Q1. Does the Fouryay Series converge on pi and -pi? Why or why not? Q2. What's the sum of 1/(1+n^2) from 1 to inf? Be sure to subscribe and let's reach 400k subs by the end of 2019, thank you!!! YayYayYayYay
1. I remember the serie expansion is non-continuous in pi and -pi, I don't exactly remember the reasoning. 2. Dirichlet theorem -> S= (pi cosh(pi) - sinh(pi))/(2 sinh(pi)).
Q1 : in the points of discontinuity of f for ex. t the fourier series of f converge to : (f(t+) + f(t-))/2 Q2: to calculate the sum you need to calculate the fourier series in pi or - pi because cos(npi)=(-1)^n=cos(-npi) because cos is an even function. f(t+)=lim(x--->t+)(f(x)) f(t-)=lim(x--->t-)(f(x))
For Q1: notice that the Fourier series will result in the same infinite series for x = π as for x = -π, yet the function it equals is different. Therefore, for at least one of the endpoints, the Fourier series does not converge. In fact, if I remember correctly, it would instead converge to the average of the function evaluated at the intervals. In other words, it would evaluate to cosh(π). This means cosh(π) = sinh(π)/π + 2•sinh(π)/π•S, where S is the series of 1/(1 + n^2) over all n starting at n = 1. Diving by sinh(π) gives coth(π) = 1/π + 2/π S, so 2S/π = coth(π) - 1/π. Then S = (π/2)coth(π) - 1/2. This answer makes sense given that the integral from x = 1 to x going to infinity of 1/(1 + x^2) is π/2 - π/4, and the sum, although we know it is bigger than this integral, is only bigger by a little bit. Meanwhile, π/4 and 1/2 are comparable in size and relatively close to one another. Also, coth(π) is indeed approximately 1. This is sufficient for us to know that that should be the correct answer. Therefore, for Q2: the answer is π·coth(π)/2 - 1/2.
Q1: Do you mean, at x = ±π? One famous property of Fourier series (FS) is that, when the limits exist, the series evaluated at any given x, equals the average of its right- and left-limits at that point. Where the FS is a continuous function, this is of course, trivial. But when there's a finite discontinuity (a saltus) at x₀ , the graph has an isolated point there, midway between where the right- and left-hand "tails" terminate. Let's first rewrite your series, factoring out of the summation, everything independent of n, and then factor things inside the sum: F(x) = [(e^π - e^-π)/2π] (1 + 2 ∑₁⁰⁰[(-1)ⁿ/(1+n²)] [cos(nx) - n sin(nx)] ) = [sinh(π)/π] (1 + 2 ∑₁⁰⁰[(-1)ⁿ/(1+n²)] [cos(nx) - n sin(nx)] ) Now the FS you've found is for eˣ, on the interval (-π, π), repeated periodically out to ±∞. At all the "junctures" - every odd multiple of π - the cos term will be (-1)ⁿ, and the sin term will be 0. So you'll have F(π) = F(-π) = F([2k+1]π) = [sinh(π)/π] (1 + 2 ∑₁⁰⁰[1/(1+n²)]) But the left- and right-limits at each of those points will be e^-π and e^π, so since the FS is everywhere bilateral-limit averaged, and since we know that the FS converges to eˣ for x ∈ (-π, π), it must be F(π) = ½(e^π + e^-π) = cosh(π) so of course, the FS can't converge to eˣ at the endpoints of your interval. And this answers Q2 (Why am I sure you planned it this way?), because we have F(π) = [sinh(π)/π] (1 + 2 ∑₁⁰⁰[1/(1+n²)]) = cosh(π) . . . so Q2: ∑₁⁰⁰[1/(1+n²)] = ½[π cosh(π)/sinh(π) - 1] = ½[π coth(π) - 1] = 1.0737242312844... [This answer agrees with PackSciences' and Angel's.] As a "sanity check," we can notice by the comparison test, that this series must lie between: 0 < ∑₁⁰⁰[1/(1+n²)] < ∑₁⁰⁰[1/n²] = ⅙π² = 1.6449340668482... and since every term of the former < the corresponding term of the latter, and the leading terms differ by ½, ∑₁⁰⁰[1/(1+n²)] < ∑₁⁰⁰[1/n²] - ½ = ⅙π² - ½ = 1.1449340668482... Fred
Pro tip: if you notice a constant is being used often, give it a name. I took (e^pi - e^(-pi))/pi to be equal to gamma and that cleaned up my work a lot! Also at the end you can notice that gamma = 2sinh(pi)/pi.
