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)
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)|).
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.
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.
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.
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.
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 :)
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
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!
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!
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.
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 :)
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
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.
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
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
ua-cam.com/video/XoYHieh_-EE/v-deo.html 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.
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
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?)
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
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 - ua-cam.com/video/dpnqhIVd-pg/v-deo.html
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.
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
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.
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
@@XanderGouws do you recommend me some book to learn that :'( I understand something about that (some book with integrals with complex numbers) English please
@@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.
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.
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!
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.
01:50 If we're allowed to memorise and use the derivative of sec(x) to solve this, why isn't it equally valid to memorise the integral if sec(x) and boom, there's your answer?
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;;
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.
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
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.
0:13 "use this on exams" Ah, when you said that, I felt like I was not supposed to be here, because I am only 13 years old. But I love mathematics, that's why I understand this calculus...
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
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!
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?
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).
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}
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"? 😎
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)
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.
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).
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.
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.
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))
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).
This just goes to show that using trig functions and memorizing their various derivatives and integrals is stupid and it would be much cleaner and easier to always express them as complex exponentials, it would also yield a single result, which can be shown to be equivalent to all 4 expressions (and to the other ones people have pointed to in the comments).
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
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?
We r getting different answers. . What does that mean?.. does that mean that graph of every one of them will have same area provided that we put same limits in all of them?
My problem with the last one is that the domain of acrtan in the complex numbers. I certainly do not know what it is, and I am guessing most people do not either
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?
I think you have a misconception about the fundamental theorem of calculus, the antiderivative is only unique up to a constant. It says nowhere that this constant cannot be imaginary, and as long as it's a constant any definite integral using this antiderivative will also be real because it cancels out. You can take the real part if you want, but it's really quite redundant to do so because it will not change the result in a meaningful way.
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)
Chrysophylax Dives wow!!!
Haha thank you sir :)!
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)|).
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.
To be honest, I have no idea. Ill try to figure it out.
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.
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.
Leonard Romano Imaginary numbers are everywhere in physics!
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!)
Ian Bornhoeft I haven't seen them anywhere except in quantum mechanics (especially in wave function)
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.
Jumpscare list:
3:25 *monster appears
19:19 *monster appears
btw nice vid
Iaago Ariel Schwoelk Lobo Sexx
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
It’s an ood from doctor who
Scary 😭
That's ben ten villan mask vilgax
I liked the partial fraction solution most. Felt more intuitive and simple. cos^2x+sin^2x=1 is an old friend by now!
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.
I quite like the last one.
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 :)
Samuel Roberts a lot, actually. I will try to do more later on. Thanks for u nice comment.
I wish UA-cam was around in 1987 when I was taking inviscid fluid flow. These videos are an excellent refresher in complex numbers.
I prefer number 2 but number 4 is interesting to know.
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.
Got a little heart attack from the mask...
Thales Freitas Macêdo It's the Ood
Deepto Chatterjee i watch dw and still everytime he puts that ood head i get jumpscared
That oodball... add last he did it ! btw, Ood are sweet
Deepto Chatterjee Just found what the Ood is... He full cosplayed it; the mask and microphone
It’s an ood
The complex result is by far the best, thanks for all.
Sec
Also the most useless
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
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
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!
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!
Use parametric formulas to express sec(x) in terms of tan(x/2), tan(x/2)=u, integrate
AHHHHHHH I FORGOT TO DO THAT!!!! >_
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.
Thanks!!! You know, I even did a vid last year on that, lol!
ua-cam.com/video/qijx9zx3HNQ/v-deo.html
The Physicist Cuber aha! the weisterass sub
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 :)
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.
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)
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
Number four is clearly the best. It's completely right, and it would have sent my teacher into conniptions!
J.J. Shank :D
Try integrating 1/sin x, it works the same way (complex definition of sin x)
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.
3:26 naniii
And @19:19 lol.
It's an Ood
It's an enemy stand?
Julián García
am
nimals ex
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
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
your videos are amazing ... of course the best method is the last one using complex expression of cosine
Thanks!
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
Ahnaf Abdullah I do!!!
Hahahaha! Ok, I'll do it soon
ua-cam.com/video/XoYHieh_-EE/v-deo.html
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.
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
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)
The complex way is the best. You started with something real and ended with a complex answer.
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?)
Loved the Ood costume haha! The microphone reminds me of them everytime
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
fingers crossed blackpenredpen notices this 😂
Yes! I know they are the same and have a proof in another video
In fact, number two is the same too
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 -
ua-cam.com/video/dpnqhIVd-pg/v-deo.html
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.
4:49 you could subsitute that sin^2(x)+cos^2(x) on the top.
Ur simile after solving a problem is great source of satisfaction😍
That Ood Mask Loool
You can also use the substitution, t = tan ½ x so that cosx =(1-t2)/(1+t2)
Excellent presentation. Wow! DrRahul Rohtak Haryana India
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
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.
