integral of sec(x), 4 results!

integral of sec(x), 4 results,
tanh^-1(sin(x))=ln|sec(x)+tan(x)|, • math for fun: tanh^-1(... ,
complex defintion of sine & cosine, • Complex definitions of... ,
4 ways to integrate sec(x),
integral of 1/cos(x),
integral of sec(x) without that trick,
blackpenredpen,
math for fun,

[@voiceinthevoid14]

If it were on a test I would go with #2 but if I knew I did really well on the test than I would go with #4 just to give my teacher a challenge in the midst of grading papers.

[@leonardromano1491]

It might also depend on the subject. I'm studying physics and the profs are really strict with their ONLY-real-solutions-policy. In physics you usually only care about the real solutions because the theories in physics are made for real numbers.

[@SchrondingersCat]

Leonard Romano Imaginary numbers are everywhere in physics!

[@rishabhdhiman9422]

If I finished my paper and had enough time, I'd write all four of them (but first they'll have to teach me calculus in school, 2 years to go!)

[@rishabhdhiman9422]

Ian Bornhoeft I haven't seen them anywhere except in quantum mechanics (especially in wave function)

[@leonardromano1491]

Of course they are everywhere. Except that you usually want your final result to be real. They appear quite often on the way though. Best example: harmonic oscillator: You get a linear combination of complex exponentials but in classical mechanics and most other theories except maybe quantum mechanics (haven't taken the course yet) you ignore the complex solutions and only take the solutions that have a real output. You may want to describe periodic thing's like plane waves and oscillators in terms of complex exponentials because the maths are easier but then again the things you are measuring (phase and amplitude) are real.

[@chrysophylaxs7208]

You could do Int(secx) dx Let u = secx du = secx tanx dx If you now form a right angled triangle with angle x, the hypotenuse will be u and the adjacent side will equal 1. By the pythagorean theorem, the opposite side will be sqrt(u^2-1). That means: du = u sqrt(u^2-1) dx For the cancellation: dx = du/(u sqrt(u^2-1)) Now substitute = Int(u · 1/(u sqrt(u^2-1))) du The u and 1/u cancel to get: = Int(1/sqrt(u^2-1)) du Which is the inverse hyperbolic cosine = cosh^-1(u) Substitute back for u = cosh^-1(secx)

[@blackpenredpen]

Chrysophylax Dives wow!!!

[@chrysophylaxs7208]

Haha thank you sir :)!

[@clawsie5543]

Hmm... I tried to use x=arctan(u) substitution and obtained arsinh(tan(x)) which is, just as you result, partially correct. The graph of arcosh(sec(x)) and arsinh(tan(x)) are kinda different from ln(|tan(x)+sec(x)|), some branches coincide but others must be reflected in order to match ln(|tan(x)+sec(x)|).

[@chrysophylaxs7208]

Yep, youre right. Actually if you solve the integral of 1/sqrt(u^2-1) using u=secθ youll be left with the standard result.

[@chrysophylaxs7208]

To be honest, I have no idea. Ill try to figure it out.

[@iaagoarielschwoelklobo6342]

Jumpscare list: 3:25 *monster appears 19:19 *monster appears btw nice vid

[@ShamimAhmed-xj6lf]

Iaago Ariel Schwoelk Lobo Sexx

[@doximanman]

It's oods from doctor who, they hold a white orb which is like their heart, and he looks like them because of his mic. lmao

[@jordanfreidel1751]

It’s an ood from doctor who

[@mr_meow_77]

Scary 😭

[@mr_meow_77]

That's ben ten villan mask vilgax

[@WildSeven19]

I liked the partial fraction solution most. Felt more intuitive and simple. cos^2x+sin^2x=1 is an old friend by now!

[@Thalesfreitas96]

Got a little heart attack from the mask...