It's interesting to see that if you ignore the sin(nx) term in the expansion, the resulting function is no longer e^x but now cosh(x) (and ignoring a_0 and cos(nx) gives sinh(x)). I guess it makes sense knowing that e^x = cosh x + sinh x and that probably also explains why lots of comments point out that expansion of e^x can be expressed very easily using the hyperbolic trig functions.
This is not a coincidence! In fact, this is a valid definition of sinh(x) and cosh(x). Every function can be uniquely decomposed as the sum of an even function and an odd function. For e^x that decomposition is e^x = cosh(x) + sinh(x) as you said. Relating that the Fourier series, the cos(x) terms of the Fourier series always describe the even part of the decomposition, and the sin(x) terms always describe the odd part. So we can use the Fourier series to calculate the even-odd decomposition of any function!
This is indeed beautiful. For one, I could never understand why some equations look comparatively more beautiful than others. I can see now that if the correct inputs are in place, the equation does the work for us! Once again though, this is indeed a fantastic series! Thank you do much! ❤🎉
YAY you made the yay video Also, notice that the term A0 is equal to sinh(π)/π. Just thought it was interesting. In fact, you can entirely factor out a multiple of sinh(π)/π from the series.
Buzea Alexandru L = lim (x -> 0+) x^(e^-1/x). log L = log [lim (x -> 0+) x^(e^-1/x)] = lim (x -> 0+) e^(-1/x)•log x, since log x is a continuous function on the interval (0, ♾). e^(-1/x)•log x = log x/[e^(1/x)], so log L = lim (x -> 0+) log x/[e^(1/x)] = lim (x -> 0+) (1/x)/[e^(1/x)•(-1/x^2)] = - lim (x -> 0+) x/[e^(1/x)] = 0, since the form is 0+/e^+Infinity = 0/infinity = 0. Hence log L = 0, so L = 1, and since L = lim (x -> 0+) x^(e^(-1/x)), this means lim (x -> 0+) x^[e^(-1/x)] = 1
Also, I know you said the Fourier series does not converged at the endpoints. But I did notice that if you let x = π, then the sin(n•x) all become 0, and cos(nπ) = (-1)^n, and if you multiply that by the coefficient (-1)^n/(1 + n^2), then the result is 1/(1 + n^2). So this would give you the sum, assuming the series were to converge to the function at the endpoints. But the problem is that it also works for x = -π,, and that seems to give a contradiction since the series would stay exactly the same, but the thing it equals would be different. It is a pity.
Angel Mendez-Rivera Yes, I actually said the exact thing when I recorded the video but decided to cut that off to shorten the length. The interesting thing is, sum of 1/(1+n^2) is equal to the average when we plug Pi and -Pi in. Crazy and I don't know why!
blackpenredpen ooh I used to know th explanation but I forgot it. It had something to with this one phenomenon that caused the oscillation to go crazy near the endpoints
No, not a pitty. The end points do converge. They just don't converge to the the values of the function at either end. Both ends converge instead to the average of the function values calculated at the end points.
I did the same thing, but for the complex version of the series. I got Sum from -infinity to infinity of sinh(pi)/pi (-1)^n e^(i*n*t) All the steps to find the complex coefficient are consistent, but Wolfram Alpha and others sums calculators says this sum diverges. Is this result consistent? I know this videos is more than 2 years old, but I didn't find the complex version of e^x series anywhere on internet.
Zabra Khan The Fourier series takes that segment and the repeats it periodically, resulting in discontinuities at every multiple of π. It repeats it at the same height.
Well i understanded proof but 1 think is not true e^-pi=lim x->-pi (f(x))=sum An*cos(nx)+bnsin(nx)=-sum An e^pi= lim x->pi(f(x))=sum Ancos(nx)+Bnsin(nx)=-sum An And this meaning e^-pi=e^pi then pi=0. Can you say my wrong?
Lol I was gonna say why didn't you use sinh (pi), then I saw that your goal was something totally differe than showing the terms of e^(x) in terms of cos and sin.