I wonder if there are complex solutions to ln/logs of negative values.
ThisCouldBeYou yes
ThisCouldBeYou If there is a way to compute them with the integrals in this video, can you show me how?
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
@@XanderGouws do you recommend me some book to learn that :'( I understand something about that (some book with integrals with complex numbers) English please
@@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.
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.
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!
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.
01:50 If we're allowed to memorise and use the derivative of sec(x) to solve this, why isn't it equally valid to memorise the integral if sec(x) and boom, there's your answer?
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;;
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.
6:37 the third way is
-$1/u²+i²
(-1/i)arctan(u/i)+c
iarctan(-iu)+c
substitute u=sinx
6:36 you have actually 3 ways tanh inverese, coth inverse and the ln situation
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
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.
Method 3 is the most intuitive one in my opinion
What are you saying at 16:15?
"If we differentiate this, the function part repeat.... "?
This is the first time I ever understood the coverup method thanks
0:13 "use this on exams"
Ah, when you said that, I felt like I was not supposed to be here, because I am only 13 years old. But I love mathematics, that's why I understand this calculus...
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
8:00 so inverse tangent of ix should just be inverse hyperbolic tangent x? Because (ix)²=-x²
Rember to divide by the derivative of ix
You must really love that fireworks.
Yes
Često te viđam ovde u komentarima. Odakle si?
Crazy Drummer
Crne Gore, što?
Recently found out about this channel. Please keep making video!~
我會的, 謝謝～
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!
from where did you get that mic..thats super cool just like your teaching....love from india...oh dude you are really awesome
I always like this dude!!!
Very useful and nice video.
Very nice presentation! #4 blew me away - never would have thought of it.
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?
david plotnik Do you mean solving for the value of the complex integral? What do you mean?
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?
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).
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}
Hahaha I got an ad as soon as the jumpscare came
lol!
btw, some students of mine didn't like the jumpscare too much bc they were too focused
Btw Im eagerly awaiting for you to upload... are u taking a break? r u ok?
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"? 😎
Yes, you definitely do!
This is a very sec(x)y video
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)
_"The integral sec(y) dy from zero to one-sixth of pi is log to base e of the square root of three times the sixty-fourth power of WHAAAT?!"_
Dear bprp, lenyűgöző volt !
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.
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).
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.
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.
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))
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).
Has this guy made video Abt differentiation of inverse functions? If yes then pls send the link..Btw great video man.. just subscribed yesterday
Saksham Garg check out my site. Www.blackpenredpen.com and resource calculus.
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
This just goes to show that using trig functions and memorizing their various derivatives and integrals is stupid and it would be much cleaner and easier to always express them as complex exponentials, it would also yield a single result, which can be shown to be equivalent to all 4 expressions (and to the other ones people have pointed to in the comments).
that was so cool. all those answers are great. great job and great brain sir
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
This is of mind blowing proportions.
Black pen red pen yeah!
I have an unrelated question: How would you calculate evenly-spaced points on a parabola?
Are all those methods available to cossecant, hyperbolic secant and hyperbolic cossecant?
Can we use integration by parts
you can also write t=tan(x/2)
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?
You can have an expression, ln | tan (x/2 + pi/4) | + C, by letting u=tan(x/2)
did you use weierstrass sub?
Maybe try Weierstrass sub?
You will get 2*arctanh(tan(x/2))+C or ln(abs((1+tan(x/2)/(1-tan(x/2))))+C if you try it. Fifth solution!!! CRAZY!!!
That Ood costume made me jump lol
We r getting different answers. . What does that mean?.. does that mean that graph of every one of them will have same area provided that we put same limits in all of them?
Yes
I have come across another way years ago but no longer have the book
Is that an OOD from Doctor Who
It is possible to come up with an answer if the substitution instead becomes, e^ix=cosx+isinx?
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?
If I may ask, what is hyperbolic tangent?
Just a little question here in one why do you even multiply by a variable can you even do that?
My problem with the last one is that the domain of acrtan in the complex numbers. I certainly do not know what it is, and I am guessing most people do not either
Weierstrass substitution makes it a trivially easy integral
man i just don't feel like loving having i in the solution
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?
How do you convert the complex definition into the standard u sub definition?
you are turning into a monster - you better eat a snickers.....
I love all of those!!!
But a bit less complex!
I'm from India but this lecture is useful for me
Good.
Please show us how to convert the last method result from complex to real value because the integeral is real
I think you have a misconception about the fundamental theorem of calculus, the antiderivative is only unique up to a constant. It says nowhere that this constant cannot be imaginary, and as long as it's a constant any definite integral using this antiderivative will also be real because it cancels out.
You can take the real part if you want, but it's really quite redundant to do so because it will not change the result in a meaningful way.
@@Kuratius yes I meant it is interesting to show that -2i×arctan(e^ix) + C= ln|secx+tanx|
I did not say that C can not be complex