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Anonymous Wed 16 Oct 16:31:19 2019 No.11064538 View Reply Original Report

Quoted By: >>11064542 >>11064546 >>11064611 >>11064712 >>11065118 >>11066307

What's the point of doing difficult integrals ? Do they have applications in real life ?

Anonymous

Anonymous Wed 16 Oct 2019 16:32:21 No.11064542 Report

Quoted By:

>>11064538
Perhaps something has some dumb complixities like this.

Anonymous

Anonymous Wed 16 Oct 2019 16:33:38 No.11064546 Report

Quoted By: >>11064627 >>11065026

>>11064538
Yes, they have applications.

Actually try and solve a freaking problem that barely anyone has thought about. Pick something ridiculously simple, yet either useless or too specific in a field for anyone who knows math to have tackled before and written and article about it.

You'll have to solve hard integrals. The more variables and functions you use to make your problem more precise, the harder the integrals get. There's no escaping from this.

You need to integrate at some point if you want to solve most differential equations.

Anonymous

Anonymous Wed 16 Oct 2019 17:15:20 No.11064611 Report

Quoted By: >>11066263

>>11064538
Just let x = arctan(t^2).
The integral becomes 2t^2 / (1+ t^4)
Which is equal to 1/(t^2 + i) + 1/(t^2 - i)
Which is equal to -i/(1 + (t^2)/i) + i/(1 + (t^2)/(-i))
Which integrates to give
(-i)sqrt(i)arctan(t/sqrt(i)) + (i)sqrt(-i)arctan(t/sqrt(-i)) + C.
Changing back to x gives
(-i)sqrt(i)arctan(sqrt(tan(x))/sqrt(i)) + (i)sqrt(-i)arctan(sqrt(tan(x))/sqrt(-i)) + C.

Safety not guaranteed. I have only done this once before.

Anonymous

Anonymous Wed 16 Oct 2019 17:25:00 No.11064627 Report

Quoted By: >>11064666 >>11064746 >>11064971

>>11064546
i call bullshit. a difficult word problem that only needs basic algebra is harder to solve than any of these stupid integrals. since it is the set-up that is tricky, not the solution. as for these retarded technics like substitution or integration by parts or memorizing complicated formulas, who the fuck needs that? there are computers for that.

Anonymous

Anonymous Wed 16 Oct 2019 17:41:22 No.11064666 Report

Quoted By: >>11064683

>>11064627
I don't remember any of the substitution rules or inverse trig stuff and just derive the formulas on the fly.
All you need to know is if f' = g(f), then the derivative of f-inverse is 1/g
Example sin' = cos = sqrt(1-sin^2)
arcsin(x)' = 1/sqrt(1-x^2)

Anonymous

Anonymous Wed 16 Oct 2019 17:47:27 No.11064683 Report

Quoted By:

>>11064666
Ye you would know, satan

Anonymous

Anonymous Wed 16 Oct 2019 17:59:24 No.11064712 Report

Quoted By:

>>11064538
having theoretic knowledge and analytic formulas is always a good thing. and when you can't get those, you have numerics for everything else

Anonymous

Anonymous Wed 16 Oct 2019 18:14:33 No.11064746 Report

Quoted By: >>11064951

>>11064627
Keep thinking that. I like having less competition.

Anonymous

Anonymous Wed 16 Oct 2019 19:22:35 No.11064951 Report

Quoted By: >>11066335

>>11064746
competition in solving integrals on paper? lol. ok, it is all you. the rest of us have wolfram

Anonymous

Anonymous Wed 16 Oct 2019 19:29:54 No.11064971 Report

Quoted By:

Highly g loaded activity; therefore, they are an excellent brainlet filter, and a fun way to pass the time.
>>11064627
You’ve never done a difficult integral before.

Anonymous

Anonymous Wed 16 Oct 2019 19:53:50 No.11065026 Report

Quoted By: >>11065031 >>11065098

>>11064546
You went full autismo there.

Real life problems are solved using numerical methods.

Anonymous

Anonymous Wed 16 Oct 2019 19:57:30 No.11065031 Report

Quoted By: >>11065148

>>11065026
you’re retarded

Anonymous

Anonymous Wed 16 Oct 2019 20:20:54 No.11065098 Report

Quoted By: >>11065148 >>11066341

>>11065026
Numerical solutions work, but are inherently inaccurate and limited in what they can explain.

Anonymous

Anonymous Wed 16 Oct 2019 20:27:01 No.11065118 Report

Quoted By:

>>11064538
>What's the point of doing difficult integrals ? Do they have applications in real life ?

YES, your solution at the end is simple as fuck to compute on a calculator all you need is tan, sqrt, log, exponentiation, addition, subtraction, multiplication, and division.

Anonymous

Anonymous Wed 16 Oct 2019 20:36:53 No.11065148 Report

Quoted By: >>11065164 >>11065173

>>11065031
he's right and you are retarded.