@@angelmendez-rivera351 if they were just definitions, then we couldnt use them to derive other things about the functions they represent. for example we wouldnt be able to say the identity he showed at the end of the video is true. we can however define them to be extensions of the domain of a function (for example e^(ix) can be evaluated using the taylor series as a definition)
nathanisbored Your comment makes absolute nonsense, considering that everything in mathematics can only be derived from axioms and definitions. The definition of the Taylor series of a function f(x) is the power series with respect to x - a with the coefficient sequence [f^(n)](a)/n!. That has nothing to do with whether there is information about the function we can derive from this series or not. The series already uses information about the function inevitably. The reason we define e^x by its Taylor series is because the function is analytic and the series converges uniformly everywhere, so it occupies the same real domain as the previous definition, but with much more convenience of usage, and this analicity gives a continuation which permits for using this for complex numbers and elements of other topological spaces. That once again has nothing to do with these series being definitions and not theorems.
adarsh kumar He already has a video on these. More specifically, he made a video on integrating the SqRt[tan(x)]. That answers half your question. As for SqRt[cot(x)], notice that cot(x) = tan(π/2 - x). As such, all you must do is perform the substitution u = π/2 - x, and your integrand will simply become a constant multiple of SqRt[tan(u)], and that answers the second part to your question.
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Q1. Does the Fouryay Series converge on pi and -pi? Why or why not?
Q2. What's the sum of 1/(1+n^2) from 1 to inf?
Be sure to subscribe and let's reach 400k subs by the end of 2019, thank you!!!
YayYayYayYay
You'll cross πmillion subscribers this year! 😉
1. I remember the serie expansion is non-continuous in pi and -pi, I don't exactly remember the reasoning.
2. Dirichlet theorem -> S= (pi cosh(pi) - sinh(pi))/(2 sinh(pi)).
Q1 : in the points of discontinuity of f for ex. t the fourier series of f converge to :
(f(t+) + f(t-))/2
Q2: to calculate the sum you need to calculate the fourier series in pi or - pi because cos(npi)=(-1)^n=cos(-npi) because cos is an even function.
f(t+)=lim(x--->t+)(f(x))
f(t-)=lim(x--->t-)(f(x))
For Q1: notice that the Fourier series will result in the same infinite series for x = π as for x = -π, yet the function it equals is different. Therefore, for at least one of the endpoints, the Fourier series does not converge. In fact, if I remember correctly, it would instead converge to the average of the function evaluated at the intervals. In other words, it would evaluate to cosh(π). This means cosh(π) = sinh(π)/π + 2•sinh(π)/π•S, where S is the series of 1/(1 + n^2) over all n starting at n = 1. Diving by sinh(π) gives coth(π) = 1/π + 2/π S, so 2S/π = coth(π) - 1/π. Then S = (π/2)coth(π) - 1/2. This answer makes sense given that the integral from x = 1 to x going to infinity of 1/(1 + x^2) is π/2 - π/4, and the sum, although we know it is bigger than this integral, is only bigger by a little bit. Meanwhile, π/4 and 1/2 are comparable in size and relatively close to one another. Also, coth(π) is indeed approximately 1. This is sufficient for us to know that that should be the correct answer.
Therefore, for Q2: the answer is π·coth(π)/2 - 1/2.
Q1: Do you mean, at x = ±π?
One famous property of Fourier series (FS) is that, when the limits exist, the series evaluated at any given x, equals the average of its right- and left-limits at that point. Where the FS is a continuous function, this is of course, trivial.
But when there's a finite discontinuity (a saltus) at x₀ , the graph has an isolated point there, midway between where the right- and left-hand "tails" terminate.
Let's first rewrite your series, factoring out of the summation, everything independent of n, and then factor things inside the sum:
F(x) = [(e^π - e^-π)/2π] (1 + 2 ∑₁⁰⁰[(-1)ⁿ/(1+n²)] [cos(nx) - n sin(nx)] )
= [sinh(π)/π] (1 + 2 ∑₁⁰⁰[(-1)ⁿ/(1+n²)] [cos(nx) - n sin(nx)] )
Now the FS you've found is for eˣ, on the interval (-π, π), repeated periodically out to ±∞.
At all the "junctures" - every odd multiple of π - the cos term will be (-1)ⁿ, and the sin term will be 0. So you'll have
F(π) = F(-π) = F([2k+1]π) = [sinh(π)/π] (1 + 2 ∑₁⁰⁰[1/(1+n²)])
But the left- and right-limits at each of those points will be e^-π and e^π, so since the FS is everywhere bilateral-limit averaged, and since we know that the FS converges to eˣ for x ∈ (-π, π), it must be
F(π) = ½(e^π + e^-π) = cosh(π)
so of course, the FS can't converge to eˣ at the endpoints of your interval.
