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Something very basic seems to slip my mind. Let's say that one wants to calculate the integral

$$\int_0^{2\pi}(1 + \sin \theta) \ d \theta$$ which is obviously equal to $2 \pi$.

Suppose that they, for whatever reason, want to use the substitution $u=\sin \theta$. If one tries to naively change the limits of integration, then both the lower and upper limits become $0$, which is wrong, since then the integral would vanish, which is not the case.

Where have they gone wrong?

Valentina
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    The substitution must be one-to-one on the domain of integration. – ThePortakal May 17 '17 at 14:48
  • The issue is not discussed properly in books. See this answer https://math.stackexchange.com/a/1780735/72031 The main criteria is that if the original integral is with variable $x$ and new integral is with variable $y$ then $dx/dy$ should be continuous. It does not matter whether $x$ is monotone function of $y$. – Paramanand Singh May 18 '17 at 05:54
  • @ThePortakal: contrary to what many believe, we don't need substitution to be one-one. See answer in my previous comment. – Paramanand Singh May 18 '17 at 05:57
  • This is also discussed in Wikipedia where we just need that $dx/dy$ should be integrable. See https://en.wikipedia.org/wiki/Integration_by_substitution – Paramanand Singh May 18 '17 at 06:05
  • Explanation in point #4 of this answer. Briefly: with the suggested substitution $u=\sin\theta,$ expressing the given integrand $(1+\sin\theta)$ in a way that satisfies the integration-by-substitution theorem's conditions is possible only piecewise, so the integral interval must first be split. – ryang Feb 10 '23 at 18:59

2 Answers2

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When substituting $\theta$ with $u = f(\theta)$, f must be a 1-1 function over the integration interval.

susami
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What happens is that the relation between $d\theta$ and $du$ is not identical throughout the range $[0, 2\pi]$.

When you let $u$ equal $\sin \theta$, you have $\frac{du}{d \theta} = \cos \theta$, which will equal $\sqrt{1 - u^2}$ or $-\sqrt{1 - u^2}$ in different subranges of $[0, 2\ \pi]$. So, assuming you wanted to perform that substitution, you would have to split your integration as follows:

$$\int_0^{2\pi} (1 + \sin \theta) \; d\theta = \int_0^{\pi \over 2} (1 + \sin \theta) \; d\theta + \int_{\pi \over 2}^{3 \pi \over 2} (1 + \sin \theta) \; d\theta + \int_{3 \pi \over 2}^{2 \pi} (1 + \sin \theta) \; d\theta \\ = \int_0^1 {{1 + u} \over \sqrt{1 - u^2}} \; du - \int_1^{-1} {{1 + u} \over \sqrt{1 - u^2}} \; du + \int_{-1}^0 {{1 + u} \over \sqrt{1 - u^2}} \; du $$

Papa Smurf
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