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I am trying to determine if the following is true:

$$ 2\pi R^2 \vec{f}(x,y,z)\int_\alpha^\beta \mathrm{\sin(\theta)d\theta} = 2\pi R^2 \vec{f}(x,y,z)\int_{\cos(\alpha)}^{\cos(\beta)} \mathrm{d(\cos(\theta))}$$

(ignore the function)
Basically if I use that, I get the correct answer in an Electrodynamics problem involving electric displacement fields and spherical symmetries where one part of a ball is filled with one permittivity up to a certain polar angle, and the rest of the ball is filled with another permittivity.

I'm familiar with integration by parts and substitution but this just seems weird. Maybe I got it wrong, maybe my notation is wrong in the equation. In the solution they go from $...\int\sin(x)dx$ to $...\int d(\cos(x))$ with changed upper and lower bounds, which is bugging me cause it seems obvious, yet I can't see it.

Can anyone show me a proof of the upper equality or disprove it?

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Yes, this is correct (EDIT: besides the missing minus sign). This is another (more or less) common way of denoting a substitution. Here, we set $u=\cos(\theta)$ from which it follows that $$\int_\alpha^\beta\sin(\theta)\,{\rm d}\theta\int_{\cos(\alpha)}^{\cos(\beta)}-1\,{\rm d}u\color{red}{=}\int_{\cos(\alpha)}^{\cos(\beta)}-1\,{\rm d}(\cos(\theta))$$ where in the last equality we just plugged in $u=\cos(\theta)$. It a nice way of keeping track what is actually going on when we enforce any kind of substitution: we change which differentials we sum up. That is why this notation is occasionally used.

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    $\begingroup$ Thank you very much. Don't know who downvoted you but I'll equalize it. $\endgroup$ Commented Aug 20, 2020 at 0:51
  • $\begingroup$ @DominikCar Happy to help! :) $\endgroup$
    – mrtaurho
    Commented Aug 20, 2020 at 0:53
  • $\begingroup$ where did the sin(theta)d(theta) disappear to? Substitution is cos(theta) and it makes sense, but where does the sine go to? $\endgroup$ Commented Aug 20, 2020 at 0:57
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    $\begingroup$ @DominikCar If we substitute $u=\cos(\theta)$ we also have to change our differential accordingly. That is $\frac{{\rm d}u}{{\rm d}x\theta}=-\sin(\theta)\Leftrightarrow{{\rm d}u}=-\sin(\theta){\rm d}\theta$ (just realized that there's a minus sign missing). $\endgroup$
    – mrtaurho
    Commented Aug 20, 2020 at 1:01
  • $\begingroup$ Holy molly yes. I'm an idiot and it's 3AM here, Thanks... so obvious..... $\endgroup$ Commented Aug 20, 2020 at 1:02

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