The technique he used is called Weierstrass Substitution. It can be used to integrate any rational function of trig bois. It is by far my favorite technique because it seems so counterintuitive but it makes integrals simplify to regular rational functions (and it is really good algebra practice).
Nah, it's really pretty intuitive when you think about it. I tried doing the normal tan x for substitution, and the reason that doesn't work nice is because it doesn't use the double angle identities from solving for cos(2*x/2) and sin(2*x/2). If its just normal tanx, then the denoms are still in sqrts. So bro thought it'd be better if he divided by 2. Quite a reasonable man, that weierstrass.
This feels like winning the game on the 40th turn rather than on the 12th when the opportunity came. I wasnt as aware of this aspect of math and general problem solving before your comment. Thx bud!
Great vid! Love your enthusiasm. The geometric meaning of the Weierstrass substitution is a change of coordinates from Cartesian to the so-called "rational parameterization" of the circle, from which (0,2π) → (0,∞): x ↦ t. The symmetry in the integrand that you have noticed, namely the phase shift x → π/2 - x is not an accident; it represents a rotation of the circle by 45 degrees. Since the rational parameterization of the circle is a homeomorphism from (0,∞) to S^1\{1}, this induces an isormophism from the field extension ℝ[cos(x),sin(x)] to the field ℝ[t], and hence also an isomorphism between the respective fractional field extensions. This is why the Weierstrass substitution can let you compute the integral of any rational function involving sin(x) and cos(x).
You can avoid the partial fractions and messy logarithms after completing the square if you know that 1/(1-u^2) integrates to arctanh(u). First factor a 1/2 out of the integral and write the denominator as 1 - [ (t-1)/sqrt(2) ]^2. Make the substitution u = (t-1)/sqrt(2) and integrate to get a simple answer of (1/sqrt(2)) arctanh (1/sqrt(2)), which is looks slightly sexier in my opinion. (The identity arctanh(x) = (1/2)ln[(1+x)/(1-x)] will return it to the form of your answer) :)
He said that he wanted to avoid "in the air" approaches like this. I know that it's a well known technique but I guess it just depends on what he really wanted to get out of the vid
The intuitive understanding you have of each alteration of your problem is super encouraging. Right now I can't intuitively follow every computation I do but this gives me hope that, even on this higher level, it's possible to actually do it.
I found an easier way to do this. Let S be the spicy integral in question. Note that by symmetry the value of the integral S does not change if we have cos^2 in the numerator instead of sin^2. Hence 2S = S + S = int (sin^2+cos^2)/(sinx+cosx) dx = int 1/(sinx+cosx) dx, and of course from 0 to pi/2. If we note that sinx + cosx = sqrt(2)*sin(x+pi/4), (which is a great trig integration trick to know) then that resulting integral is not difficult to evaluate. It is basically just the integral of csc(x). So evaluate it, and then divide by 2 to get the value of S. This approach is much simpler and agrees with value found by Weierstrass substitution.
You can write sin(x) + cos(x) as sqrt(2) * sin(x + pi/4). So 2 * sqrt(2) * I becomes csc(x + pi/4) from 0 to pi/2 which is simply ln|(csc(pi/4) + cot(pi/4))/(csc(3pi/4) + cot(pi/4)). Thus, we get 2 sqrt(2) I = ln|(sqrt(2) + 1)/(sqrt(2) - 1)| = 2 ln|sqrt(2) + 1| which gives us I = ln(sqrt(2) + 1)/sqrt(2).
Great video. I tried to answer the integral on my own and found an alternative method. When arriving at: 2I=int( 1/(sin(x)+cos(x)) ) from 0 to pi/2 You may just plug-in the identity sin(x)+cos(x)=sqrt(2)*cos(x-pi/4) (which can be derived by just analyzing the period and amplitude of sin(x)+cos(x). The problem simplifies to the integral of sec(x-pi/4) which is a lot easier to integrate and the same answer follows.
In India we are taught the method of taking log and after log applying a property. But you are really daddy of all the integrals.This one was quite typical.
integral at 4:50 can be easily computed with compound angle formula 1/(sqrt(2)cos(x-pi/4)), which then saves about 15 mins of computation. That being said, the partial fraction is sooo satisfying.
Incredibly looking video. Great integral right here! Very inspiring. Next time I’m gonna suggest you some chilly bois I thought about these days. Not only quality content, but also entertainment on a spicy subject. Good luck with the channel man, you deserve it.
