Mathematics - 2015-09-02

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12:00 AM

Long live King @Robjohn!

I for one welcome our new Bruin overlords

@KevinDriscoll we rule through kindness and ~~cattle prods~~ generosity

1 hour later…

1:25 AM

I've got a good one for you

:math.stackexchange.com/questions/1417772/…

Alexander Gruber

1:41 AM

@MikeMiller gimme a cool topology paper

2 hours later…

3:23 AM

@AlexanderGruber: maths.ed.ac.uk/~aar/papers/exotic.pdf

4:11 AM

summer lightning, first sign of summer ending :(

4:30 AM

Lol only in a place like here can one simply request for a cool topology paper

@TedShifrin sup mr professor dude?

@TedShifrin how are you? I just got back from Toronto on vacation with my fam

that's not true, you could ask for a cool topology paper anywhere

in a bar, on reddit, via email to obama

question in case , how "cool" is that

user61230

...I wonder what would happen if enough people emailed Obama asking for an interesting topology paper.

humanity would be much nicer if they are more interested in maths than invading or offending each other nation

user61230

that's probably true

5:06 AM

@Emrakul not much, it happens every day I'm sure

5:28 AM

Hi @Stan ... Welcome home. I'm now travelling in No Cal, currently at Stanford. Have a good time?

1 hour later…

6:56 AM

@TedShifrin: Still up?

@TedShifrin It was great! I hadnt been able to physically travel for a while, but I am fortunately healthy these days and this is my first trip in about 4-5 years. It went great! Everyone enjoyed themselves. We had a blast. Played like 24 games of pool everynight with my bro. He loves it, not a math guy lol. I hadn't been out of the country before tho. Things didn't seem all that different. I kinda like how everyone in Canada uses the metric system. :D made me feel right at home

7:25 AM

Let $f: D\rightarrow \mathbb{R}$ and $g: D\rightarrow \mathbb{R}$ be functions ($D$ nonempty). Suppose $f(x)\leq g(y)$ for all $x,y\in D$. Is this already enough to say $\sup_{x\in D}f(x)\leq g(y)$?

Okay I'm convinced I guess. I just feel like I must be missing something from this question

2 hours later…

9:30 AM

Hello!! Could you take a look at the proof of the Theorem 3 and tell me at which point we use Lemma 6? I have read the proof several times but I haven't understood how this Lemma helps us...
The Lemma and the Theorem 3 are the following:

7

Q: $F[t]$ has undecidable positive existential theory in the language $\{+, \cdot , 0, 1, t\}$

Consider the ring $F[t, t^{-1}]$ (the polynomials in $t$ and $t^{-1}$ with coefficients in the field $F$). Theorem 1. Assume that the characteristic of $F$ is zero. Then the existential theory of $F[t, t^{-1}]$ in the language $\{+, \cdot , 0, 1, t\}$ is unecidable. Theorem 2. Assu...

number-theory logic computability formal-languages

2 hours later…

11:36 AM

Hello@Balarka

hi

What have you been up today @Balarka

nothing very fun.

Oh.

i'm mostly doing technical things, nothing conceptual. i have a few ideas regarding something, but those are still at the level of vague philosophy.

11:42 AM

Hmm.. @Balarka How much of algebra would be good if you want to start algebraic topology ?

as much as possible. all of what's in Dummit-Foote would be required at some point of time, except perhaps the commutative algebra part and representation theory.

barely 2 months off this chatroom and topology is still the main discussion subject here

@Agawa001 not really

much discussion have been done on algebraic geometry, representation theory and functional analysis too.

It has changed to algebraic geometry now

geometry is cool

11:45 AM

do you mean to say euclidean geometry? that's hardly related to algebraic geometry,

lol

people ,dont you think that "lol" is like a bowling ball missing two skittles ?

As Balarka states " You require loads and loads of algebra to do algebraic geometry" @Agawa001

really, i think they are same

what are the same?

11:49 AM

@Balarka I remember you telling me Algebraic geometry does not work with set theory . What did you mean by that?

I never said that.

I don't even know what that means, in fact.

Said something similar ... I might have forgotten . Let me check

Sorry @Balarka it was not set theory .

Jul 29 at 13:09, by Balarka Sen

ZFC is a set theory, @Remember. It's very weak in the sense that you can't do algebraic geometry in it. Type theory is something close to the hypothetical objects that "unifies foundations of mathematics". The techinique of incorporating homotopy theory grew out of trying to take account to, say, Grothendieck universes in a unified logical system. At least, this is the motivation, as I understand it.