[@deeptochatterjee532]

Thales Freitas Macêdo It's the Ood

[@yoavcarmel1245]

Deepto Chatterjee i watch dw and still everytime he puts that ood head i get jumpscared

[@clauskutsche918]

That oodball... add last he did it ! btw, Ood are sweet

[@Thalesfreitas96]

Deepto Chatterjee Just found what the Ood is... He full cosplayed it; the mask and microphone

[@jordanfreidel1751]

It’s an ood

[@mythicmansam1082]

Hey blackpenredpen I love your videos, I check everyday to see if you posted new content. The types of math problems you walk us through tickles my brain such an awesome way! Btw method number 4 was by far my favorite. It makes me wonder what other types of intergrals can have complex results :)

[@blackpenredpen]

Samuel Roberts a lot, actually. I will try to do more later on. Thanks for u nice comment.

[@Whizzer]

I quite like the last one.

[@ktuluflux]

You can also use the tangent half angle substitution. Also (erroneously) called the Weierstrass substitution, you can write any trig function in terms of tan(x/2) There is a general case for integral (1/(a+bcosx)). Just put b = 1 and a= 0 in the final answer there.

[@chrysophylaxs7208]

Now you could also write secx as cscx times tanx That will get you Int(cscx tanx) dx Now let u = tanx du = sec^2(x) dx Do the same trick again, create a right-angled triangle with acutw angle x. The opposite side will be equal to u, and the adjacent side will be equal to 1. By Pythagoras, the hypotenuse is equal to sqrt(u^2+1). Now, sec^2(x) = u^2+1 cscx = sqrt(u^2+1)/u dx = du/(u^2+1) Now substitute: = Int( sqrt(u^2+1)/u · u · du/(u^2+1)) 1/u and u will cancel: = Int( sqrt(u^2+1)/(u^2+1) ) du = Int( 1/sqrt(u^2+1) ) du Which is the inverse hyperbolic sine! = sinh^-1(u) Substitute back for u = sinh^-1(tanx) + C

[@pituitlechat3807]

I prefer this answer because for x=[pi/2,3pi/2], ln|sec + tan| = -sinh^-1(tan) and the slope (= sec x) must be >0 in this interval

[@PackSciences]

I prefer number 2 but number 4 is interesting to know.

[@TheLucidDreamer12]

I know this is old, but 1/2(arctanh(sqrt(sin x) - arctan(sqrt (sin x) ) + K is also pretty fun. Use the German K to really mess with them.

[@jimstiles26287]

I wish KZbin was around in 1987 when I was taking inviscid fluid flow. These videos are an excellent refresher in complex numbers.

[@victorlacerda9659]

The complex result is by far the best, thanks for all.

[@shyamalshil4309]

Sec

[@pbj4184]

Also the most useless

[@Kuratius]

Another cool thing: the same approach as in #4 also works for the integral of 1/sin x : you get -2*artanh (e^ix) as the antiderivative. I hadn't thought about that, thank you!

[@gorillajock]

I was a math geek in high school and college, but in my work I ended up never needing anything much beyond algebra and probability, never the calculus that I spent years learning, except to tutor a friend for a year at work who was having trouble with basic calc derivatives, and basically forgot all about it (that was 30+ years ago, even the tutoring was 20 years ago). I never suspected how much I still love this stuff until I started watching your videos (and Mathologers too) a couple months ago. I always try to solve your integrals, but am almost never able to - I don't even remember if we ever did integrals like the ones you show (and I went to a hard core math/science/engineering institute in California, studying calculus and advanced math for four years, plus a couple years in high school), and I am always amazed and binge watch your videos for hours on end. I'll never use it, but the pure beauty of it outweighs any practical use it might have or not. A BIG THANKS!

[@JulianGarcia-gx2wg]

3:26 naniii

[@johnneumann8878]

And @19:19 lol.

[@michaelhyh5552]

It's an Ood

[@shadowkiller0071]

It's an enemy stand?