And this answers Q2 (Why am I sure you planned it this way?), because we have
F(π) = [sinh(π)/π] (1 + 2 ∑₁⁰⁰[1/(1+n²)]) = cosh(π) . . . so
Q2:
∑₁⁰⁰[1/(1+n²)] = ½[π cosh(π)/sinh(π) - 1] = ½[π coth(π) - 1] = 1.0737242312844...
[This answer agrees with PackSciences' and Angel's.]
As a "sanity check," we can notice by the comparison test, that this series must lie between:
0 < ∑₁⁰⁰[1/(1+n²)] < ∑₁⁰⁰[1/n²] = ⅙π² = 1.6449340668482...
and since every term of the former < the corresponding term of the latter, and the leading terms differ by ½,
∑₁⁰⁰[1/(1+n²)] < ∑₁⁰⁰[1/n²] - ½ = ⅙π² - ½ = 1.1449340668482...
Fred
Pro tip: if you notice a constant is being used often, give it a name. I took (e^pi - e^(-pi))/pi to be equal to gamma and that cleaned up my work a lot! Also at the end you can notice that gamma = 2sinh(pi)/pi.
You have been helping me understand math since calculus and I'm in PDE's now. You're a true hero
You have no idea how much I needed this. Thanks. May Chen Lu be with you.
Just a minor nitpick: (e^(pi) - e^(-pi))/2 can be abbreviated sinh(pi).
It's interesting to see that if you ignore the sin(nx) term in the expansion, the resulting function is no longer e^x but now cosh(x) (and ignoring a_0 and cos(nx) gives sinh(x)). I guess it makes sense knowing that e^x = cosh x + sinh x and that probably also explains why lots of comments point out that expansion of e^x can be expressed very easily using the hyperbolic trig functions.
That’s awesome, I didn’t even know that. I’m just learning about Fourier series and it’s really cool, I wish I learned about all this earlier.
This is not a coincidence! In fact, this is a valid definition of sinh(x) and cosh(x). Every function can be uniquely decomposed as the sum of an even function and an odd function. For e^x that decomposition is e^x = cosh(x) + sinh(x) as you said. Relating that the Fourier series, the cos(x) terms of the Fourier series always describe the even part of the decomposition, and the sin(x) terms always describe the odd part. So we can use the Fourier series to calculate the even-odd decomposition of any function!
This is indeed beautiful. For one, I could never understand why some equations look comparatively more beautiful than others. I can see now that if the correct inputs are in place, the equation does the work for us! Once again though, this is indeed a fantastic series! Thank you do much! ❤🎉
This was on my Calculus II exam back in 2011. Those were some good times.
Yay!!!
YAY you made the yay video
Also, notice that the term A0 is equal to sinh(π)/π. Just thought it was interesting. In fact, you can entirely factor out a multiple of sinh(π)/π from the series.
Yup, it is! : )
yaya yay yay yay series aka fouryay yay ya y ay yatyb series!!!!!!!!!
This was really cool, I enjoy anything Fourier so seeing this was great. Thanks!
This is so nice. Proud to see a Nigerian youth be so creative ❤
I wish you success in all you are about to embark on.
Great Sir, i never understand fourier series till i watched this video, Bravoo
Will you be doing Fourier Transform?
Hello sir I'm from India
This was extremely good...
Thank you
Thank you sir....This sum was asked in my internals exam....Now i got this sum clear
Hands down. Me too! 😂😅
Thanks helped a lot
thanks, nice trick on the integral
Bro you are great yaar
Just keep working
I appreciate your work 👍👍
that was an amazing result looking forward to more awesome results thanks for sharing and keep it up
Thanks for the video!
you're a life saver BPRP #yay
You're welcome!
Math for its own sake is a beautiful thing, gotta say ❤️
Thankyou sir,
B. Engineering
Learning from India .
thank you sir
Amazing!
Thank you so much.... You are like God to me
I like the way you teach .....
Excellent!
Thanks bro u make my day
Thank you
Intense
it scared me when those children started yaying
This is amazing ... thank you
Ты великолепный человек, спасибо большое!