Here what do you mean when you say that you cant do algebraic geometry in ZFC @Balarka

There are things, I have heard, in Grothendieck-type algebraic geometry that you can't do in ZFC. Probably the word is topoi.

As I mentioned above, it was my understanding of what the motivation of HoTT is, so it might as well be false. You shouldn't take everything I saw so literally.

I don't know anything about all this.

Oh okay.

I now actually (probably) know what the real motivation of HoTT is, after having a discussion with Qiaochu. But I still think you can do algebraic geometry inside HoTT, hearing everything they say about how "HoTT is the internal language of a topos", etc.

Anyway, I don't know anything. (That is, I am abandoning this discussion)

11:59 AM

24

Q: The unification of Mathematics via Topos Theory

When the paper The unification of Mathematics via Topos Theory by Olivia Caramello, says "one can generate a huge number of new results in any mathematical field without any creative effort." is this an exaggeration, and if not is this a new idea or has it always been thought that topos theory co...

ct.category-theory ag.algebraic-geometry lo.logic topos-theory computer-algebra

12:11 PM

@Balarka While doing revision of subspace topology today I was stuck at this thought which came to my mind:
If $M$ is a subspace of the space $X$ and we have a mapping of $M$ from the space $Y$ can I extend this map to a mapping from the space $X$ to the space $Y$ .

For this I thought of the example . Lets say $M=S^1$, $X=\Bbb{R}^2$ and $Y=I_2$ where $I_2$ denotes the unit square. We know there is a mapping from the unit circle to the unit square (You gave me this exercise long back). But again can we extend that map to $\Bbb{R}^2$

google tietze extension theorem

it's not true in general.

Does this only work with $\Bbb{R}$. I mean the tietze extension theorem

Tietze extension theorem

In topology, the Tietze extension theorem (also known as the Tietze–Urysohn–Brouwer extension theorem) states that continuous functions on a closed subset of a normal topological space can be extended to the entire space, preserving boundedness if necessary. == Formal statement == If X is a normal topological space and is a continuous map from a closed subset A of X into the real numbers carrying the standard topology, then there exists a continuous map with F(a) = f(a) for all a in A. Moreover, F may be chosen such that , i.e., if f is bounded, F may be chosen to be bounded (with the same bound...

Here

yeah, well, closed subsets of $\Bbb R$ in general.

but it's false for many cases other than $\Bbb R$.

ok, gotta run

So can we somehow classify the spaces on which this can work?By classify I mean to say that all these spaces that have this kind of property have some kind of property in similar (except being topological spaces) @Balarka

12:27 PM

@Remember dunno.

@Rememberme oh, I didn't see the second part of the question. no, it's not possible to extend that homeomorphism $S^1 \to \Bbb I_2$ to all of $\Bbb R^2$. the proof can be done with a bit of homotopy theory : if it was possible, then restricting to the disc bounded by $D^2$ would give you a nullhomotopy of the map $S^1 \to \Bbb I_2$. this is impossible, as that map is precisely a generator of $\pi_1(S^1) \cong \Bbb Z$.

i am not sure if there is any easier proof. i can't think up one off the top of my head, at least.

1:35 PM

I am asked to show that a complete subset S of R^n is closed, but does this not just follow trivially? I mean, if every Cauchy sequence in S converges to some point in S, then S contains all of its limit points and is hence closed.

2 hours later…

3:44 PM

Hello@TedShifrin

hi @TedShifrin

primary decomposition is nice.

@MikeMiller I have a vague philosophical question which is probably going to sounds very stupid. Do you want to hear it, or should I think about it for a few days and try to rigorously state it?

user105491

In case anyone knows of a reference, I'm linking my recent question here.

user105491

0

Q: Lucas's proof of a special case of Beal's conjecture

While studying the properties of a certain elliptic curve, I came across the equation $x^4+y^4=z^3$. There is no solution of this equation in relatively prime integers, and this is a special case of Beal's conjecture. According to this source, the proof that $x^4+y^4=z^3$ has no solutions in rel...

number-theory reference-request diophantine-equations

hi @Sanath.

@Balarka You are free ?