[@shahriarparvez1041]

Julián García am

[@shahriarparvez1041]

nimals ex

[@Green_Eclipse]

If you stop method 4 at 2/i * integral of 1/(u^2+1)du then that equals 2/i * integral of 1/(u^2-i^2)du. Factor denomintor with difference of squares and use partial fraction. You end up with ln((e^(i*x) + i)/(e^(i*x) - i)) + C

[@dakkoshite]

13:26 Do you have to use absolute value ln when both 1+sinx and 1-sinx will never return negative? (i.e. is 1/2 ln[(1+sinx)/(1-sinx)] + C also correct, or does it have to be 1/2 ln|(1+sinx)/(1-sinx)| + C?)

[@utuberaj60]

Very nice derivation, especially the last one using i and the one using stadard hyperbolic tanx. But there is one more method in text books using the substitution u=tan(x/2), which reduces the trig function into rational function and is integrated in the stadard way we use for this class of functions. The result is I think = ln | tan(x/2+π/4) |+C. This method is I learnt recently known as WEIRSTRASS substitution.

[@thephysicistcuber175]

Use parametric formulas to express sec(x) in terms of tan(x/2), tan(x/2)=u, integrate

[@blackpenredpen]

AHHHHHHH I FORGOT TO DO THAT!!!! >_

[@thephysicistcuber175]

blackpenredpen lol, I've got you covered. Also, I've learnt that as a standard method to deal with rational functions of trigs, so naturally I would consider it.

[@blackpenredpen]

Thanks!!! You know, I even did a vid last year on that, lol!

[@blackpenredpen]

kzbin.info/www/bejne/p5rNqWywrZh7hLM

[@ishaan1]

The Physicist Cuber aha! the weisterass sub

[@RodrigoCastroAngelo]

Loved the Ood costume haha! The microphone reminds me of them everytime

[@andreguimaraes9347]

Put every single one in a plot, and got the exact same thing... Amazing! (had to use only the real part for the last one though)

[@frederickm9823]

If you love doing integrals the hard way, you can do the following: Take the second method and replace the 1 in the nominator with sin^2(x)+cos^2(x). After a lot of work you'll end up with ln(sin(x/2)+cos(x/2)) - ln(cos(x/2)-sin(x/2)) Don't ask me how to get there, a friend of mine told me that :)

[@jackdoherty762]

Fredo Lucroy when he said that sin^2(x) + cos^2(x) = 1, that was my first instinct. Maybe I'll give that one a try some day.

[@adilsonfranciscoquissai3888]

You'd have 1/cos(x)=(cos^2(x/2)+sin^2(x/2))/(cos^2(x/2)-sin^2(x/2))=(1+tan^2(x/2))/(1-tan^2(x/2)) and now you can use u=tan(x/2) where x=2tan^-1(u)

[@donmoore7785]

Very nice presentation! #4 blew me away - never would have thought of it.

[@abd-elrahmanmohamed9839]

your videos are amazing ... of course the best method is the last one using complex expression of cosine

[@blackpenredpen]

Thanks!

[@shandyverdyo7688]

I always like this dude!!! Very useful and nice video.

[@calyodelphi124]

This made me want to have a little bit of fun with MathJax tonight. So I took the four solutions you demonstrated in the video, plus the three others I found lurking in the comments, and compiled them all together into a single document as a little bit of practice with LaTeX and MathJax. My brain is thoroughly fried from having to basically learn the math side of LaTeX from pretty much nothing. O..o;;

[@farukhsandhu1066]

Ur simile after solving a problem is great source of satisfaction😍

[@user-wf3oj2vv2w]

That Ood Mask Loool

[@MatthewBrown1227]

If you combine the recent video on weierstrass substitution, you pretty easily get the integral for 2tanh-1(t) which after re-substituting t gives 2(tanh-1(tan(x/2)))+C

[@General12th]

Number four is clearly the best. It's completely right, and it would have sent my teacher into conniptions!

[@Kuratius]

J.J. Shank :D Try integrating 1/sin x, it works the same way (complex definition of sin x)

[@elumbraarjay1005]

that was so cool. all those answers are great. great job and great brain sir

[@AhnafAbdullah]

Guys I'll make a video of this where I'll differentiate all of these functions to get back sec(x), assuming anyone wants to see lol

[@blackpenredpen]

Ahnaf Abdullah I do!!!