Instead of using half angle formula for converting 1/(sinx + cosx ) dx in tan(x/2) .... We can directly multiply and divide by √2 (constant) in (sinx+cosx) !! sin pi/4=cos pi/4=1/√2 ...pretty simple it's forming sin(x+pi/4) and √2 comes out of integral 😅 now we can easily integrate ***(this is the shortcut)*** Agree : hit like 😁
Fun work but using simple trigonometric identities, we can reduce this to just the integral of sec(x)/sqrt(2) from 0 to pi/4 (1 + sin^2(x) - cos^2(x)) / (2(sin(x) + cos(x))) = 1/(2(sin(x)+cos(x))) + (sin(x)-cos(x))/2 Integral of (sin(x)-cos(x)) is zero from 0 to pi/2, 1/(sin(x)+cos(x)) becomes cos(x-pi/4) sqrt(2) From that we just get the answer to be ln((sec(pi/4)+tan(pi/4))/(sec(0)+tan(0)))/sqrt(2) = ln(1+sqrt(2))/sqrt(2)
@ 7:10 Much easier to draw a right triangle with angle (x/2) and sides opposite and adjacent being t and 1 respectively, then the hypotenuse is sqrt(1+t^2) by pythagorus. Then read off sin(x/2) and cos(x/2) as oppo/hypot and adj/hypot from the triangle. sin(x) = 2 sin(x/2)cos(x/2) = 2 * t/sqrt(1+t^2) * 1/(sqrt(1+t^2)) = 2t/(1+t^2). Similar idea for cos(x) = cos(x/2)^2 - sin(x/2)^2 = (1/sqrt(1+t^2))^2 - (t/sqrt(1+t^2))^2 = (1-t^2)/(1+t^2)
I would've used harmonic addition, in that \sin x + \cos x = \sqrt{2} \cos \left( \frac{\pi}{4} - x ight) that gave a secant integral which is quite easy to evaluate as \frac{\sqrt{2}}{2} \ln \left( 1 + \sqrt{2} ight) edit: I used the expression at 4:32 but everything after that's a bit overkill, you really only need one page after that cool tho
As opposed to using the t substitution, you can use cos(x) + sin(x) = Rcos(x-a). It works out much faster if you know the integral of sec x. Nice one nevertheless
21:50 I think after that a more "visually pleasing" step would be to use the property of logarithms to take out the square to the outside, canceling out with the 2 on the denominator, leaving you with [1/sqrt(2)] * ln[sqrt(2)+1] But my god, what a journey. Great video
instead of the substitution of tan^2(x/2) , u can divide the denominator by √2 which will make the denominator as cos(x-π/4) and integral of sec(x-π/4) ...we all know!
At like 5:00 you can use harmonic addition theorem/Rcos(x+a) method and instantly something integratable. It jumps you straight to the part at 22:00 since you get 1/sqrt(2) *ln|sec(x-pi/4)+tan(x-pi/4)| from 0 to pi/2.
Your video is great! I want to share with you an easier approach: When you had to solve for the integral of 1/(sinx+cosx), we could makr the transformation sinx+cosx=√2 cos(x-π/4). Then the integral becomes the integral of sec(x-π/4)/√2. That is in fact ln|sec(x-π/4)+tan(x-π/4)|/√2 When you plug that result we get the exact same answer as you got in your video. I look forward to your answer.
Or as ArcCoth(Sqrt(2)). Or at 22:15 cancel the prefactor of 1/2 with the exponent of 2 in the logarithm. But the hyperbolic inverse functions are even sexier.
The nice thing about the hyperbolic inverse tangent is that the argument is the same as the prefactor, that's why I mentioned that one. But as you say, there are many possiblities.
Great video! An other way to solve it was to let u=x/4 so that your lower and uper bounds to be from 0 to 2π and from there you could use complex analysis, all and all, a very good job!
19:50 I would've used the -dv to change the bounds of the left integral and then since both integrals would have the same function (1/v and 1/w) and the same bound (square root of 2) I would combine them to an integral from sqrt(2)-1 to sqrt(2)+1 of 1/x dx 22:00 here I would've just brought the 2nd power as a factor outside of the logarithm and cancel it with the fraction But if I'm being truly honest, I would've substituted again the t-1 at 15:00 and made an easier PFD
There is another shorter method too use a+b-x Then add both u will end up with 1/cos x + sin x (identical to what u did) write cos x as sin 90-x now we can write sin (x) + sin (90-x)= 2 sin 45 sin (x-45) Now use u substitution on x-45 U will end up with integral of cosec x (u should know that) And ur done, way shorter method
Instead of solving it by the partial fractions method at 15:12 you could've just used the formula: ∫1/(a²−x²) dx = (1/2a) log(|(a+x)/(a−x)|) Where a → 2^(½) and x → (t - 1) And finally substitute the value of t at the end. Great video though!
@@gamma_dablam I guess there are two ways to prove it, and I know I'm late. But this formula is also just the formula for integrating arctanh(x). Memorizing the derivatives of the inverse trig/hyp functions is so useful for these cases.
