Let’s say for example you want to evaluate this integral: $$\int_0^{\pi/2} \sin(x)\cos(x)\,dx$$ The best way to do that is to use a substitution, namely you will let $u=\sin(x) \implies du=\cos(x)dx$, And you have to change the bounds of integration, so the integral will look like this : $$\int_0^1 u\,du=\frac{1}{2}$$

But there is a way to do this integral without changing the bounds of integration: $$\int_0^{\pi/2} u\,du=\frac{u^2}{2} \bigg\rvert_0^{\pi/2} $$ Now instead of calculating it directly you should go back to the original variable, namely $u=\sin(x)$ : $$\frac{u^2}{2}\bigg\rvert_0^{\pi/2} = \frac{\sin^2(x)}{2} \bigg\rvert_0^{\pi/2}=\frac{1}{2} $$

My question is which is better changing the bounds of integration first or doing this method? I think that this method is quite nice because if you have a hard substitution and you don’t know how to solve for $u$ To get the new bounds, you can just use this method. Is that correct?

  • $\begingroup$ It's generally best to change the bounds first. $\endgroup$ Apr 11, 2021 at 10:53
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    $\begingroup$ If you're going to do the second method, make sure to write your limits as $x=0$ and $x=\pi/2$. Otherwise, you'll lose track of what you're doing and end up with an answer of $\pi^2/8$. $\endgroup$
    – Teepeemm
    Apr 11, 2021 at 22:14
  • $\begingroup$ Not an answer, but you may enjoy comparing the two strategies with $$\int_{0}^{\pi} \frac{e^{e^{\sin x}} \sqrt{\log(1 + \sin^{10}x)}}{\sqrt[5]{4 + \sin^{3}x}} \cos x\, dx,$$ or some such. $\endgroup$ Apr 11, 2021 at 22:52

4 Answers 4


The reason why you wont see your second method (not changing the bounds) presented as often is that it is not correct to say

$$ \int_0^{\pi/2}\sin(x)\cos(x)\,dx = \int_0^{\pi/2} u \,du$$ while it is correct to say

$$ \int_0^{\pi/2}\sin(x)\cos(x)\,dx = \int_0^{1} u \,du.$$

Note that on the right hand side, $u$ is a dummy variable of integration and the substitution made before is irrelevant. One could just as easily write $\int_0^{\pi/2}v \,dv$ and the meaning doesn't change. So if you want to give a clear demonstration of your work given by a chain of equalities relating one quantity to the next, the 2nd method makes it difficult to do that.

In other words, there is no concise mathematical notation for "then go back and substitute $\sin(x)$ for $u$."

For working mathematicians, the two methods are `the same' (no difference in intellectual content) and therefore they are more likely to defer to the one that can be cleanly and correctly written down, even if they might still work it out in their mind using the second method.

So there really is a difference in the way people write vs how they think, especially in publications like textbooks with high editorial standards.

You should continue to think however you like, but make sure to use correct statements and equations in your writing.


Both methods are correct, but whether you change the bounds or wait until the very end is often, in my experience, very much to do with question.

For example, if you perform a $u$-substitution and then convert everything back into $x$ and this final result in $x$ is very messy, it can often simply save time and possible arithmetical errors by sticking with the integral in terms of $u$ and changing the bounds beforehand. I particularly think this is a good idea if your substitution is of the form $u=f(x)$ for some function $f(x)$, as to find the bounds for $u$ will then be very simple.

On the other hand, if we were making a weird or complicated substitution like $u+\frac{1}{u}=f(x)$ it may be easier just to do the substitution without worrying about the bounds, as finding the bounds for $u$ will be a little complicated.

I don't think I can say much more really; in summary I'd weigh up how relevant the following points are when deciding whether to work out the $u$ bounds:

  • Is writing the integral in terms of $x$ messy/time consuming?
  • Are the bounds for the $u$ integral easy/hard to find?
  • Does working out the $u$ bounds result in 'nicer' bounds that are easier to work with for that particular integral (eg if the original bounds for $x$ were $\arctan 5$ and $0$ then making the substitution $u=\tan x$ the $u$ bounds are just $5$ and $0$ respectively, which is neater)?
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    $\begingroup$ if you did multiple substitutions without changing the bounds of integration you have to back to your original variable $x$ and this would very hard to do. so this is another reason to change the bounds of integration. $\endgroup$
    – PNT
    Apr 11, 2021 at 13:43
  • $\begingroup$ @Yassir Agreed, good point, I didn't think of that. $\endgroup$ Apr 11, 2021 at 17:24

Here's a more mathematical way to do something like the second method. If the problem is to evaluate the definite integral $$\int_0^{\pi/2} \sin(x)\cos(x)\,dx,$$ you can choose first to evaluate the indefinite integral $$\int \sin(x)\cos(x)\,dx.$$ Through the U-substitution $u = \sin(x)$ you find that $$\int \sin(x)\cos(x)\,dx = \int u\,du = \frac{u^2}{2} + C = \frac{\sin^2(x)}{2} + C.$$

Putting this in words rather than equations, $\frac12 \sin^2(x)$ is an antiderivative of $\sin(x)\cos(x)$ and $\sin(x)\cos(x)$ is the derivative of $\frac12 \sin^2(x)$, so we can put $f(x) = \sin(x)\cos(x)$ and $F(x) = \frac12 \sin^2(x)$ in the Fundamental Theorem of Calculus and conclude that

$$\int_0^{\pi/2} \sin(x)\cos(x)\,dx = \frac{\sin^2\left(\frac\pi2\right)}{2} - \frac{\sin^2(0)}{2}, $$

or in typical notation for the solutions of indefinite integrals,

$$\int_0^{\pi/2} \sin(x)\cos(x)\,dx = \left.\frac{\sin^2(x)}{2} \right\rvert_0^{\pi/2}. $$

To put it simply, rather than writing incorrect limits of integration (limits that would give a result different from the one you were looking for), don't put any limits of integration at all. You will legitimately and correctly find an antiderivative of the original integrand and you can use that to solve the definite integral.


I think either method is acceptable. Often when doing difficult (definite) integrals with multiple substitutions it is convenient to not worry about the bounds at all, do the indefinite integral and apply the original bounds at the very end.

You will get the same correct answer if you use either method, as long as you make sure to use either the old bounds with the old variable, or the new bounds with the new variable.


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