[@AhnafAbdullah]

Hahahaha! Ok, I'll do it soon

[@AhnafAbdullah]

kzbin.info/www/bejne/jqC8eZybncRge6c I'm a bit late but here it is! I'm also planning to differentiate the integral of cube root of tan x, since the first guy did it on scratchpad, it's very unclear.

[@saxbend]

When integration a complex expression doesn't the constant have a potential to be a complex number as well? In which case maybe it's better to write plus z instead of plus c.

[@yaxeenrahman]

@11:22 can we use (sqrt U)^2 and make the inverse hyperbolic tangent and normal inverse tangent? integration of 1 / 1+(sqrt U)^2 = arctan (sqrt U) also integration of 1 / 1-(sqrt U)^2 = inv.Hyp.tan (sqrt U) ?

[@Kuratius]

What are you saying at 16:15? "If we differentiate this, the function part repeat.... "?

[@serouj2000]

While you're at 6:38, you could have noticed it is in the form 1/1-k with k=u^2, you can turn that to an infinite series in this case, sum (u^2n) {n,0,infinity} After integrating, it would be sum ((1/(2n+1))u^(2n+1)){n,0,infinity}, then replace u by sin(x) which gives sum ((1/(2n+1))sin^(2n+1)x)){n,0,infinity}+c

[@Thalesfreitas96]

Are all those methods available to cossecant, hyperbolic secant and hyperbolic cossecant?

[@robertowusu4188]

You can also use the substitution, t = tan ½ x so that cosx =(1-t2)/(1+t2)

[@dominicellis1867]

How do you convert the complex definition into the standard u sub definition?

[@user-ge2py5uf3i]

There is the fifth way for integral of sec(x) d(x) Put t = tan(x/2) Then sec(x) = (1+t^2)/(1-t^2) dx = 2dt/(1+t^2) integral sec(x) = integral (1+t^2)/(1-t^2) × 2dt/(1+t^2) By simplification by (1+t^2) and separation on 2 fractions we get integral sec(x)= integral of (2dt/(1-t^2) = Ln/(1+t)/(1-t)/ +C Cordially yours

[@moonman239]

So, if I understand #1, we're really just trying to manipulate secant of x into an expression that we can then use u-substitution to "cancel out" the secant of x, which then leaves us with just u. As it happens, secant^2(x) + secxtanx / secant(x) + tan(x) is just an expression, because when u = secant(x) + tan(x) we get secant^2(x) + secxtanx.

[@imme3024]

We can write 1/cos(x) as tan(x)/sin(x), by putting u = sin(x) we get sqrt(1-u^2) = cos(x) and therefore tan(x) = u/sqrt(1-u^2), and dx = du/sqrt(1-u^2). By simplyfing with "u", we get the integral of 1/[sqrt(1-u^2)*sqrt(1-u^2)], which is exactly what was found with the partial fraction or the inverse hyperbolic tangent

[@j5v133]

In the 1/cosx method you can also sub cosx=(1-tan²(x/2))/(1+tan ²(x/2))..😬 Then tan(x/2)=t and dt=1/2(sec²(x/2))...which is the num so the integral becomes sort of int(dt/(1-t²)

[@ab-tt7hf]

How can multiple different functions have the same derivative? (Assuming no constant terms in the original functions) I tired to graph the first and third results on Desmos and both seemed to be very similar to each other. Are they actually identical? Or do they just have different domains and ranges? I thought there are only one anti-derivative to elementary functions. Can somebody correct me.

[@IvyANguyen]

This was so cool to see!

[@JBaker452]

This is of mind blowing proportions. Black pen red pen yeah!

[@LearnWithFardin]

I love all of those!!! But a bit less complex!