Why do we have to integrate a none periodic function in the interval [-pi, pi]?
Thanku bro
Your videos about calculus are really cool man. I learned a lot from them. Can you do the limit as x approaches 0+ of x^(e^-1/x) in your next video?
Buzea Alexandru L = lim (x -> 0+) x^(e^-1/x). log L = log [lim (x -> 0+) x^(e^-1/x)] = lim (x -> 0+) e^(-1/x)•log x, since log x is a continuous function on the interval (0, ♾). e^(-1/x)•log x = log x/[e^(1/x)], so log L = lim (x -> 0+) log x/[e^(1/x)] = lim (x -> 0+) (1/x)/[e^(1/x)•(-1/x^2)] = - lim (x -> 0+) x/[e^(1/x)] = 0, since the form is 0+/e^+Infinity = 0/infinity = 0. Hence log L = 0, so L = 1, and since L = lim (x -> 0+) x^(e^(-1/x)), this means lim (x -> 0+) x^[e^(-1/x)] = 1
I ended up getting a 4 on the AP calculus ab
Step jumps are there
Also, I know you said the Fourier series does not converged at the endpoints. But I did notice that if you let x = π, then the sin(n•x) all become 0, and cos(nπ) = (-1)^n, and if you multiply that by the coefficient (-1)^n/(1 + n^2), then the result is 1/(1 + n^2). So this would give you the sum, assuming the series were to converge to the function at the endpoints. But the problem is that it also works for x = -π,, and that seems to give a contradiction since the series would stay exactly the same, but the thing it equals would be different. It is a pity.
Angel Mendez-Rivera
Yes, I actually said the exact thing when I recorded the video but decided to cut that off to shorten the length. The interesting thing is, sum of 1/(1+n^2) is equal to the average when we plug Pi and -Pi in. Crazy and I don't know why!
blackpenredpen ooh I used to know th explanation but I forgot it. It had something to with this one phenomenon that caused the oscillation to go crazy near the endpoints
No, not a pitty. The end points do converge. They just don't converge to the the values of the function at either end. Both ends converge instead to the average of the function values calculated at the end points.
It's called Gibbs phenomenon. Look it up
Daniel Auto Yeah, I actually realized that before you commented, and I posted in my long response to the comment in he asked the questions
お疲れ様でした
Yes! Say! Fouriyay!
Could you please solve this equation x^x=e
I did the same thing, but for the complex version of the series.
I got
Sum from -infinity to infinity of sinh(pi)/pi (-1)^n e^(i*n*t)
All the steps to find the complex coefficient are consistent, but Wolfram Alpha and others sums calculators says this sum diverges.
Is this result consistent? I know this videos is more than 2 years old, but I didn't find the complex version of e^x series anywhere on internet.
Great💚
Dear Sir , Can you explain how you get the integration of a0 and a n to the power n ??? please
tq
nice video :) what happens if its e^(x-pi) how to calculate this?
How would you visualize this when looking at the graphs of both e^x and the Fourier series of e^x?
Zabra Khan The Fourier series takes that segment and the repeats it periodically, resulting in discontinuities at every multiple of π. It repeats it at the same height.
@@angelmendez-rivera351 Right, except the discontinuities are at every odd multiple of π.
Fred
ffggddss Correct, that is what I meant to say.
Pls do a video on the summation of
x+ x^2 +x^4 + x^8+x^16 ... (inf)
Yayayayayayayayayayayayayaya......👍👍👍👍👌👌👌👌this video deserves 💐💐💐💐
I get it, you're assuming that it has a form then extracting the form using linear operators.
sugarfrosted He is assuming it can be expressed as a linear combination of the basis space of waves.
COOL!
hello brother i have a question Can I replace e ^ (π) -e ^ (- π) with sinh (π) ?????
200K (almost) subs #YAY
Among the last of the good creators at the heart of this platform #YayYayYayYay
:)))
Pi is the best 😁😁😁😁😁thank you!!!
blackpenredpen 😁😁
I concur that this deserves 4 yay's!
Fred
Thanks you just saved my life lol
is this on a open interval from -pi to pi????