@@gamma_dablam of course! I always derive my formulas I hardly memorize anything. You just memorize them naturally when you use them so much haha. The logarithmic expressions of the inverse hyperbolic functions are a pain to derive though. Also it's amazing that papa sees the comments on old videos, you're awesome keep it up! Edit: I remember when I was first introduced to the inverse hyperbolic functions and I was told I'd have to memorize the derivatives and I didn't understand why, since they take like 20s to derive. But I learned the hard way that I need to memorize them for integration, for obvious reasons lmao, you're not going to guess which of the 12 it is you're integrating. Just a calc 2 student by the way don't judge me if I'm being dumb xd
There’s an easier way to use the t=tan1/2x substitution in this situation where the x is between 0 and pi/2 you could just write tanx in terms of t using the double angle formula and draw the triangle that has angle x and get the hypotenuse using Pythagoras and then you can get the sin and cos from the triangle
I think a better way to think about and explain Weierstrass sub is with x = 2arctan( t ) rather than t = tan( x/2 ) because of the geometric ways of explaining nested trig( inverse trig ) functions makes far more sense than extending trig functions to be in terms of tan( x/2 )
What an interesting way to find sin(x) and cos(x) in terms of t! I would have used a right triangle with angle x/2 but your way was unexpected and came together very nicely
When you get to 1/(sinx+cosx) multiply and divide by sqrt2 and you get 1/sqrt2 .1/(sqrt2/2sinx+sqrt2/2cosx) =1/sqrt2 1/sin(x+pi/4),and 1/sin(x+pi/4)=sin(x+pi/4)/sin^2 (x+pi/4)=sin(x+pi/4)/(1-cos^2(x+pi/4)) and you do cos(x+pi/4)=u du=-sin(x+pi/4)dx and then it becomes way easier
cant you just use the other ln - property in the very end and bring the ^2 inside the ln to the outside as a factor (which would cancel out with the 2 in the denominator of the ln's pre-factor) which would yield ln(sqrt(2)+1)/sqrt(2) in the end?😅😅
I'm finding it personally more convenient to use hyperbolic substitution for the difference of squares and then just deriving the inverse hyperbolics on the spot lol. I like partial fractions but sometimes it's just too messy for me. And there should be a class on manipulating integrals in terms of itself.
I thought this looked really familiar, and then I realised it's because they teach the Weierstrass substitution identities during high school in Australia! We just called them the t-results and only used them to prove trig equations though.
I still remember the time when my friend and I went straight to another trigonometric substitution as we arrived the step at 15:05. Clumsy algebra resulted.
You can write a * sin x + b * cos x in terms of just sin or cos: I'll write sqrt(a²+b²) as c just to make the following look simpler: Define d such that cos d = a/c. Since (a/c)²+ (sin d)² = 1 a²+c²(sin d)²=a²+b² sin d =± b/c Choose a d such that sin d = b/c Now multiply the original expression by c/c c( (a/c)(sin x) + (b/c)(cos x)) c( (cos d)(sin x) + (sin d)(cos x)) c(sin(d+x)) You can write this in terms of cos by defining cos d = b/c and -sin d = a/c or by writing c(sin(d+x)) as c(cos(pi/2-(d+x)))= c(cos(d+x-pi/2)) This is usually useful because the solution to a second order linear differential equation usually contains a * sin x + b * cos x where a and b are arbitrary constants and you can write that as c(sin(d+x)) where c and d are arbitrary constants, but in this integration after 1/(sin x +cos x) you could write that as 1/(sqrt(2)sin(x+pi/4)) which is just the integral of the cosecant.
at 4:57, the denominator was simply \sqrt{2}\times \cos(\pi/4-x) and with a change of variable to u=\pi/4-x you would get integral of \sec(x) over -pi/4 to +\pi/4 and that was it ! (= \ln(\sec(x)+\tan(x)))
That seemed like a long way around. I paused and had a go before looking at your solution. In this case just use sin(x) + cos(x) = sqrt(2) sin (x+pi/4) then x = t+pi/4 to get an integral of (1/cos(t) + 2 sin(t) ) over [-pi/4,pi/4]. The second term vanishes and then u = sin(t) --> integral (1/(1-u^2) ), then partial fractions give syou your logs.
At 13:12 you could "complete the square" on the denominator to get 2-(t-1)^2. Then use a substitution sqrt(2)v=(t-1) and it turns into a standard integral that you can do by recognition/inspection. Great video though, and your full method doesn't require a backlog of memorised standard integrals! Edit was to correct the substitution :)
Rather than adding the two I's together to get 2I you could have subtracted them. We'd get 0 equal to the integral from 0 to pi/2 of (cos^2 - sin^2)/(cosx + sinx) and you would've had the most epic proof that the integral from 0 to pi/2 of cosx - sinx is equal to 0
I solved this using a different method. I started by considering this integral to be A, and the integral of cos^2x/(sinx+cosx) from 0 to pi/2 to be B. I then added and subtracted them, found the values and converted into a linear system of equations. I found the integral of this AND cos^2x/(sinx+cosx) using this
in this question at 1/cos(x)+sin(x) this expression we can divide and multiply by one by root two then we will left with sin(pi/4 +x) in the denominator then we can easily integrate cosec(pi/4 +x).
At 4:48 i would replace sin(x) +cos(x) by sqrt(2)*cos(x-pi/4) So 1/sin(x)+cos(x) is equal to 1/sqrt(2)*cos(x-pi/4) =(1/sqrt(2))*sec(x-pi/4) and i do integration...