[@demetrisgiovanniedwards1984]

What If you have a definite integral. How would I resolve tanh^-1(sinx)

[@tanqs789]

Recently found out about this channel. Please keep making video!~

[@blackpenredpen]

我會的, 謝謝～

[@k_wl]

if you look closely the partial fraction one is just ln|secx+tanx| + c as (1+sinx)/(1-sinx) = (secx + tanx)^2 (trig identitiy), and the square goes outside and the abs value remains, the 1/2 and 2 cancel out and were left with ln|secx+tanx| + c

[@WarpRulez]

I have an unrelated question: How would you calculate evenly-spaced points on a parabola?

[@butterflyspinart]

If I may ask, what is hyperbolic tangent?

[@CLeonard484]

The complex way is the best. You started with something real and ended with a complex answer.

[@jacobharris5894]

That last one is so cool and it seems to be the shortest way to do it. It would be nice though if you could convert to a real answer at the end if that’s possible.

[@voiceinthevoid14]

I wonder if there are complex solutions to ln/logs of negative values.

[@blackpenredpen]

ThisCouldBeYou yes

[@Kuratius]

ThisCouldBeYou If there is a way to compute them with the integrals in this video, can you show me how?

[@XanderGouws]

Take a complex number z=re^iθ ln(z) = ln(r*e^iθ) = ln(r) + iθ * ln(e) {By log properties} This simplifies to: ln(z) = ln(r) + iθ If you plug in a negative number, θ=π+2πk for k ε Z r = | z | Ofcourse, you can restrict your possible θ values to -π < θ

[@gustavoalexanderma8587]

@@XanderGouws do you recommend me some book to learn that :'( I understand something about that (some book with integrals with complex numbers) English please

[@XanderGouws]

@@gustavoalexanderma8587 One of my favorite general maths books is "mathematical methods for physics and engineering" by Riley, Hobson, and Bence. I (and a couple others) have also made some videos on those topics :) Hope this helps.

[@ronycb7168]

This is the first time I ever understood the coverup method thanks

[@user-kg3ps6zy7t]

Hi, many thanks for this clear presentation! I still have a question, how is possible that this integral has 4 solutions? they look different...isn' the solution supposed to be unique?

[@jc2801]

You are incredible. The best.

[@davidkippy101]

Is there a way to set the complex integral equal to a different integral, plug in an x, and then solve for the value of i?

[@Kuratius]

david plotnik Do you mean solving for the value of the complex integral? What do you mean?

[@davidkippy101]

Kuratius I'm saying that you can set the result of that very last integral equal to the noncomplex result of the other integration methods before it, then solve for i?

[@Kuratius]

david plotnik They aren't equal, they differ by the constant c=-i π/2 . Anti derivatives do not have to be equal. They just have to differ by a constant (i.e. they change in the same way=they have the same derivative).

[@dr.rahulgupta7573]

Excellent presentation. Wow! DrRahul Rohtak Haryana India

[@satishkumarsingh2823]

from where did you get that mic..thats super cool just like your teaching....love from india...oh dude you are really awesome

[@orvenpamonag2234]

It is possible to come up with an answer if the substitution instead becomes, e^ix=cosx+isinx?

[@PUBGgamer-kf5vz]

Great work sir (thank you )

[@zizo-ve8ib]

Just a little question here in one why do you even multiply by a variable can you even do that?

[@zerversinvaders]

Technically, the first and third answers are the same. Let's take 1/2*ln|(1+sinx)/(1-sinx)| which is the third answer. multiply numerator and denominator by 1+sinx 1/2*ln|(1+sinx)^2/(1-sin^2(x))| denominator becomes cos^2(x), and since we have 1/2 outside the ln, we raise the inside fraction by 1/2 ln|(1+sinx)/cosx| separate into two fractions 1/cosx and sinx/cosx, which gives us ln|secx+tanx| which is the first answer

[@zerversinvaders]

fingers crossed blackpenredpen notices this 😂

[@blackpenredpen]

Yes! I know they are the same and have a proof in another video

[@blackpenredpen]

In fact, number two is the same too

[@Ni999]

By definition, tan¯¹u = ½[ln(1+u) - ln(1-u)] And that's 12:43 and so forth. Edit - bprp video linked at the end is even better - kzbin.info/www/bejne/mqHRopt_i8lgpsk

[@sakshamgarg8938]

Has this guy made video Abt differentiation of inverse functions? If yes then pls send the link..Btw great video man.. just subscribed yesterday

[@blackpenredpen]

Saksham Garg check out my site. Www.blackpenredpen.com and resource calculus.