4:00 PM

Hi people! (especially them who know boolean arithmetics ;)

Just got a simple question from a friend about boolean XOR.
**Is it possible to get a, b, c when you have a⊕b, b⊕c, a⊕c ?**
where ⊕ is the boolean XOR (exclusive OR) operator

@Sanath: I did not want to link to gossip on main, but here is a link to some people who appear to be professional mathematicians arguing against the name "Beal's conjecture".

user105491

Hello @BalarkaSen!

@Balarka, Tietze works fine for $\Bbb R^n$-valued functions.

goodnight @MikeM

user105491

@MikeMiller Not that I disagree with you; I've just heard the term used more often.

@Sanath How're you doing?

4:08 PM

I wasn't aware that Noam Elkies used the term; that counts one professional mathematician I know that uses it.

user105491

I'm doing well, how are you?

I tend to agree with the arguments you can find at that link.

@TedShifrin ok, that's nice.

Morning @Ted.

Hi @Remember

4:09 PM

@Sanath I'm alright. What kind of math have you been doing?

user105491

I've been studying algebraic number theory.

user105491

Elliptic curves and modular forms.

cool, dunno anything about them.

one of my former students asked for help on a GRE question. It is cute: How many ways can you write 100 as the sum of 2 or more positive integers, counting reordering the numbers?

Why is that cute?

4:11 PM

@MikeMiller did you see my earlier ping? i just want to make sure whether you're interested in hearing the question or not. if not, i'll get back to you later with a formalized question instead of philosophy.

Do as you will.

Well, as cute as anything combinatorial can be ... This was for the young'uns, not you. :)

user105491

@MikeMiller I'm not sure I downloaded the right book. On page 630, Lucas's name is mentioned, but the equation $x^4+y^4=z^3$ is not (not is any variant of it).

@MikeMiller ok, let me write up the question. it'll take some time.

@Ted I have a question

4:14 PM

Have there been editions since then? The reference I have does not cite an edition, so maybe it's the first one.

@Sanath let me know when you understand how galois representations arise from modular forms. i could never make head or tails of that.

user105491

I downloaded the first edition (it's apparently in the second volume)

user105491

@Balarka Haha, sure. I'll be sure to send you an email (or a ping).

sure, you can do either.

user105491

@MikeMiller The second PDF document you linked to seems to go into some more detail; I'm checking it out now.

4:16 PM

@Sanath: If you can't find any reference I'll delete my post.

Another place to check is Mordell's "Diophantine equations". I don't know if this is available online.

user105491

No, please don't; your answer is great.

user105491

I see how Lucas's work relates.

chat.stackexchange.com/transcript/message/23828450#23828450 Here the homeomorphism map from the unit circle to the unit square is is $\frac{r}{|r|}$ . Balarka did give me a proof using homotopy. Can you help me or give me some idea on how to prove it using point set topology@TedShifrin

user105491

@MikeMiller The page in Dickson's book provides a result of Lucas on some other problem, but it is the key step to proving that $x^4+y^4=z^3$ has no solutions. The relation between Lucas's work and this is in the PDF document you linked.

OK, so you can put these together and form a complete answer?

user105491

4:18 PM

Sure!

People can you prove that this answer is correct?
 puzzling.stackexchange.com/a/20836/15786
It seems plausible but is the OP missing something?

P.S. the question is related to boolean arithmetics

4:46 PM

@Sanath: If you care in general about (classically solved) Diophantine equations, you should find a copy of Mordell's book. Unfortunately it doesn't have the equation by Lucas you're looking for, but it has plenty of others, and an index that points out where each is solved. (It is very classical and doesn't contain any of the modern, more scheme-theoretic techniques.)

Is there anyone who can help me with this? math.stackexchange.com/questions/1418541/…

@Remember: It really is not point-set topology. It's more subtle than that. It's a special case of absolute retracts, which is the beginning of algebraic topoligy.

5:04 PM

If you're given $f(x) = x^2-5x+2$, and $f(2) = f(3) = -4$,
How can you easily find the abscisse of the minimum of the function ?

@MikeMiller OK, let's see. I was asking whether the equivalence $[-, K(G, n)] = H^n(-; G)$ of functors hints towards an existence of a general framework for homotopy theory, i.e., a category where you can study homotopy types without caring about the redundant data of topological spaces. I've thought about this a bit, but essentially everything I did was by writing down the proof of the equivalence and throwing away all the facts about topological spaces I didn't need.

Anyway, enough philosophy : define a "suspension system" $\{X_i, f_i\}$ to be a collection of topological spaces $X_i$ and homeomorphisms $f_i : \Sigma X_i \to X_{i+1}$ between them. Examples include $\{\Sigma^i X\}$ for some topological space $X$.

