First thing that came to mind was multiplying the top and bottom by (sinx-cosx) and using double angle formulas, no substitution needed. Cool method tho!
I used the method of replacing sinx + cosx by sin(x+π/4)/sqrt(2) and take u = x+π/4, we obtain by developing the numerator and separating the fraction 1/2 integral of (1 + cos(u)/sin(u)), and finish with the same result. Have a good day 😊
@@leif1075 because sine and cosine of π/4 are the same, with the formula of the sin of a sum, you get 1/√2sin(x) + 1/√2cos(x), you just multiply by √2 and you obtain sin(x) + cos(x)
I only occasionally remember this method and saw someone win an integration in person doing this. All the math professors were confused, but eventually it was verified to be correct lol
I had a really different way of solving it. It involves a transform I defined that I called λ. I first defined it as λ{x} = x+x'+x''+x''' ... Essentially, just a sum of derivatives of x(t). It turns out this has a nice property with integration. int(x/λ{x}) = t - ln|λ{x}| + c I'm gonna pull a Fermat and not write the proof, but for now, at least take it as conjecture. And with some more trickery, λ{sin(t)} = 1/2*(sin(t)+cos(t)). Sin(t) makes less sense in the original derivation, but with some e^(i*pi) trickery you can figure out sin and cos. Turns out, despite the fact that these don't make sense with the original interpretation, they DO work out in the integral. So, I happened to recognize that integral was x/λ{x} int[sin(t)/(1/2*(sin(t)+cos(t)))] = 2*int[sin(t)/(sin(t)+cos(t))], which by my formula earlier is equal to t - ln|1/2*(sin(t)+cos(t))| + C. Dividing both sides by 2 then simplifying the ln, we arrive at the same answer! I hope that wasn't too hard to follow...I can post a better proof on a google doc or something. A YT comment is a hard place to put a lot of heavy math.
Damn! Very interesting to know, why this makes sense when λ{sin(t)} does not. Btw what is the origin of this transform? Quite a peculiar property it has
@@mixam7711 Although I am sure it exists somewhere outside of my notes, I made this independently. My motivation was to construct something analogous to a geometric series.
Another method is multiplying the top and bottom by the conjugate of sinx + cosx which is sinx - cosx. Some creatuve trigonometric identities and some u-sub would give the same answer.
This method can be used to solve any integeral in the form (Asinx+Bcosx)/(C sinx + D cosx) By finding k1 and k2 Where (Asinx+Bcosx)/(C sinx + D cosx) = k1 (Csinx+Dcosx)/(C sinx + D cosx) + k2 (Ccosx-D sinx)/(C sinx + D cosx)
I love your techniques how you just logically solve the diophantine equations I am inspired by you after watching your videos I feel more lovely towards mathematics Thanks!!☺☺☺👌👌👌👏👏👏🙏🙏🙏
@@leecherlarry ok let's do this Find Sum from n = 1 to infinity (nπ Si(2nπ) + (1 - nπ^2)/2) If your compi says it's divergent than your compi is breaking maths as It's simple to prove this
@@pardeepgarg2640 i have input *Sum[n \[Pi] Sin[2 n \[Pi]] + (1 - n \[Pi]^2)/2, {n, 1, \[Infinity]}]* and compi says "Sum::div: Sum does not converge." So which result should be the answer of the sum?
How I solved it: Multiply top and bot by (cos(x) - sin(x)) Write in terms of cos(2x) and sin(2x), and pull out constant term to integrate separately. This part gives 1/2 x Set u = sin(2x). Remaining integral simplifies as 1/4 int((u-1)/(1-u^2)du), which evaluates to -1/4 ln|sin(2x) +1| = -1/4 ln|2sin(x)cos(x) + sin^2(x) + cos^2(x)| = -1/2 ln|sin(x) + cos(x)|. Add a constant of integration and you get the full answer.