[@renerpho]

Here is a fun expression: 2*[tanh^-1((sqrt(2)+1)*tan(x/4))-tanh^-1((sqrt(2)-1)*tan(x/4))]+C We start from the identity cos²(x/2)=(1+cos(x))/2. Write this as cos(x)=2*cos²(x/2)-1, and replace 1/cos(x) under the integral sign by 1/(2*cos²(x/2)-1) = 1/2*[1/(cos²(x/2)-1/2)]. The denominator can be factorized: 1/2*[1/(cos²(x/2)-1/2)] = 1/2*[1/[(cos(x/2)-1/sqrt(2))*(cos(x/2)+1/sqrt(2))]]. Partial fraction (using your cover up method): 1/2*[(sqrt(2)/2)/(cos(x/2)-1/sqrt(2))-(sqrt(2)/2)/(cos(x/2)+1/sqrt(2))]. Factor out sqrt(2)/2: sqrt(2)/4*[1/(cos(x/2)-1/sqrt(2))-1/(cos(x/2)+1/sqrt(2))]. Now, tackle these as two seperate integrals. We use the standard trick how to solve integrals of the form R(sin(t),cos(t)), where R is a rational function: The substitution u=tan(t/2) changes the integral into a rational function of sin(t)=2u/(1+u²), cos(t)=(1-u²)/(1+u²) and dt=2*du/(1+u²) (this always works). In our case, we have a rational function of cos(x/2) in each of the two integrals, but the trick still works (we just have to be careful with the chain rule). We substitute u=tan(x/4) to get cos(x/2)=(1-u²)/(1+u²) and dx=4*du/(1+u²). The calculation for the integral of sqrt(2)/4*1/(cos(x/2)-1/sqrt(2)) goes as follows (the other one involves essentially the same steps): sqrt(2)/4*1/[(1-u²)/(1+u²)-1/sqrt(2)]*4/(1+u²) du =sqrt(2)/[((1-u²)*sqrt(2)-(1+u²))/(sqrt(2)*(1+u²)]*1/(1+u²) du =2/[sqrt(2)*(1-u²)-(1+u²)] du =2/[(sqrt(2)-1)-u²*(sqrt(2)+1)] du =2/[(sqrt(2)-1)*(1-u²*(sqrt(2)+1)/(sqrt(2)-1))] du =2/(sqrt(2)-1)*1/[1-(sqrt((sqrt(2)+1)/(sqrt(2)-1))*u)²] du Substitute s=sqrt((sqrt(2)+1)/(sqrt(2)-1))*u, du=sqrt((sqrt(2)+1)/(sqrt(2)-1)) ds 2/(sqrt(2)-1)*sqrt((sqrt(2)+1)/(sqrt(2)-1))*1/(1-s²) ds =2/(1-s²) ds We can integrate this, using the tanh^-1 function. The integral is 2*tanh^-1(s). Substitute back for s: 2*tanh^-1(sqrt((sqrt(2)+1)/(sqrt(2)-1))*u). Notice that sqrt((sqrt(2)+1)/(sqrt(2)-1))=sqrt(2)+1: 2*tanh^-1((sqrt(2)+1)*u). Substitute back u=tan(x/4) to get the final result (same for the second integral). Voila!

[@williamwarren9397]

what if you integrate up to the point with sec(x)=1/cos(x) and still changed 1 to sin^2(x )+cos^2(x), but you separate the two sides of the numerator into integral (sin^2(x)/ cos(x)) + integral (cos^2(x)/cos(x)) and got sec(x) +sin(x) +c?