What about writing the denominator as R*sin(x-a), for some R and a. Then make the substitution u=x-a, and you get something reasonable on the denominator. Then use v=-cos(u) and you're left with something reasonable to solve.
I used sinx+cosx=sqrt2*sin(pi/4+x), then substitution pi/4+x=y......., finnaly you will get 1+2/(4sqrt2) * integral cos^2 y/ sin y, after the substitution cos y= t you will get the same result.This method is more universal, for others intervals.
After you got the original difference of two squares, a u=1-t substitution would have made it much nicer. After working it out, it would return the same integral, only it would replace the t-1 with just a t.
There is two "formulas" for the tangent derivative 1+tan^2 and sec^2, and it is so much imporant for integrals that everyone know it so why don't use the right formula at the right moment?? When t=tan(x/2), it is clear that dt=(1+tan^2(x/2))dx=(1+t^2)dx ! Nice integral by the way! Also I want to share an other way without the half angle sub: you can transform cosx+sinx in √2cos(x-π/4) and use or redemonstrate the int of sec!
Just apply King's property to find 2I=int(1/(sin(x)+cos(x)))dx = sqrt(2)/2 *int(csc(x+pi/4)dx And we know that integral of csc(pi/4+x)=ln| tan(x/2+pi/8)|
I have a few questions on what went on at 17:05 when says that the right side of the equation must have degree 0. For this reason he focuses on the t terms, but why them and nothing else?
![A pretty awesome integral! Exploiting some symmetry bois [ Papa Flammy's V2 ]](http://i.ytimg.com/vi/RZr7RytBQHs/mqdefault.jpg)
![A pretty awesome integral! Exploiting some symmetry bois [ Papa Flammy's V2 ]](/img/tr.png)
don't do drugs kids
Mathematicians: Don't drink and derive
Please don't do that get some help
The technique he used is called Weierstrass Substitution. It can be used to integrate any rational function of trig bois. It is by far my favorite technique because it seems so counterintuitive but it makes integrals simplify to regular rational functions (and it is really good algebra practice).
BPRP calls it "Weierstraps" ^^
Now it's my favourite tooo
i didnt know it was called that way, i just know it like u=tan(x/2) lol
Nah, it's really pretty intuitive when you think about it. I tried doing the normal tan x for substitution, and the reason that doesn't work nice is because it doesn't use the double angle identities from solving for cos(2*x/2) and sin(2*x/2). If its just normal tanx, then the denoms are still in sqrts. So bro thought it'd be better if he divided by 2. Quite a reasonable man, that weierstrass.
22:05 NoooooOOOOoo log property!! Log property! Use a property of a logarithm to get rid of the power!
Yeah I think that would make the final result prettier. But I guess it's just a matter of taste.
Wow I never noticed that haha true story
I was screaming that at the monitor
This feels like winning the game on the 40th turn rather than on the 12th when the opportunity came. I wasnt as aware of this aspect of math and general problem solving before your comment. Thx bud!
something else that would've been nice was take the 2 out of the ln so that 2 and 1/2√2 would cancel out to 1/√2
my exact thoughts
Thought of the saym thing
Exactly the same thought.
That’s what I thought as well!
Agreed
in your last step i probably would just have pulled the square out of the log to cancel the two in the denominator.
log(1+sqrt(2))/sqrt(2)
Great vid! Love your enthusiasm.
The geometric meaning of the Weierstrass substitution is a change of coordinates from Cartesian to the so-called "rational parameterization" of the circle, from which (0,2π) → (0,∞): x ↦ t. The symmetry in the integrand that you have noticed, namely the phase shift x → π/2 - x is not an accident; it represents a rotation of the circle by 45 degrees. Since the rational parameterization of the circle is a homeomorphism from (0,∞) to S^1\{1}, this induces an isormophism from the field extension ℝ[cos(x),sin(x)] to the field ℝ[t], and hence also an isomorphism between the respective fractional field extensions. This is why the Weierstrass substitution can let you compute the integral of any rational function involving sin(x) and cos(x).
You can avoid the partial fractions and messy logarithms after completing the square if you know that 1/(1-u^2) integrates to arctanh(u). First factor a 1/2 out of the integral and write the denominator as 1 - [ (t-1)/sqrt(2) ]^2. Make the substitution u = (t-1)/sqrt(2) and integrate to get a simple answer of (1/sqrt(2)) arctanh (1/sqrt(2)), which is looks slightly sexier in my opinion. (The identity arctanh(x) = (1/2)ln[(1+x)/(1-x)] will return it to the form of your answer) :)
I would have used integral of sec(x) wrt x= ln|sec x +tan x| and used that sin x + cos x = sqrt 2 * cos (x-pi/4)
Well, I tried doing that to avoid using sub-t but got stuck in the integral of sec(x-pi/4) from 0 to pi/2. Any idea how to integrate that?