Define morphisms of these systems in the obvious way : $\{X_i, f_i\} \to \{Y_i, g_i\}$ is defined to be a collection of maps $h_i : X_i \to Y_i$ such that each diagram consisting of vertices $\Sigma X_i, X_{i+1}, \Sigma Y_i, Y_{i+1}$ commutes, with row maps being the suspension map $\Sigma h_i$, and vertical maps are $f_i, g_i$. I think these constitutes a category.

I am not sure what the correct idea of homotopy is here, but whatever it is, homotopy classes of maps $[X_i, Y_i]$ (between topological spaces, not these systems!) would be a group, because $[X_i, Y_i] = [\Sigma X_{i-1}, Y_i] = [X_{i-1}, \Omega Y_{i}]$ which gets a group structure from the $H$-space structure of $\Omega Y_i$. This indicates homotopy of maps between these suspension systems should give information about standard homotopy of maps at each level.

The next problem is to define a good notion of homology theories in this context. I'm sure that having a notion of homotopy and homology of these sequences would give me an isomorphism $H^n(\{X\}; G) \cong [\{\Sigma^i X\}, \{K(G, i)\}]$ or something like it. That's what I am gunning for, at least.

OK, that's it.

You invented spectra.

Congrats @BalarkaSen

2

Well, not quite. You don't want those maps to be homeomorphisms.

@MikeMiller ... i thought ... i thought they were category theoretic foundation of homotopy theory or something?

5:14 PM

I don't know why you thought that.

@MikeMiller How would you give group structure on the homotopy classes of maps at each level, then?

@BalarkaSen: Note that you already don't have homeomorphisms like that for $X_i = K(G,i)$.

You have an inclusion $\Sigma(X_i) \to X_{i+1}$ that's an inclusion into a subcomplex.

eh, right. hmm.

I think you can get a perfectly equivalent homotopy theory of spectra even if you don't demand anything about the structure maps $\Sigma X_i \to X_{i+1}$, but I don't know much about this.

By the way, I should mention that I was also wondering if you can do this at a more abstract level. In a more general (homotopy) category, say, with the $\Sigma$ functor replaced by a functor which is adjoint to some functor $\Omega$ which takes things to group objects or something.

And maybe you can also require commutativeness of the diagram to be modulo homotopy instead of up to nose.

5:24 PM

I'm no homotopy theorist.

This is all pipe-dream, anyway.

wtf is algebra

LOL

oh I forgot to email you back

D:

Shrug :)

5:28 PM

What time would be best for you?

I think I only have class from 11-12 on Friday, and I should finish my homework before then.

Now I think I have lunch plans in SF for Fri, so later in the afternoon ....

Alright. So like three?

Two?

Haven't confirmed yet ... Getting hard to juggle everyone, transportation, etc, but 3 should be safe. I'll let you know when I know more ...

So wtf is algebra?

Alright, no sweat. I'll be around. And yeah, wtf is algebra!

Andrew Thompson

How have you been doing lately, Ted? :)

5:32 PM

This class has been so strange-part of it is because he assumes we've had exposure, but we haven't covered the stuff in class that's on the homework, which is fine because it's probably in the reading, but then like the homework is so weird

Care to make that more specific, Anthony?

Algebra just feels so clean as opposed to analysis

idk

I'm just a babbling brook

It can be somewhat formal/symbolic, which I don't like so much.

Hi @Andrew.

Isn't analysis also formal?

I don't even know what I'm saying.

not to me ...

i try to teach algebra more geometrically/conceptually ...

5:34 PM

Actually I think I get what you mean.

I see.

Math is weird, man.

sure :)

Andrew Thompson

Hm, never been too much of a fan of geometric approaches to algebra, although it is slowly growing on me.

What's an example of a geometric approach to algebra?

so what homework are you stuck on?

Actually the closest I've gotten to geometry is topology lol

5:36 PM

symmetries, group actions

@Anthony Thinking of groups by studying group action on geometric objects, e.g., symmetry

"Erlangen Programme"

@TedShifrin Oh I just need to think about a problem, just marching through the problem set...

Andrew Thompson

@Anthony, some lecturers like to draw orbits of group actions. Also studying rings like $k[X, Y]/(XY)$ can lead to some pretty geometric pictures, I believe Miles Reid's book has an exposition on that.