>>11065098
>Numerical solutions work
that's it. full stop.

Anonymous

Anonymous Wed 16 Oct 2019 20:43:53 No.11065164 Report

Quoted By: >>11065179

>>11065148
>t. CS major

Anonymous

Anonymous Wed 16 Oct 2019 20:46:51 No.11065173 Report

Quoted By: >>11065179

>>11065148
No, both of you are idiots and don’t know what you’re talking about.

Anonymous

Anonymous Wed 16 Oct 2019 20:48:52 No.11065179 Report

Quoted By:

>>11065164
>>11065173
no, if you want to see two idiots here they are:
>>11065164
>>11065173

Anonymous

Anonymous Thu 17 Oct 2019 04:20:49 No.11066263 Report

Quoted By: >>11066454

>>11064611
yes, it's incredibly simple if you allow yourself to integrate a complex valued function (totally valid) but getting the result to "look" real at the end is not easy. or maybe it is.

Anonymous

Anonymous Thu 17 Oct 2019 04:43:57 No.11066307 Report

Quoted By:

>>11064538
Often having an analytical solution is better than having a numerical one. Particularly if you are doing something where you have to keep recomputing that integral.

Anonymous

Anonymous Thu 17 Oct 2019 05:02:49 No.11066335 Report

Quoted By:

>>11064951
agreed. This is like someone bragging about the arithmetic they could do right after the computer calculator was invented. The methods and techniques are interesting and good to know, but that guy is way overstating how valuable his niche skill is.

Anonymous

Anonymous Thu 17 Oct 2019 05:05:54 No.11066341 Report

Quoted By:

>>11065098
if a numerical method would be inherently inaccurate, the integral in question would produce an irrational number anyways, which, guess what, will be solved numerically.

as for proofs and theories, 100% agree.

Anonymous

Anonymous Thu 17 Oct 2019 06:02:14 No.11066454 Report

Quoted By: >>11066497

>>11066263
Using arctan(x) = (i/2)log((1-ix)/(1+ix)) and a bit of algebra makes it work. No cleverness and a bunch of real/imaginary parts.

(-i)sqrt(i)arctan(sqrt(tan(x))/sqrt(i)) + (i)sqrt(-i)arctan(sqrt(tan(x))/sqrt(-i)) + C
= Re[2(-i)sqrt(i)arctan(sqrt(tan(x))/sqrt(i))] + C (LOOKS REAL)
= Re[sqrt(i)log((1-sqrt(i)sqrt(tan(x)))/(1+sqrt(i)sqrt(tan(x))))] + C
= (sqrt(2)/2)Re[log((1-sqrt(i)sqrt(tan(x)))/(1+sqrt(i)sqrt(tan(x))))] - (sqrt(2)/2)Im[log((1-sqrt(i)sqrt(tan(x)))/(1+sqrt(i)sqrt(tan(x))))] + C
= (sqrt(2)/2)log(|(1-sqrt(i)sqrt(tan(x)))/(1+sqrt(i)sqrt(tan(x)))|) - (sqrt(2)/2)Arg((1-sqrt(i)sqrt(tan(x)))/(1+sqrt(i)sqrt(tan(x)))) + C

log(|(1-sqrt(i)sqrt(tan(x)))/(1+sqrt(i)sqrt(tan(x)))|)
= log(|(1-(sqrt(2)/2)sqrt(tan(x))-i(sqrt(2)/2)sqrt(tan(x)))/(1+(sqrt(2)/2)sqrt(tan(x))+i(sqrt(2)/2)sqrt(tan(x)))|)
= (1/2)log(|(1-sqrt(2tan(x))+ tan(x))/(1+sqrt(2tan(x))+tan(x))|)

Arg((1-sqrt(i)sqrt(tan(x)))/(1+sqrt(i)sqrt(tan(x))))
= Arg((1-sqrt(i)sqrt(tan(x)))(1+ sqrt(-i)sqrt(tan(x)))/(1+sqrt(2)sqrt(tan(x))+tan(x)))
= Arg((1-sqrt(i)sqrt(tan(x)))(1+ sqrt(-i)sqrt(tan(x))))
= Arg(1-tan(x)-(i)sqrt(2)sqrt(tan(x)))
= arctan(sqrt(2tan(x))/(tan(x)-1))

The final result is
-(sqrt(2)/2)arctan(sqrt(2tan(x))/(tan(x)-1)) + (sqrt(2)/4)log(|(1-sqrt(2tan(x))+ tan(x))/(1+sqrt(2tan(x))+tan(x))|) + C
which agrees with OP.
It looks nasty but it's not too bad on paper.

Anonymous

Anonymous Thu 17 Oct 2019 06:20:51 No.11066497 Report

Quoted By:

>>11066454
now tex that shit like a good boy

Anonymous

Anonymous Thu 17 Oct 2019 06:21:39 No.11066501 Report

Quoted By:

There is no point. It's like art. You push yourself to create a masterpiece.

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