I got it right now, I was using a wrong value for integral of sec(x) - my bad - using ln(secx+tanx) gives the right result spot on.
He said that he wanted to avoid "in the air" approaches like this. I know that it's a well known technique but I guess it just depends on what he really wanted to get out of the vid
15:52 i spot a wild euler!
“It might not look really spicy at first sight”
The intuitive understanding you have of each alteration of your problem is super encouraging. Right now I can't intuitively follow every computation I do but this gives me hope that, even on this higher level, it's possible to actually do it.
I found an easier way to do this. Let S be the spicy integral in question. Note that by symmetry the value of the integral S does not change if we have cos^2 in the numerator instead of sin^2. Hence 2S = S + S = int (sin^2+cos^2)/(sinx+cosx) dx = int 1/(sinx+cosx) dx, and of course from 0 to pi/2. If we note that sinx + cosx = sqrt(2)*sin(x+pi/4), (which is a great trig integration trick to know) then that resulting integral is not difficult to evaluate. It is basically just the integral of csc(x). So evaluate it, and then divide by 2 to get the value of S.
This approach is much simpler and agrees with value found by Weierstrass substitution.
You can write sin(x) + cos(x) as sqrt(2) * sin(x + pi/4).
So 2 * sqrt(2) * I becomes csc(x + pi/4) from 0 to pi/2 which is simply ln|(csc(pi/4) + cot(pi/4))/(csc(3pi/4) + cot(pi/4)).
Thus, we get 2 sqrt(2) I = ln|(sqrt(2) + 1)/(sqrt(2) - 1)| = 2 ln|sqrt(2) + 1| which gives us I = ln(sqrt(2) + 1)/sqrt(2).
Great video. I tried to answer the integral on my own and found an alternative method.
When arriving at: 2I=int( 1/(sin(x)+cos(x)) ) from 0 to pi/2
You may just plug-in the identity sin(x)+cos(x)=sqrt(2)*cos(x-pi/4) (which can be derived by just analyzing the period and amplitude of sin(x)+cos(x).
The problem simplifies to the integral of sec(x-pi/4) which is a lot easier to integrate and the same answer follows.
did the same thing :P
also did the same thing
In the end, why didnt you put square infront of natural log? 2's would cancel out, would be prettier result i think :D
Jon Pustavrh Mičović jaaaaws, i noticed the same😂
@@hwendt idem 🙃
Bruh when he forgot to do that I entered depression
CONGRATULATIONS!!!!!!!!!!!!
PAPA FLAMMY, HERE'S TO 15K MORE!!!!
Watching others work and solve problems without needing supervision.
papa flemings reads all the comments
In India we are taught the method of taking log and after log applying a property. But you are really daddy of all the integrals.This one was quite typical.
Leonhard *Boi* ler
boiler?
Koray Acar Euler is pronounced “Oi ler”. Replace “Oi” with “Boi” XD
integral at 4:50 can be easily computed with compound angle formula 1/(sqrt(2)cos(x-pi/4)), which then saves about 15 mins of computation. That being said, the partial fraction is sooo satisfying.
At the end you could get the power 2 in front of the log. Answer would be 1/sqrt(2) × ln(1+sqrt(2))
Thanks! My first heart here :)
can I get a triple? :P
This was sooo good I watched the partial fraction decomposition twice!!
I did it this way,
cosx+sinx=sqrt(2)cos(x-pi/4)
Then end up with integral of sec(x-pi/4)
You my friend take integrations quite seriously!!
Love the channel by the way
Incredibly looking video. Great integral right here! Very inspiring. Next time I’m gonna suggest you some chilly bois I thought about these days. Not only quality content, but also entertainment on a spicy subject. Good luck with the channel man, you deserve it.
You deserve a 100-fold of your current subscribers
Instead of using half angle formula for converting 1/(sinx + cosx ) dx in tan(x/2) .... We can directly multiply and divide by √2 (constant) in (sinx+cosx) !! sin pi/4=cos pi/4=1/√2 ...pretty simple it's forming sin(x+pi/4) and √2 comes out of integral 😅 now we can easily integrate ***(this is the shortcut)***
Agree : hit like 😁
this channel deserves more subscribers
22:00 You could also move the 2 from the square to the front (log properties) and then it cancels with the 2 in the denominator from 1/(2*sqrt(2))
Fun work but using simple trigonometric identities, we can reduce this to just the integral of sec(x)/sqrt(2) from 0 to pi/4
(1 + sin^2(x) - cos^2(x)) / (2(sin(x) + cos(x))) = 1/(2(sin(x)+cos(x))) + (sin(x)-cos(x))/2
Integral of (sin(x)-cos(x)) is zero from 0 to pi/2, 1/(sin(x)+cos(x)) becomes cos(x-pi/4) sqrt(2)
From that we just get the answer to be ln((sec(pi/4)+tan(pi/4))/(sec(0)+tan(0)))/sqrt(2) = ln(1+sqrt(2))/sqrt(2)
Your chalkbord is always so clean, boi! 😍
An easier way to express cos^2(x/2) is 1/sec^2(x/2) or 1/(1+tan^2(x/2))
@ 7:10
Much easier to draw a right triangle with angle (x/2) and sides opposite and adjacent being t and 1 respectively, then the hypotenuse is sqrt(1+t^2) by pythagorus. Then read off sin(x/2) and cos(x/2) as oppo/hypot and adj/hypot from the triangle.
sin(x) = 2 sin(x/2)cos(x/2) = 2 * t/sqrt(1+t^2) * 1/(sqrt(1+t^2)) = 2t/(1+t^2).
