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Suppose I want to calculate

$$\int_{0}^{2\pi} f(x) \, dx$$

for some continuous $f: \mathbb{R} \to \mathbb{R}$. Then if I make the change of variable $u = \sin x$, I get:

$$\int_{u(0)}^{u(2\pi)} f(u(x)) \cdot \, u'(x) \, dx = \int_{\sin 0}^{\sin 2\pi} f(\sin x) \, \cos x \, dx = \int_{0}^{0} f(\sin x) \, \cos x \, dx = 0.$$

Obviously, something's going wrong here -- it's not true that $\int_{0}^{2\pi} f(x) \, dx = 0$ for all $f$. But I can't seem to figure out where the flaw is! As far as I'm aware, $u = \sin x$ is a perfectly legal change of variable. Can anyone shed some light on what's going on? Any suggestions would be greatly appreciated.

A-Level Student
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Adithya Chakravarthy
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  • See some of the excellent answers to my question : https://math.stackexchange.com/questions/3841637/why-doesnt-trig-substitution-work-for-definite-integrals – A-Level Student Nov 06 '20 at 16:00

1 Answers1

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In order to make a change of variable, you need your function diffeomorphism, so it needs to be a one to one function on $[0,2\pi] $ and the $\sin$ function is not, and the change of variable you made was $u=\sin{x} $, which means that $du=\cos x dx$ so $\int \frac{f(\arcsin(u))}{\cos(\arcsin(u))}du=\int \frac{f(\arcsin(u))}{\sqrt{1-u^2}}du$

NHL
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