$$3\int_{0}^{2\pi} \sin(t) \cos(t) \,{\rm d}t$$

My calculus is a bit rusty and I can not find where I get it wrong. Setting $u= \sin(t)$, I get ${\rm d} u=\cos(t) \,{\rm d} t$ and, thus,

$$3\int_{u=0}^{u=0}u \,{\rm d}u=0$$

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    $\begingroup$ Excellent. You only need to add $=0$. The substitution is perfect. $\endgroup$ Aug 25, 2020 at 10:03
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    $\begingroup$ $\sin t \cos t=\frac 1 2 \sin (2t)$. $\endgroup$ Aug 25, 2020 at 10:03
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    $\begingroup$ @Gribouillis The substitution should be invertible. Just because the result is correct it does not imply the method is. $\endgroup$
    – Miguel
    Aug 25, 2020 at 10:12
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    $\begingroup$ This one $\endgroup$ Aug 25, 2020 at 10:13
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    $\begingroup$ It is absolutely ridiculous to have closed this question with a reference to a completely different question. Here it is true that the integral is zero and the method used by newhere is excellent. It is a shame for math.stackexchange.com to close this question. It means that people who don't understand mathematics rule this site. $\endgroup$ Aug 25, 2020 at 18:38

4 Answers 4


The substitution is correct. If $f$ is a function with an antiderivative $F$, one has by the fundamental theorem of calculus \begin{equation} \int_a^b f(u(t))u'(t)dt = \int_a^b(F(u(t)))' dt = F(u(b)) - F(u(a)) = \int_{u(a)}^{u(b)}f(x) d x \end{equation} A sufficient hypothesis is that $u'$ is continuous in $[a,b]$ and that $f$ is continuous on an interval that contains $u([a,b])$.

In your case $u(t) = \sin(t)$ and $f(t)=t$.

I'm amazed that so many people are puzzled by this simple application of the fundamental theorem of calculus.

Let us take the case of the integral \begin{equation} I = \int_0^\pi \sin(x) d x \end{equation} and let $u(x) = - \cos(x)$ as suggested in the comments, with $f(x)= 1$. The above substitution formula gives \begin{equation} I = \int_{-\cos(0)}^{-\cos(\pi)} dx = \int_{-1}^1 d x = 2 \end{equation} which is the correct result.

There is no contradiction with this counterexample because in the counterexample, the invalid substitution is $u(x) = \sin(x)$. It would imply $f(u) = \pm\frac{u}{\sqrt{1-u^2}}$ and the original integral must be split at $\pi/2$ to choose between $+$ and $-$. It does not invalidate the above formula with the continuity condition on $f$.

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    $\begingroup$ @YvesDaoust The point is that you need to express $\cos(t)$ in terms of $\sin(t)$ and that depends on the interval. You would have $f(u) = \pm \sqrt{1-u^2}$ and you need to cut the integral at $\pi/2$ and $3\pi/2$ to choose between + and -. $\endgroup$ Aug 25, 2020 at 14:16
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    $\begingroup$ Invertibility is required in indefinite integration, where you are required to go back to the original variable $\endgroup$ Aug 25, 2020 at 15:00
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    $\begingroup$ @Miguel The main thing to stress in the non-invertible case is the difference between $u([a, b])$ and $[u(a), u(b)]$. Also note that the continuity conditions could most certainly be relaxed. $\endgroup$ Aug 25, 2020 at 15:12
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    $\begingroup$ @Riemann'sPointyNose, of course you're right, I was implicitly thinking of the second case you shown $\endgroup$ Aug 25, 2020 at 15:20
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    $\begingroup$ @Riemann'sPointyNose This reminds me of the conditions for integrability. Continuity is not necessary, but in practice it is much easier to break the interval into parts where the function is continuous, than to introduce the Lebesgue integral and argue that the set of discontinuouities has measure zero---of course I am exaggerating a bit :) $\endgroup$
    – Miguel
    Aug 25, 2020 at 15:40

I thought I'd create a separate answer to just try and explain the confusion people have a bit more. @Gribouillis has given you a really good answer, so please give him the solution check-mark.

It's not true that the substitution needs to be injective. You can see for yourself on the Wikipedia article the conditions required for $u$ substitution, and injectivity is not one of those requirements: https://en.wikipedia.org/wiki/Integration_by_substitution. In this case - your solution for this integral is completely valid (it is indeed $0$) and there is nothing wrong with your method.

There are some cases where if you blindly apply substitution, you can get the incorrect result. For example, consider


Let's substitute ${u=\sin(x)}$. Then the bounds become ${\int_{0}^{0} ...du=0}$. Does that mean the original integral is $0$? NO!. We have incorrectly applied the theorem of Integration by Substitution. All substitution says is that


${\int_{0}^{\pi}\sin(x)dx}$ does not match this form. You can make it happen - so let's try it. We can write


Now here's the problem - we need to write the bottom ${\cos(x)}$ in terms of ${\sin(x)}$ - we know ${\cos(x)=\pm\sqrt{1-\sin^2(x)}}$ - that's the problem - the ${\pm}$. If ${x \in \left[0,\frac{\pi}{2}\right)}$, we have ${\cos(x)=\sqrt{1-\sin^2(x)}}$, and if ${x \in \left(\frac{\pi}{2},\pi\right]}$ then ${\cos(x)=-\sqrt{1-\sin^2(x)}}$. So you will have to split up the integral into two halves, like so:


And now we can apply substitution to the two individual integrals to get


So as you can see - there are certain subtleties with using substitution - but injectivity is not a direct condition required for it. Your solution to the integral is absolutely fine.