Similar idea for cos(x) = cos(x/2)^2 - sin(x/2)^2 = (1/sqrt(1+t^2))^2 - (t/sqrt(1+t^2))^2 = (1-t^2)/(1+t^2)
I would've used harmonic addition, in that \sin x + \cos x = \sqrt{2} \cos \left( \frac{\pi}{4} - x
ight)
that gave a secant integral which is quite easy to evaluate as \frac{\sqrt{2}}{2} \ln \left( 1 + \sqrt{2}
ight)
edit: I used the expression at 4:32 but everything after that's a bit overkill, you really only need one page after that
cool tho
As opposed to using the t substitution, you can use cos(x) + sin(x) = Rcos(x-a). It works out much faster if you know the integral of sec x. Nice one nevertheless
Would a complex substitution work if I then integrated a quarter of the way around the unit circle?
A nice trick for this one is to rewrite sin(x)+cos(x) as sqrt(2)sin(x+pi/4) making the integral relatively easy, but learning Weierstrass was fun too.
You are going so far and so fast Herr Werhner von Braun would not be able to keep up!
hello cant we just combine sin + cos = SIN(X+pi/4) ....... like what grow up bois
4/root5(ln(2+root5/2-root5))
Applied king and then tanx=t substitution and completing the square followed by partial fractions :)
21:50 I think after that a more "visually pleasing" step would be to use the property of logarithms to take out the square to the outside, canceling out with the 2 on the denominator, leaving you with [1/sqrt(2)] * ln[sqrt(2)+1]
But my god, what a journey. Great video
I just realized that I am 1 year too late, oh well. Guess I've been on this rabbit hole for way to long
instead of the substitution of tan^2(x/2) , u can divide the denominator by √2 which will make the denominator as cos(x-π/4) and integral of sec(x-π/4) ...we all know!
multiply divide by root 2 and get sin(x+(pi/4))in the denominator, its cosec can then be directly intrgrated , log(cosec-cot)
At like 5:00 you can use harmonic addition theorem/Rcos(x+a) method and instantly something integratable. It jumps you straight to the part at 22:00 since you get
1/sqrt(2) *ln|sec(x-pi/4)+tan(x-pi/4)| from 0 to pi/2.
Found in my recommendations...
Not disappointed
Your video is great! I want to share with you an easier approach:
When you had to solve for the integral of 1/(sinx+cosx), we could makr the transformation sinx+cosx=√2 cos(x-π/4). Then the integral becomes the integral of sec(x-π/4)/√2. That is in fact ln|sec(x-π/4)+tan(x-π/4)|/√2
When you plug that result we get the exact same answer as you got in your video.
I look forward to your answer.
For a more enjoyable answer, you can write (1/2) Log(3+2 sqrt(2)) = ArcTanh(1/sqrt(2)).
Or as ArcCoth(Sqrt(2)). Or at 22:15 cancel the prefactor of 1/2 with the exponent of 2 in the logarithm. But the hyperbolic inverse functions are even sexier.
The nice thing about the hyperbolic inverse tangent is that the argument is the same as the prefactor, that's why I mentioned that one. But as you say, there are many possiblities.
This was integral bukake
Great video! An other way to solve it was to let u=x/4 so that your lower and uper bounds to be from 0 to 2π and from there you could use complex analysis, all and all, a very good job!
ousome
19:50 I would've used the -dv to change the bounds of the left integral and then since both integrals would have the same function (1/v and 1/w) and the same bound (square root of 2) I would combine them to an integral from sqrt(2)-1 to sqrt(2)+1 of 1/x dx
22:00 here I would've just brought the 2nd power as a factor outside of the logarithm and cancel it with the fraction
But if I'm being truly honest, I would've substituted again the t-1 at 15:00 and made an easier PFD
i love writing results like this in the form of artanh like this one is just 1/sqrt(2)*artanh(1/sqrt(2)) its just chefs kiss
There is another shorter method too
use a+b-x
Then add both u will end up with 1/cos x + sin x (identical to what u did)
write cos x as sin 90-x
now we can write sin (x) + sin (90-x)= 2 sin 45 sin (x-45)
Now use u substitution on x-45
U will end up with integral of cosec x (u should know that)
And ur done, way shorter method
In the particular integral here, one can write the bottom as 1/sqrt(2) times sin(x+pi/4), so the integral is 1/sqrt(2) times csc(x+pi/4) dx
Oh shit sqrt(2) times sin(x+pi/4)
Instead of solving it by the partial fractions method at 15:12 you could've just used the formula:
∫1/(a²−x²) dx =
(1/2a) log(|(a+x)/(a−x)|)
Where a → 2^(½) and x → (t - 1)
And finally substitute the value of t at the end.