I rewrote the equation as 1/(cotx+1) then substituted u=cotx+1 Made it waaay more complicated than it had to be but at least I got more practice out of it
Hi, love your videos! Could you please make a video showing the proof of integrating dx/x? It seems a bit weird and I was wondering if you could show a clean proof, as you always do
Καλές οι λύσεις αυτές, αλλά δέστε μια απίθανη άλλη λύση: 1. Διαιρούμε πάνω και κάτω με συνχ 2. Στον αριθμητή την μονάδα τη γράφουμε στεμ^2χ-σφ^2χ 3. Χωρίζω το κλάσμα σε 2 4. Το 1ο είναι το ln|1+σφχ| 5 Στο 2ο αφαιρώ και προσθέτω 1 6. Το χωρίζω σε άλλα 2 κ.ο.κ.
Its very triggy I think.Also cool content lol. Edit 1:Nevermind its really easy,but I wanna try It with harmonic addition theorem. Edit 2:I solved it and got the same answer,guess we didnt need algebra here,but I respect ur method anyways. Edit 3:I checked my answer and I got the negative of It,but the harmonic addition theorem still applies to other cases.
@@SyberMath hey I just done my graduation exam few days ago, it went well, some of my problems are similar to others of yours, I love to have crossed this channel.
You obviously do not know Bioche rules: 1. If the integrand f(cos x,sin x) dx is invariant by -x, t=cos x 2 If it is invariant by pi-x, t=sin x 3. If it is invariant by pi+x, t=tan x 4. Otherwise t = tan (x/2) Your integrand is in case 3; so t=tan x will give you the result with no effort.
Substitute u = x + π/4. Then, sin(u = x+π/4)=(sin x + cos x)/√2 and sin(x = u - π/4) = (sin u - cos u)/√2 and du = dx. Then integrate (sin u - cos u) / (√2 √2 sin u) du = 1/2 [ u - ln sin u] + c = 1/2 [ x + π/4 - ln | sin x + cos x | + ln √2 ] + c = x - ln | sin x + cos x | + C [ where C = c + π/8 + 1/2 ln 2 ].
Άλλη λύση έξυπνη : 1. Πολαπλασιάζω πάνω και κάτω με συνχ - ημχ. Άρα έχω ολοκλ. ημχ . συνχ - ημ^2 χ / συν2χ 2. Το χωρίζω σε 2 ολοκληρώματα: Το πρώτο 1/2 .ημ2χ / συν2χ και το δεύτερο 1/2 . (1 -συν2χ) / συν2χ. 3. Το πρώτο ολοκλ. 1/2 . εφ2χ και το δεύτερο το χωρίζω πάλι σε 1/2 τεμ2χ - 1/2. 4. Άρα έχω -1/4.ln|συν2χ| +1/4 .ln| τεμ2χ+εφ2χ| - 1/2. χ +c.
Διορθώνω την λύση που έστειλα: 1. Διαιρώ πάνω και κάτω με ημχ Έχω ολοκλ. 1/ 1+σφχ 2. Την μονάδα την γράφω στεμ^2 χ _ σφ^2 χ Έχω ολοκλ. στεμ^2 χ _σφ^2 χ/ 1+σφχ 3. Το χωρίζω σε 2 ολοκλ. Το πρώτο -ln|1+σφχ| και στο δεύτερο αφαιρώ 1 και προσθέτω 1 4. Τι σφ^2 χ -1=(σφχ-1)(σφχ+1) κ.ο.κ.
Not really, the log without a subscript means common log, or log base 10. log base e is the same as ln. I think it's clear just to write ln instead of log. Otherwise, it would be marked wrong all time.
No this is the toughest one....because you it will give you only a constant which value can never be found because it will not have any variable term so that we can put the variable value and get constant value
@@paragswarnkar6858 Do you see any limits on the integral proposed? No. So the constant is C. If the integral had limits its evaluation would be zero, and why do mathematical constructs need to be practical anyway?
@@davidbrisbane7206 I am not talking about limit 🙄 I am talking about boundry conditions and maths is for practical use that's why I am taking about practical condition.....and these constant values either found by boundry conditions or by experiments...but in case when some entity is constant then it can only be found by experiments
I would suggest changing the numerator of the right hand side as follows: x^2-1 = (x^2+1)-2, and then separate those two terms into separate fractions. You should be left with 1-(2/(x^2+1)) I’ll let you do the rest. If you need help though, do ask
![Infinite tetration of rad2 = 2 [tetration]](/img/n.gif)
We can also use (1/2)(sinx + cosx + sinx - cosx) to ease our integral.