@AndrewThompson I got so hang of geometric approach to everything that I'm having a hard time playing with cohomology.

That didn't work.

Andrew Thompson

5:38 PM

@BalarkaSen Oh, I'm kind of glad, if that means algebraic topology will be less geometric after a while.

My brain doesn't swing that way.

Most of algebraic topology is non-geometric.

Hatcher's first few chapters are just basic stuff + Hatcher is a geometric topologist

user105491

@BalarkaSen I just got out of the shower, but this looks like the foundation for stable $\infty$-categories which generalize the $\infty$-category of spectra.

Hmm. I still don't really know explicitly what you guys mean by geometric. Like I have some vague notion of why people say symmetries are geometric, and I can believe that it's called geometric if you draw out pictures, but is there some deeper meaning to a geometric interpretation of things?

eek, that kind of went over my head. if you'll explain, I can try to listen.

Andrew Thompson

Yes. I am not too much a fan of his exposition on CW complexes.

user105491

5:40 PM

Sure, I'll try (I'll change and be back)

@Anthony sure, there are examples where geometric interpretations run pretty deep

for example, if you think of groups by drawing cayley graphs, you'll soon run into geometric group theory which is a pretty deep subject

user105491

I'm back

user105491

The $\infty$-category of spectra admits a closed monoidal structure

user105491

in particular, it's enriched over itself

user105491

so if $X$ and $Y$ are spectra then $Map(X,Y)$ is also a spectrum

5:43 PM

I'll mumble some stuff at you in person, @Anthony.

user105491

In addition, there are functors $\Omega$ and $\Sigma$ adjoint to each other which are equivalences on the $\infty$-category of spectra.

:D

user105491

A stable $\infty$-category is a higher category which is enriched over spectra and has maps $\Omega$ and $\Sigma$ both of which are equivalences. These things are studied extensively in Lurie's book Higher Algebra

that sounds interesting, @Sanath. so what is this $\infty$-category of spectra? I mean, how do you define it?

alright I gotta get to class

hasta luego everyone

5:45 PM

Bubye

user105491

There are many models for it

user105491

For example, S-modules, orthogonal spectra, symmetric spectra, etc.

user105491

S-modules are the simplest to think about (imho)

user105491

S here is the sphere spectra (so the maps $S^n\to S^{n+1}$ given by smashing with $S^1$)

user105491

The smash product of spectra (cf the wikipedia article for eg)

5:46 PM

ok, what's the definition of S-modules?

user105491

allows one to define modules over a (commutative) spectra

go on

user105491

the category of S-modules is a model for the category of spectra

user105491

but a nice model-independent definition is given by Jacob Lurie: the \infty-category of spectra is free stable \infty-category with colimits generated by a single object (the sphere spectrum)

user105491

This essentially says the same thing as S-modules but in a higher categorical fashion

user105491

5:49 PM

This is all very modern tech.

user105491

Do you know about quasicoherent sheaves?

I don't even know what a free stable \infty-category is

@Sanath I know what sheaves are.

user105491

cool; so a quasicoherent sheaf is like an analogue of modules

user105491

if R is a ring, then a quasicoherent sheaf on Spec(R) is equivalent to the data of an S-module

can you give me the precise definition of a quasicoherent sheaf?

I'm curious.

user105491

5:51 PM

In mathematics, especially in algebraic geometry and the theory of complex manifolds, coherent sheaves are a specific class of sheaves having particularly manageable properties closely linked to the geometrical properties of the underlying space. The definition of coherent sheaves is made with reference to a sheaf of rings that codifies this geometrical information. Coherent sheaves can be seen as a generalization of vector bundles, or of locally free sheaves of finite rank. Unlike vector bundles, they form a "nice" category closed under usual operations such as taking kernels, cokernels and finite...

user105491

so you can use the smash product of spectra to define commutative/associative ring spectra

user105491

actually there's a whole host of structures in the middle which say commutativity up to kth homotopy (you can think of this as being possible because a spectrum is a sequence of top. spaces)

ok, good. coherent sheaves seem exactly like the analog of vector bundles, which in turn correspond to projective modules.

user105491

right

user105491

you can define the spec of a ring spectrum

5:53 PM

right, you have mentioned that before.

one can build an algebraic geometry over them an all.