Great video though!
That formula is derived from the partial fractions method
@@gamma_dablam I guess there are two ways to prove it, and I know I'm late. But this formula is also just the formula for integrating arctanh(x). Memorizing the derivatives of the inverse trig/hyp functions is so useful for these cases.
@@nourgaser6838 knowing how to get them is essential as well as knowing them
@@gamma_dablam of course! I always derive my formulas I hardly memorize anything. You just memorize them naturally when you use them so much haha. The logarithmic expressions of the inverse hyperbolic functions are a pain to derive though.
Also it's amazing that papa sees the comments on old videos, you're awesome keep it up!
Edit: I remember when I was first introduced to the inverse hyperbolic functions and I was told I'd have to memorize the derivatives and I didn't understand why, since they take like 20s to derive. But I learned the hard way that I need to memorize them for integration, for obvious reasons lmao, you're not going to guess which of the 12 it is you're integrating. Just a calc 2 student by the way don't judge me if I'm being dumb xd
@@nourgaser6838 log forms of hyperbolic inverses are definitely a pain to get. Arcosh and arsech especially.
There’s an easier way to use the t=tan1/2x substitution in this situation where the x is between 0 and pi/2 you could just write tanx in terms of t using the double angle formula and draw the triangle that has angle x and get the hypotenuse using Pythagoras and then you can get the sin and cos from the triangle
I think a better way to think about and explain Weierstrass sub is with x = 2arctan( t ) rather than t = tan( x/2 ) because of the geometric ways of explaining nested trig( inverse trig ) functions makes far more sense than extending trig functions to be in terms of tan( x/2 )
this dude's energy is in another level
What an interesting way to find sin(x) and cos(x) in terms of t! I would have used a right triangle with angle x/2 but your way was unexpected and came together very nicely
Excellent video Pappa Flammy. I can't wait to see what you have next!
3:27 JO MAMA is a variable.............heheheheheh
When you get to 1/(sinx+cosx) multiply and divide by sqrt2 and you get 1/sqrt2 .1/(sqrt2/2sinx+sqrt2/2cosx) =1/sqrt2 1/sin(x+pi/4),and 1/sin(x+pi/4)=sin(x+pi/4)/sin^2 (x+pi/4)=sin(x+pi/4)/(1-cos^2(x+pi/4)) and you do cos(x+pi/4)=u du=-sin(x+pi/4)dx and then it becomes way easier
cant you just use the other ln - property in the very end and bring the ^2 inside the ln to the outside as a factor (which would cancel out with the 2 in the denominator of the ln's pre-factor) which would yield ln(sqrt(2)+1)/sqrt(2) in the end?😅😅
That’s what I was going to say.
Came to the comments just to say that
You had one job, Flammy.
I'm finding it personally more convenient to use hyperbolic substitution for the difference of squares and then just deriving the inverse hyperbolics on the spot lol. I like partial fractions but sometimes it's just too messy for me. And there should be a class on manipulating integrals in terms of itself.
I thought this looked really familiar, and then I realised it's because they teach the Weierstrass substitution identities during high school in Australia! We just called them the t-results and only used them to prove trig equations though.
That's some spicy trigonometry.
You could have pulled the squared factor out of the ln instead of squaring it within the ln though.
I still remember the time when my friend and I went straight to another trigonometric substitution as we arrived the step at 15:05. Clumsy algebra resulted.
You can write a * sin x + b * cos x in terms of just sin or cos:
I'll write sqrt(a²+b²) as c just to make the following look simpler:
Define d such that cos d = a/c.
Since (a/c)²+ (sin d)² = 1
a²+c²(sin d)²=a²+b²
sin d =± b/c
Choose a d such that sin d = b/c
Now multiply the original expression by c/c
c( (a/c)(sin x) + (b/c)(cos x))
c( (cos d)(sin x) + (sin d)(cos x))
c(sin(d+x))
You can write this in terms of cos by defining cos d = b/c and -sin d = a/c or by writing
c(sin(d+x)) as c(cos(pi/2-(d+x)))= c(cos(d+x-pi/2))
This is usually useful because the solution to a second order linear differential equation usually contains a * sin x + b * cos x where a and b are arbitrary constants and you can write that as c(sin(d+x)) where c and d are arbitrary constants,
but in this integration after 1/(sin x +cos x) you could write that as 1/(sqrt(2)sin(x+pi/4)) which is just the integral of the cosecant.
at 4:57, the denominator was simply \sqrt{2}\times \cos(\pi/4-x) and with a change of variable to u=\pi/4-x you would get integral of \sec(x) over -pi/4 to +\pi/4 and that was it ! (= \ln(\sec(x)+\tan(x)))
The answer IS 1/√2argth(-1/√2)
I understand it is for demonstration but I think the easiest approach would be to realize sin(x)+cos(x) = sqrt(2)sin(x+pi/4) and then from there.