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    $\begingroup$ I have to think about how to teach this. The emphasis on the form $\int_{a}^{b}f(\phi(x))\phi'(x)dx$ is not useful in practice. If you identify $f$ then you do not need substitution to begin with and the integral is immediate. $\endgroup$
    – Miguel
    Aug 25, 2020 at 15:33
  • $\begingroup$ @Miguel this is true, I completely agree - which is why usually it's helpful to combine these type of substitutions with tricks, like I did with ${\int_{0}^{\pi}\sin(x)dx=\int_{0}^{\pi}\frac{\sin(x)}{\cos(x)}\cos(x)dx=2\int_{0}^{1}\frac{u}{\sqrt{1-u^2}}du}$ (although I think we'll both agree this is a super silly example haha). But yeah, ultimately the statement for Integration by Substitution is actually rather a simple one - and in practice usually people don't worry too much about the rigorous little bits. But when something goes wrong - usually you have to go back to it, unfortunately $\endgroup$ Aug 25, 2020 at 15:40
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    $\begingroup$ @Riemann'sPointyNose Some people have decided to close the question because they want to promote a completely different question that satisfy their ideology. I find this so shameful that I'm considering to withdraw very soon my membership from math.stackexchange. Thank you for your support in this question. There is definitely something wrong in the decision system on these sites. $\endgroup$ Aug 25, 2020 at 19:23
  • $\begingroup$ @Gribouillis we definitely need to get this reopened. The question does not get answered in the linked post, and the substitution works just fine $\endgroup$ Aug 25, 2020 at 19:41
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    $\begingroup$ @Gribouillis I have silently followed this question and also your answer from -1 to +3, as people realise you're correct. Please don't lose faith in the site. I think you can flag the question for moderator attention to explain why it should not be closed. If that still doesn't work, I'd love to see this debated on Meta. $\endgroup$ Aug 25, 2020 at 19:50

I'd just like to point out that there is a simpler method to finding the definite integral than by substitution. We have the identity $$\sin2x=2\sin x \cos x$$ so dividing by $2$ gives $$\frac{1}{2}\sin2x=\sin x \cos x$$ so this allows us to find the value of your definite integral by integrating $\sin t$, which is much easier (as $\int \sin{t} dt=-\cos t +c.$)

I hope that's useful.


The result is correct, because the function $x \mapsto \sin(x) \cos(x)$ is $\pi$-periodic. However, this was more of a lucky strike here, because the sub is not appropriate.

Another (bad) example would be the integral $\displaystyle\int_0^\pi \sin(x) \mathrm{d}x$. If you set $u = \sin(x)$, then you get $\mathrm{d}x = \dfrac{\mathrm{d}u}{\sqrt{1 - u^2}}$, and after substitution, you end up with $\displaystyle\int_0^0 \dfrac{\mathrm{d}u}{\sqrt{1 - u^2}} = 0$, which is clearly false. So something went wrong somewhere.

The main problem, in both cases, is that the inverse sine function is only defined to give you outputs in the "rightmost" part of the trigonometric circle (in other words, the image of $\sin^{-1}$ is $[-\pi/2, \pi, 2]$).

So, since in your example, your range is $[0, 2\pi]$, it messes up as soon as $x$ goes beyond $\pi/2$.

That's why you need to be careful when making trig subs. To do it properly, you'd have to split your intervals in parts where $\sin$ is injective, so that you can get a well defined inverse.

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    $\begingroup$ I'm sorry but this is a different problem. The integral $\int \sin(x) d x$ is not under the form $\int f(u(x)) u'(x) d x$ with $f\in C^0$ and $u\in C^1$. It does not invalidate @newhere 's calculation $\endgroup$ Aug 25, 2020 at 10:28
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    $\begingroup$ The integral is not in the same form indeed, but why would the transformation be valid in one case, and not in the other? In both cases, $\sin^{-1}$ is not well-defined over the range. (I may very well be wrong, but I don't understand why) $\endgroup$
    – Azur
    Aug 25, 2020 at 10:32
  • $\begingroup$ @Gribouillis I agree with you but I am not sure if nowhere had checked the form. So unless one is sure, it may be better to just integrate, get back to the original form by removing substitution and then apply interval values. $\endgroup$
    – Math Lover
    Aug 25, 2020 at 10:33
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    $\begingroup$ @Nala If you want to write it under the same form with $u(x) = \sin(x)$, you need to take $f(x) = \frac{x}{\sqrt{1-x^2}}$ but this function is not continuous on $u([0, \pi]) = [0, 1]$, hence it fails. $\endgroup$ Aug 25, 2020 at 11:09
  • $\begingroup$ This is incorrect. The substitution is actually fine. It conforms to the requirements of $u$ substitution: en.wikipedia.org/wiki/Integration_by_substitution $\endgroup$ Aug 25, 2020 at 13:25

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