24

The Problem is, you have to exactly specify the bijection (one-to-one and onto) used for the substitution.

$$u(\theta) = a \sin(\theta)$$ maps bijective to $[-a,a]$ for several choices of ${\rm Dom}_u$. The most straight-forward of these is $$u: \left [-\frac{\pi}{2}, \frac{\pi}{2}\right ] \to [-a,a]$$ But the only limitations on the domain are

  1. $\left.\sin\right|_D$ must be a bijection
  2. If $D = [d_0, d_1]$ then $\sin(d_0) \in \{-1,1\} \ni \sin(d_1) \neq \sin(d_0)$ so that the endpoints $u(d_0), u(d_1) \in \{-a,a\}$ and by intermediate value theorem $u(D) = [-a,a]$ since $\sin(D) = [-1,1]$

So to find the range of theta start with a maximum at $\sin(d_0)$, i.e. $$\cos(d_0) = 0 \Leftrightarrow d_0 = \frac{\pi}{2} + k\pi, \qquad k\in \mathbb{Z}$$ and take an interval of length $\pi$, so that $\sin(d_1) = \sin(d_0 + \pi) = -\sin(d_0)$. You get $$D = [\frac{\pi}{2} + k\pi, \frac{\pi}{2} + (k+1) \pi]$$ In your above notation $"n-k" = (k+1)-k = 1$ so your two solutions are the same (correct).

  • $\begingroup$ but k=0 yields $D = [\pi/2, 3\pi/2]$ and that's wrong, isn't it? $\endgroup$ – Severo Raz Sep 12 '13 at 5:08
  • $\begingroup$ Nope, that's perfectly fine, but the integral will transform to $$\int_{\frac{3\pi}{2}}^\frac{\pi}{2} \ldots$$ Which is where the sign reveses to the correct result ;-) $\endgroup$ – AlexR Sep 12 '13 at 10:56

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