That seemed like a long way around. I paused and had a go before looking at your solution. In this case just use sin(x) + cos(x) = sqrt(2) sin (x+pi/4) then x = t+pi/4 to get an integral of (1/cos(t) + 2 sin(t) ) over [-pi/4,pi/4]. The second term vanishes and then u = sin(t) --> integral (1/(1-u^2) ), then partial fractions give syou your logs.
At 13:12 you could "complete the square" on the denominator to get 2-(t-1)^2. Then use a substitution sqrt(2)v=(t-1) and it turns into a standard integral that you can do by recognition/inspection. Great video though, and your full method doesn't require a backlog of memorised standard integrals!
Edit was to correct the substitution :)
An arguably easier way is to write the integral 1/(sinu+cosu) as 1/sqrt2*sin(x+pi/4) and then integrate using the well-known secant integration.
Rather than adding the two I's together to get 2I you could have subtracted them. We'd get 0 equal to the integral from 0 to pi/2 of (cos^2 - sin^2)/(cosx + sinx) and you would've had the most epic proof that the integral from 0 to pi/2 of cosx - sinx is equal to 0
Very good. But I expected you to take that 2 out of the ln at the end for a slightly simpler answer.
Papa flammy's integral fevah week. I was unaware it was called weierstrass sub. What a GOAT.
I solved this using a different method.
I started by considering this integral to be A, and the integral of cos^2x/(sinx+cosx) from 0 to pi/2 to be B.
I then added and subtracted them, found the values and converted into a linear system of equations. I found the integral of this AND cos^2x/(sinx+cosx) using this
in this question at 1/cos(x)+sin(x) this expression we can divide and multiply by one by root two then we will left with sin(pi/4 +x) in the denominator then we can easily integrate cosec(pi/4 +x).
Instead of converting cos^2 to 1/(1+tan^2) you could just let the cos^2 from the differential cancel it out.
At 4:48 i would replace
sin(x) +cos(x) by sqrt(2)*cos(x-pi/4)
So 1/sin(x)+cos(x) is equal to
1/sqrt(2)*cos(x-pi/4) =(1/sqrt(2))*sec(x-pi/4) and i do integration...
You also can use : sin(x)+cos(x)=sqrt(2)cos(x-pi/4). Its easier
15:28 where is the arcsin boi?
We have (from the timestamp):
=arcsin(x-1/sqrt(2))](0,1)
=arcsin(-1/sqrt(2))-arcsin(0)
=-pi/4
What about writing the denominator as R*sin(x-a), for some R and a. Then make the substitution u=x-a, and you get something reasonable on the denominator. Then use v=-cos(u) and you're left with something reasonable to solve.
At 15:35, could you have factored out a -1, and then got something in terms if inverse hyperbolic tangents?
I used sinx+cosx=sqrt2*sin(pi/4+x), then substitution pi/4+x=y......., finnaly you will get 1+2/(4sqrt2) * integral cos^2 y/ sin y, after the substitution cos y= t you will get the same result.This method is more universal, for others intervals.
algebraic manipulation has to be the coolest shit ever
those two juijitzu moves to get to 2I were absolutely off the charts!!..
Don't fuck around! Integrate! Just do it! Don't let your math dreams be dreams. Just do it.
cool I was just brushing up on Weierstrass sub! For this problem however, King's Principle is *much* quicker imho
As somebody who hasn't learned calculus to this level yet, I wonder how many different ways there are to solve problems like this.
After you got the original difference of two squares, a u=1-t substitution would have made it much nicer. After working it out, it would return the same integral, only it would replace the t-1 with just a t.
There is two "formulas" for the tangent derivative 1+tan^2 and sec^2, and it is so much imporant for integrals that everyone know it so why don't use the right formula at the right moment??
When t=tan(x/2), it is clear that dt=(1+tan^2(x/2))dx=(1+t^2)dx !
Nice integral by the way!
Also I want to share an other way without the half angle sub: you can transform cosx+sinx in √2cos(x-π/4) and use or redemonstrate the int of sec!
Just apply King's property to find 2I=int(1/(sin(x)+cos(x)))dx = sqrt(2)/2 *int(csc(x+pi/4)dx
And we know that integral of csc(pi/4+x)=ln| tan(x/2+pi/8)|
A very clean Weierstrass-Substitution. Indeed. 👌
I have a few questions on what went on at 17:05 when says that the right side of the equation must have degree 0. For this reason he focuses on the t terms, but why them and nothing else?
I don't see any complex problem when factorise -t^2+2t+1. The result is [1+sqrt(2)-t][1-sqrt(2)-t] and the difference from yours is just a minus sign.
Papa Flammy, I was just wondering; is substitution still equivalent if the limits of the integral are different from the original?
sure