Can somebody walk me through how the steps $(10)$ and $(11)$ were carried out?

$(10)$ What happened with $\sin \theta$ after the substitution?

$(11)$ What's the name of the theorem which allows $f = \frac{d}{dx} \int f dx$ because I think that's what have been carried out in this step (Leibnitz Integral rule?)

link: http://physweb.bgu.ac.il/COURSES/PHYSICS_ExercisesPool/33_Electric_Field/e_33_2_132_s_TeX.pdf

  • $\begingroup$ In step $(10)$ he substituted $t=\cos x$ thus $dt = \sin x dx$. $\endgroup$ – Tolaso Feb 19 '18 at 16:32
  • $\begingroup$ Dang. I could have worked that out. Thanks. $\endgroup$ – mathnoob123 Feb 19 '18 at 16:33
  • $\begingroup$ Can you also answer for 11? $\endgroup$ – mathnoob123 Feb 19 '18 at 16:34
  • $\begingroup$ Please change the title.It’s not good enough for other readers who would come to this site to check out this question.Change it to a more suitable ones. $\endgroup$ – user517784 Feb 19 '18 at 16:38
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    $\begingroup$ The theorem of (11) is simply the Fundamental Theorem of Calculus: en.wikipedia.org/wiki/… $\endgroup$ – Emilio Novati Feb 19 '18 at 16:46

The substitution $\cos \theta = t$ is applied. This gives that $-sin \theta d\theta = dt$, making the $\sin \theta$ disappear in equation (10), but adding a minus in front of the integral. Because of the substitution, $t$ runs from $\cos 0 = 1$ to $\cos \pi = -1$. Using the minus from $-\sin\theta d\theta$, the integration bounds are switched to their usual order: from $-1$ to $1$.

In line (11), we compute the derivative of $1/\sqrt{R^2 + z^2 - 2Rzt} = (R^2 + z^2 - 2Rzt)^{-1/2}$ with respect to $z$. This derivative equals $$-\frac{1}{2}\frac{1}{\sqrt[3]{R^2 + z^2 - 2Rzt}}(2z - 2Rt)$$ and then derivative and integral are switched.

The switching of derivative with respect to $z$ and integral with respect to $t$ is possible because of Leibniz Integral rule, since the integration bounds are independent of $z$.

  • $\begingroup$ Can you explain how the switching was done and the theorem that allowed it? $\endgroup$ – mathnoob123 Feb 19 '18 at 16:42
  • $\begingroup$ @mathnoob123 I have added a link. The name of the theorem was the one you suspected :) $\endgroup$ – Student Feb 19 '18 at 16:44
  • $\begingroup$ Can you please again check step 10? It looks like they have omitted the negative sign that is supposed to come due to the transformation of infinitesimal? $\endgroup$ – mathnoob123 Feb 19 '18 at 16:48
  • $\begingroup$ @mathnoob123: this is the first part of my answer: indeed, the substitution gives the following integral: $-\int_{1}^{-1}\ldots$. Then use the minus to switch the integration domain: $-\int_{1}^{-1}\ldots = \int_{-1}^{1}$. Was this not clear from the first part of my answer? If not, I will rewrite it. $\endgroup$ – Student Feb 19 '18 at 16:50
  • $\begingroup$ Oh yes, thanks. It is perfect. Apologies. $\endgroup$ – mathnoob123 Feb 19 '18 at 16:52

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