[@shubhamide]

you can also write 1=(sinx)^2+(cosx)^2 in the numerator and then write sin squarex by cosx plus cos squarex by cosx.... and then we will get integration of tanxsecx+cosx and then sinx +secx +c i wnat to know if this is correct

[@silentclashing423]

Can we use integration by parts

[@LuisRodriguezPalomo_]

you are amazing!

[@leonardromano1491]

You could do the substitution u=it where you integrate 1/(1-u^2) so the result becomes -i*arctan(i*sin(x))+c how cool is that?

[@seegeeaye]

A short cut, multiply secx by a fraction of secx + tanx over secx + tanx, the integral becomes d(secx + tanx)/(secx + tanx) which results ln(secx + tanx) + C

[@YogeshKumar-rf5ys]

Sir what is the complex solution of |x| = some -ve number?

[@vkillion]

What if you do partial fraction decomposition on the complex result instead of using the inverse tangent? Does partial fraction decomposition work for complex numbers? I worked it out based on what I remember of calculus (it's been a while since I've used it) and the answer I got was: ln | (e^(ix) - i) / (e^(ix) + i) | + c Let me know if this is a valid result or not.

[@Kuratius]

Vincent Killion Compute Wolfram Alpha says yes: I think you dropped a minus somewhere, I added it in. '- d/dx ln((e^ix - i)/(e^ix + i))' with the Wolfram|Alpha website (www.wolframalpha.com/input/?i=-+d%2Fdx+ln%28%28e%5Eix+-+i%29%2F%28e%5Eix+%2B+i%29%29) or mobile app (wolframalpha:///?i=-+d%2Fdx+ln%28%28e%5Eix+-+i%29%2F%28e%5Eix+%2B+i%29%29).

[@vkillion]

Kuratius That's entirely possible. I was doing it quickly and probably forgot to copy one down. Thanks for checking it. I wasn't sure if Wolfram Alpha would handle it correctly.

[@Kuratius]

Keep in mind, I think in this case taking the absolute value as the argument for the logarithm is wrong. Wolfram returns a different result in that case.

[@Kuratius]

Vincent Killion Here's a plot of your antiderivative: I suspect it's actually the same one as #4, I'll check in a minute. www.wolframalpha.com/input/?i=-ln((e%5Eix+-+i)%2F(e%5Eix+%2B+i))

[@Kuratius]

Vincent Killion I checked, it is actually a different antiderivative! They differ by i*π. Compute ' ln((e^ix - i)/(e^ix + i)))-2*i*arctan(e^ix) for x=0' with the Wolfram|Alpha website (www.wolframalpha.com/input/?i=+ln%28%28e%5Eix+-+i%29%2F%28e%5Eix+%2B+i%29%29%29-2*i*arctan%28e%5Eix%29+for+x%3D0) or mobile app (wolframalpha:///?i=+ln%28%28e%5Eix+-+i%29%2F%28e%5Eix+%2B+i%29%29%29-2*i*arctan%28e%5Eix%29+for+x%3D0).

[@baskard8018]

i like the last one(complex version

[@ali_aljwahery]

Thaaaanx a lot 💚💚

[@green3487]

That Ood costume made me jump lol

[@alfonsogarcia3855]

Method 3 is the most intuitive one in my opinion

[@Wyrmlingbait]

It's possible to manipulate sec x to show that it's equal to (sec^2 x + tan x sec x)/(sec x + tan x). Not easy in a comment to show it though. 1/cos x =cos x / cos^2 x = cos x / (1- sin^2 x) = cos x / (1 + sin x) * 1 / (1-sinx) = 1/ (sec x + tan x) * 1 / ( 1-sin x) * (1+sin x)/(1+ sin x) = 1 / (sec x + tan x) * (1+sin x) / (1-sin^2 x) = 1 / (sec x + tan x) * (1+sin x) / cos^2 x = 1/ (sec x + tan x) * (sec^2 x + sec x tan x) = (sec^2 x + sec x tan x)/ (sec x + tan x) integrate it.

[@MathAdam]

My favourite is the one that uses tanh^(-1). This integral illustrates so many connections between various functions-- logs, trig, complex, hyperbolic... Do I get to say "#first comment in 3 years"? 😎

[@blackpenredpen]

Yes, you definitely do!