Right a0/2
There are different definition for a0. The one bprp uses contains the factor 1/2.
azy j’ai lâché un like juste pour le début chu un gamin mdrr
yaaaaaay
Why is cos(-nπ)=(-1)^n and cos(nπ)=(-1)^n
Well i understanded proof but 1 think is not true
e^-pi=lim x->-pi (f(x))=sum An*cos(nx)+bnsin(nx)=-sum An
e^pi= lim x->pi(f(x))=sum Ancos(nx)+Bnsin(nx)=-sum An
And this meaning e^-pi=e^pi then pi=0.
Can you say my wrong?
Dude,
ty!! when i was practicing, i thought that -pi in Sen(kx)-->(Senk-pi) it could not use the identity :c
Viva!
Could be possible that you made a mistake on the a0?. Shouldn't it be a0 = (e^pi - e^-pi)/pi ?
There are different definition for a0. The one bprp uses contains the factor 1/2.
@@Bayerwaldler Thanks!
Tetration of yays!
Sir, can't we use the hyperbolic functions in this problem?
Kar sake hai
Lol I was gonna say why didn't you use sinh (pi), then I saw that your goal was something totally differe than showing the terms of e^(x) in terms of cos and sin.
is cosnπ is equal to cos(-nπ) ,
How to deal if the interval is between say any two numbers a and b rather than - pi to pi?
it gets ugly
Definitely.
How get 1 power n
How do you test anything for convergence?
the problem is why we need to approximate the e?
I actually have been wanting to do that sum of (-1)^n/(1+n^2) so that's why I do these.
But the formula for fourier serious also includes 1/2 a0 ...which you didn't write
There are different definition for a0. The one bprp uses contains the factor 1/2.
Could you make a video proving the 1/pi approxiamation?
Intuitive proof for taylor and fourier series?
GLaDOS What exactly do you mean by proving them? They are definitions.
@@angelmendez-rivera351 they are not definitions, and he just proved them in a couple previous videos
nathanisbored He did not prove them. He explained what they are, why we want them, and how to get them.
@@angelmendez-rivera351 if they were just definitions, then we couldnt use them to derive other things about the functions they represent. for example we wouldnt be able to say the identity he showed at the end of the video is true. we can however define them to be extensions of the domain of a function (for example e^(ix) can be evaluated using the taylor series as a definition)
nathanisbored Your comment makes absolute nonsense, considering that everything in mathematics can only be derived from axioms and definitions. The definition of the Taylor series of a function f(x) is the power series with respect to x - a with the coefficient sequence [f^(n)](a)/n!. That has nothing to do with whether there is information about the function we can derive from this series or not. The series already uses information about the function inevitably. The reason we define e^x by its Taylor series is because the function is analytic and the series converges uniformly everywhere, so it occupies the same real domain as the previous definition, but with much more convenience of usage, and this analicity gives a continuation which permits for using this for complex numbers and elements of other topological spaces. That once again has nothing to do with these series being definitions and not theorems.
How did (1+n)^2 become 1+n^2?
It didn't. 1/(1 + n²) came from the factor of 1/(a² + b²) in the integrals, ∫eªˣ sin(bx) dx and ∫eªˣ cos(bx) dx; with a=1, b=n.
Fred
I like foyay
13:15 what else would i like? i
you got some big arms
Do you even lift bro
Hello ,did you the exact integral of (root over tan(X) +root over cot(X))dx . Please help me guy i'm in trouble
adarsh kumar What is that expression supposed to mean? More specifically, what exactly is “root over tan(x)”?
No friend , it is integral of (√(tan(x))+√(cot(x)) dx which is to be determined
adarsh kumar He already has a video on these. More specifically, he made a video on integrating the SqRt[tan(x)]. That answers half your question. As for SqRt[cot(x)], notice that cot(x) = tan(π/2 - x). As such, all you must do is perform the substitution u = π/2 - x, and your integrand will simply become a constant multiple of SqRt[tan(u)], and that answers the second part to your question.
@@angelmendez-rivera351 thanks a lot
@@angelmendez-rivera351 I've one more question that is integral of 1over (sin^4(x) +sin^2(X) cos^2(X) +cos^4(x)) whole multiple dx
:o this is more cute than the mess I got XD
I didn't simplify those cos(nπ) and sin(nπ), but I transformed those exponential into sinh(π)
: )))
#yay
I am a wannabe mathfreak!
I'm the happiest... You included:
#yay #yay #yay
Yay!!!!!
Can you integrate (ln(1/x)+sqrt(x))/(x^2+sin(1/x^2)) ?
I am fairly certain this is not integrante using elementary functions.
wrong