This seems like a really basic calculus question, which is a tad embarrassing since I'm a graduate student, but what does it mean when a substitution in a definite integral makes the bounds the same? For example, if we have some function of $\sin(x)$:

$$\int_0^{\pi} f(\sin(x)) \,\mathrm{d}x$$

If we make the substitution $u = \sin(x)$, then $du = \cos(x)\,\mathrm{d}x$, we find

$$\int_{\sin(0)}^{\sin(\pi)} \frac{f(u)}{\cos(x)} \,\mathrm{d}u = \int_0^0 \frac{f(u)}{\sqrt{1-u^2}} \,\mathrm{d}u$$

This would imply that the integral is zero. Is this always the case? For another example (more relevant to the problem I'm actually trying to solve) consider

$$\int_{-b}^{b} \frac{1}{\sqrt{x^2 + a^2}}\,\mathrm{d}x$$

Clearly this can be solved using a trigonometric substitution to get $2\operatorname{arcsinh}(b)$, but what if I substituted $u = \sqrt{x^2 + a^2}$? Then

$$\mathrm{d}u = \frac{x\,\mathrm{d}x}{\sqrt{x^2 + a^2}} \implies \mathrm{d}x = \frac{u\,\mathrm{d}u}{x} = \frac{u\, \mathrm{d}u}{\sqrt{u^2 - a^2}},$$

so the integral becomes

$$\int_{\sqrt{b^2 + a^2}}^{\sqrt{b^2 + a^2}} \frac{1}{\sqrt{u^2 - a^2}}\,\mathrm{d}u$$

This integral seems to be zero, which is not the case for the integral before the substitution. What's going on here? Does this just mean that these substitutions are not valid?

  • 2
    $\begingroup$ It is a problem that arises when $u$ is not 1-1 over the region of integration. If you set your limits as $-\sqrt{b^2 + a^2}$ to $\sqrt{b^2 + a^2}$, you would arrive at the correct answer. And, I am having a hard time articulating exactly why. $\endgroup$
    – Doug M
    Commented Oct 12, 2016 at 19:43
  • $\begingroup$ Ok. Do substitutions just need to be bijections? I don't recall ever learning that (certainly not in high school calculus), but it sort of makes sense. $\endgroup$
    – Klein Four
    Commented Oct 12, 2016 at 19:45
  • $\begingroup$ If you do a u-sub $sinx=u$, you should take a look t the graph of the function. In layman's terms, you do a u-sub to make a term "go away" so that integration becomes easier. In case of the sine, that function involved is very often odd, and so the net area is then zero. $\endgroup$
    – imranfat
    Commented Oct 12, 2016 at 19:50
  • $\begingroup$ Yes they do... at least in theory. Sometimes, you can get away with it when they are not. But, that, I suppose, separates the mathematicians from the engineers. I don't remember it coming up in high-school either. It comes up whet there is some result that just doesn't seem to be making sense. $\endgroup$
    – Doug M
    Commented Oct 12, 2016 at 19:50
  • 1
    $\begingroup$ My answer here explains (giving various examples) that the integration by substitution theorem doesn't inherently require injectivity, and that whenever injectivity is called for, the culprit is never the theorem's conditions per se. (This is not to say that mandating injectivity is a bad idea.) Implicit substitutions (x=g(t) instead of u=g(x)) are required to be invertible, and this is an implicit consequence of the theorem. $\endgroup$
    – ryang
    Commented Feb 8, 2023 at 16:37

2 Answers 2


For the second integral, note that the substitution $u=\sqrt{x^2+a^2}$ implies: $$ u\ge 0 \quad \mbox{and}\quad x=\pm\sqrt{u^2-a^2} $$ so:

$$ dx=\frac{udu}{\sqrt{u^2-a^2}} \mbox{for}\quad x \ge 0 $$

$$ dx=\frac{udu}{-\sqrt{u^2-a^2}} \mbox{for}\quad x < 0 $$

and the integral splits in two parts as $\int_{-b}^0 +\int_0^b$. This gives the correct result.

We have an analogous situation for the first integral with the substitution $$ u=\sin x \qquad \cos x=\pm \sqrt{1-u^2} $$

  • 1
    $\begingroup$ Ok. So is my problem specific to these examples? Or is it a general principle that substitutions need to be injective? $\endgroup$
    – Klein Four
    Commented Oct 12, 2016 at 20:25
  • 1
    $\begingroup$ Really I never thought of this as a general principle. but I know that we must be care when a substitution is not one-one. $\endgroup$ Commented Oct 12, 2016 at 20:32
  • $\begingroup$ If $f$ is the identity function, the first integral doesn't equal $0$, it equals $2$ $\endgroup$
    – Emilio
    Commented Oct 16, 2016 at 5:19
  • $\begingroup$ @Emilio: You are right! Thank you. I 'll change my answer. $\endgroup$ Commented Oct 20, 2016 at 8:02

The first integral is NOT zero! Let $f$ be the identity function, for example.

In the second integral it is wrong to say that $x=\sqrt{u^2-a^2}$ for all values of $x$. In the first integral, the same: $cos(x)=\sqrt{1-u^2}$ is not true for all values of $x$.

When the substitution is not injective, problems arise when you try to express the integrand in terms of the new variable, as it can be seen from this examples. So always split the integration domain so that there is injectivity in each part.

  • $\begingroup$ So the substitutions are not valid... Is something else unclear? If in the first integral you let $f$ be the identity function, the integral has value $2$ $\endgroup$
    – Emilio
    Commented Oct 19, 2016 at 23:03

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