[@Mohsinkhan-xq2gy]

Great job

[@ahmadkalaoun3473]

Last one was the best 😍😍😍😍

[@jackkalver4644]

To derive the secret in #1, you can differentiate sec x to get sec x tan x, divide by the original function to get tan x, differentiate to get sec^2 x, then divide sec x tan x+sec^2 x by sec x+tan x to get sec x. If f’(x)+g’(x)=g(x)sec x+f(x)sec x, then d/dx ln[f(x)+g(x)]=sec x.

[@aashsyed1277]

4:49 you could subsitute that sin^2(x)+cos^2(x) on the top.

[@komalshekar6998]

Thank you for this amazing video...................

[@fountainovaphilosopher8112]

You must really love that fireworks.

[@blackpenredpen]

Yes

[@alanturingtesla]

Često te viđam ovde u komentarima. Odakle si?

[@fountainovaphilosopher8112]

Crazy Drummer Crne Gore, što?

[@roderictaylor]

My favorite method is to use the substitution tan x = sinh u from which it follows that sec x = cosh u and dx = 1/cosh u

[@manuelodabashian]

I have come across another way years ago but no longer have the book

[@bro_vega_1412]

The differentiation of coth^-1(x) is also 1/(1-x^2),right?I remember you have made a video of this. BTW,tanh^-1(x)=1/2*ln((1+x)/(1-x)),coth^-1(x)=1/2*ln((x+1)/(x-1)).You may have some same results. Anyway,good video!

[@PUBGgamer-kf5vz]

Great ... video

[@KeyMan137]

There's a great Wikipedia article: en.wikipedia.org/wiki/Integral_of_the_secant_function You could mention how the integral of sec(x) = tanh^(-1)(sin(x)) or sinh^(-1)(tan(x)) or cosh^(-1)(sec(x)) even though these functions are not equivalent. Another common form is ln(abs(tan(π/4+x/2))). Also, the integral of the secant function defines the inverse of the Gudermannian function (which relates the circular functions and hyperbolic functions without explicitly using complex numbers). Another interesting integral is secant cubed: en.wikipedia.org/wiki/Integral_of_secant_cubed {integral sec^3(x) dx = 1/2 sec(x) tan(x) + 1/2 ln(abs(sec(x)+tan(x))) + constant}

[@BenjaminKeilty]

Integral of secx dx = int 1/cosx dx =int cosx/(1-sin^2x)dx Let u=sinx =int 1/(1-u^2) du We know that d/dx arctanx = 1/(1+x^2), so by the chain rule, d/dx arctan(ix)=(1/(1+(ix)^2))(i)=i/(1-x^2) Back to the integral, we multiply by 1 to get: int (-i)(i)(1/(1-u^2))du =(-i)*int (i/(1-u^2))du =(-i)*arctan(isinx) Not sure if this is meaningfully different from the other answers, or even right, but once you went down the path of #2 that's what I saw.

[@avelkm]

I would naturally go for 3d method, but 4th is just more beautiful:)

[@francis6888]

Why not change sec x to 1/(sin(x+pi/2)) and integrate from there?

[@xrisku]

love the fourth!

[@blackpenredpen]

Me too!!!

[@xrisku]

blackpenredpen love your videos. they are always so helpful and you alwaus make me smile. thank you so much for your efforts. peace!

[@ahmedkhalil9917]

very helpful thanks

[@MrRyanroberson1]

8:00 so inverse tangent of ix should just be inverse hyperbolic tangent x? Because (ix)²=-x²

[@JensenPlaysMC]

Rember to divide by the derivative of ix

[@lalitverma5818]

Genius

[@Gabrichai]

Hi, I have another solution. You could use the universal change and name cosx = (1-t^2)/(1+t^2), dx=2dt/(1+t^2), t=tg(x/2). Then, after simplify, you go to a rational integral: 2dt/1-t^2. And the solution is ln(1+tg(x/2))/(1-tg(x/2))