user105491

right

user105491

so there's the notion of a quasicoherent sheaf as well

user105491

now a quasicoherent sheaf on Spec(S) where S is the sphere spectrum

user105491

is simply an S-module

user105491

which is a spectrum

user105491

5:53 PM

So you use algebraic geometry to study spectra

this is quickly getting more and more technical. maybe you should tell me about this when I already know something about spectra and algebraic geometry :)

user105491

haha sure

and that'd be quite a few months later, presumably.

user105491

but let me just mention that this application of algebraic geometry to studying spectra is the subject of a recent field known as chromatic homotopy theory

@Sanath You are using the same word to define two totally distinct things. It frustrates me.

user105491

5:55 PM

I'm sorry, what two distinct things?

spectrum

user105491

well that's standard terminology

haha, good point, @PVAL!

user105491

you could say PRIME spectrum over and over again but that eventually becomes frustrating (even more so than saying spectrum for both objects)

user105491

the spectrum of a spectrum

5:56 PM

i wonder if the derived algebraic geometers get confused by that terminology.

user105491

probably lol

user105491

i mean it's usually pretty clear.

6:13 PM

hi all - how is everybody :)

6:56 PM

good evening (or day/night for those who live in America/Asia)

those who live in africa too

7:27 PM

@MikeMiller So I haven't ate lunch yet and I need to. Im gonna have to back out on the asking Freed about your question (this week).

7:52 PM

@PVAL: No worries. I will proably just post it on MO.

8:12 PM

@PVAL: I did.

Thanks, whoever's removing those stars.

salut @Hippalectryon

@Agawa001 o/

Guys I'm somehow stuck on a very very basic question :c how do I show that $\lim_{n\infty}\sum_1^n\frac{1}{\sqrt{n^2+k}}=1$ ?

8:36 PM

nvm

you dont speak french ?

I do :P but since more people here speak English when I ask questions I do so in English

Je suis Français

8:52 PM

@Hippalectryon puisque tu es français, connais tu la difference entre ce que signifie ton pseudonyme et le pegase ?

enfin ils sont les deux hybrides

@Agawa001 Le pégase est un cheval ailé avec de grandes ailes type oiseau volant, et c'est plus de la mythologie grecque si je ne m'abuse

L'hyppalectryon est mi cheval mi coq et vient de la mythologie étrusque

non, tellement vrai

Et puis le pégase est blanc alors que l'Hippalectryon est rouge-jaune-orange-..

le pegase est meme plus attirant

j'ai l'air d'oublier la licorne, elle est aussi splendide

Mais personne ne connait l'hippalectryon :(

8:57 PM

oui, apparemment

9:12 PM

Hi :)

✌

9:43 PM

✌

10:36 PM

Hi @Hippalectryon :)

10:48 PM

@Chris'ssistheartist \int_0^{\pi/2}\frac{cos^2x}{a\cos^2x + b\sin^2x}dx ?????

Hi @Kaj.

Chris's sis the artist

@M.S.E It's a trivial question. Besides that now I prepare a new question for american monthly that I wanna send tomorrow morning.

@M.S.E Use wisely the symmetry.

Denote $$I= \int_0^{\pi/2}\frac{\cos^2x}{a\cos^2x + b\sin^2x}dx $$ $$J= \int_0^{\pi/2}\frac{\sin^2x}{a\cos^2x + b\sin^2x}dx $$

All you need is to make up a clever system of question in integrals.

equations

If there is a peak of extremely hard work to me, then I think in the last hours I was exactly there.

I'm out.

11:22 PM

Salut @Hippa

@TedShifrin Haha it's so weird seeing you on Facebook.

I just saw the post you commented on through Ribet.

(I guess Ribet liked the post)

Chris's sis the artist

11:42 PM

Back to add one more thing (I just recollect)

@M.S.E another way is to consider the integral of the form $$\int \frac{1}{\tan^2(x)-\tan^2(a)} \ dx$$ that can be evaluated in a very clever way.

Welcome back, @Ted

Chris's sis the artist

Stilll for the integral you mentioned, that is $\displaystyle\int_0^{\pi/2}\frac{\cos^2x}{a\cos^2x + b\sin^2x}dx$, I don't see a more elegant way than the one I mentioned by a system of equations.

Note that: $1)$ $a I+b J =\pi/2$ and $2)$ $\displaystyle I+J=\int_0^{\pi/2}\frac{1}{a\cos^2x + b\sin^2x}dx$

where the integrals in the right side can be finished extremely easily (factorize $\cos^2(x)$).

Honestly, is there a faster, nicer way I'm not aware of?

@M.S.E ^^^

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