piskolforji 0 Posted March 8 hi. how to solve the integral that i attached in this post? i tried to solve it but seems its too hard for me 😕 plz help me 0 Share this post Link to post Share on other sites

taeto 74 Posted March 8 You could begin by substituting \(r = a\cdot s,\) that surely helps. What is the upper integration bound for \(\varphi,\) it looks like \(\gamma_\pi?\) 0 Share this post Link to post Share on other sites

piskolforji 0 Posted March 8 8 minutes ago, taeto said: You could begin by substituting r=a⋅s, that surely helps. What is the upper integration bound for φ, it looks like γπ? The upper integration bound for fi is 2 pi 0 Share this post Link to post Share on other sites

taeto 74 Posted March 8 10 minutes ago, piskolforji said: The upper integration bound for fi is 2 pi Great. After substituting \(r\) do you see \(1 + \cos \varphi\) somewhere, and does that remind you of something, like as if it could be another expression squared? 0 Share this post Link to post Share on other sites

piskolforji 0 Posted March 8 (edited) 16 minutes ago, taeto said: Great. After substituting r do you see 1+cosφ somewhere, and does that remind you of something, like as if it could be another expression squared? 1+cosfi= 2 (cosfi/2)^2 But the denominator is a+r cosfi Edited March 8 by piskolforji 0 Share this post Link to post Share on other sites

taeto 74 Posted March 9 12 hours ago, piskolforji said: 1+cosfi= 2 (cosfi/2)^2 But the denominator is a+r cosfi Indeed. I misread the integrand, sorry about that. But re-reading it now, I wonder if you copied it correctly from its source. The differentials \(dr\) and \(d \varphi\) are in the wrong order with respect to the order of the integration symbols. You have to know which integral is the "outer" and which is the "inner", since in this special case you are not free to swap them around as you please. The integrand is not defined at \(r = a\) and \(\varphi = \pi,\) and that makes it harder than usual. 0 Share this post Link to post Share on other sites

piskolforji 0 Posted March 9 i could solve it by Mathematica but i need to know how to solve it.☹️ 0 Share this post Link to post Share on other sites

taeto 74 Posted March 9 (edited) Good, that helps. Now the integral over \(r\) is of standard form \(\int \frac{r}{a+br}dr,\) with \(b=\cos \varphi.\) An antiderivative is \( \frac{r}{b} - \frac{a}{b^2}\log | a+br|.\) So we get \(\int_0^a \frac{r}{a+br}dr = \frac{a}{b} (1-\frac{\log (1+b)}{b} ) \) when \(b \not\in \{0,-1\}.\) Agree? Edited March 9 by taeto 0 Share this post Link to post Share on other sites

piskolforji 0 Posted March 9 (edited) did u use integral table for that integral? can u give me a reference plz ? Edited March 9 by piskolforji 0 Share this post Link to post Share on other sites

taeto 74 Posted March 9 en.wikipedia.org/wiki/List_of_integrals_of_rational_functions 0 Share this post Link to post Share on other sites

piskolforji 0 Posted March 9 8 hours ago, taeto said: en.wikipedia.org/wiki/List_of_integrals_of_rational_functions But it gives a more difficult integral for fi 0 Share this post Link to post Share on other sites

taeto 74 Posted March 10 Maybe use \(\cos(2\pi - \varphi) = \cos \varphi\) to change the integral from \(0\) to \(2\pi\) to two times an integral from \(0\) to \(\pi.\) Then use \(\cos(\pi - \varphi) = -\cos \varphi\) to change this integral to an integral from \(0\) to \(\pi/2.\) Finally substitute \(s = \sin \varphi\) to get an integral from \(0\) to \(1\) without any trigonometric functions (but a \(\log s\) ). I will try to look if that can be done. 0 Share this post Link to post Share on other sites