We can also do this way for the ease of calculation:-
1/2(2sinx/sinx + cosx)dx
Then,
1/2{ (sinx + cosx) + (sinx - cosx)/ sinx + cosx}dx
Considering (sinx + cosx) = t
dt = (cosx - sinx)dx
1/2{ (sinx + cosx) - (cosx - sinx)/ sinx + cosx }dx
1/2{1}dx - 1/2{dt/t}
x/2 - ln( sinx + cosx)/2
Thanks for reading !!
Wow, u r really good at mathematics,as well as choosing good problems, u literally converted an integral into a system of equations 😱,great work 👍
Thank you! 💖
First thing that came to mind was multiplying the top and bottom by (sinx-cosx) and using double angle formulas, no substitution needed. Cool method tho!
Excellent!
I used the method of replacing sinx + cosx by sin(x+π/4)/sqrt(2) and take u = x+π/4, we obtain by developing the numerator and separating the fraction 1/2 integral of (1 + cos(u)/sin(u)), and finish with the same result. Have a good day 😊
Wherecdid you get pi/4 from?
@@leif1075 because sine and cosine of π/4 are the same, with the formula of the sin of a sum, you get 1/√2sin(x) + 1/√2cos(x), you just multiply by √2 and you obtain sin(x) + cos(x)
I only occasionally remember this method and saw someone win an integration in person doing this. All the math professors were confused, but eventually it was verified to be correct lol
@@djridoo Ah thank you I wasnt thinking of law of sines sum formula..
Nice work!
I had a really different way of solving it. It involves a transform I defined that I called λ. I first defined it as λ{x} = x+x'+x''+x''' ... Essentially, just a sum of derivatives of x(t). It turns out this has a nice property with integration.
int(x/λ{x}) = t - ln|λ{x}| + c
I'm gonna pull a Fermat and not write the proof, but for now, at least take it as conjecture.
And with some more trickery, λ{sin(t)} = 1/2*(sin(t)+cos(t)). Sin(t) makes less sense in the original derivation, but with some e^(i*pi) trickery you can figure out sin and cos. Turns out, despite the fact that these don't make sense with the original interpretation, they DO work out in the integral.
So, I happened to recognize that integral was x/λ{x}
int[sin(t)/(1/2*(sin(t)+cos(t)))] = 2*int[sin(t)/(sin(t)+cos(t))], which by my formula earlier is equal to t - ln|1/2*(sin(t)+cos(t))| + C. Dividing both sides by 2 then simplifying the ln, we arrive at the same answer!
I hope that wasn't too hard to follow...I can post a better proof on a google doc or something. A YT comment is a hard place to put a lot of heavy math.
Very cool approach 👍
Wow!!!
Nice
Damn! Very interesting to know, why this makes sense when λ{sin(t)} does not. Btw what is the origin of this transform? Quite a peculiar property it has
@@mixam7711 Although I am sure it exists somewhere outside of my notes, I made this independently. My motivation was to construct something analogous to a geometric series.
Another method is multiplying the top and bottom by the conjugate of sinx + cosx which is sinx - cosx. Some creatuve trigonometric identities and some u-sub would give the same answer.
that's how i did it
sin x+cos x=(2/sqrt2)* sin(x+pí/4), then we use substitution y=x+pí/4
Yes. The integration is not too hard and can be shown to be equivalent to the answer given.
Muy interesante, muchas gracias. Desde luego este modo de resolver la integral es mejor que cualquier otro. Gracias, desde Madrid.
Nice approach. I wrote sin(x)+cos(x) = root(2) sin(x+pi/4) then substituted t = x + pi/4. Then you just expand out sin(t-pi/4) up top.
Very good!
Why would anyone ever think of that method?? I don't see why.
This type of manipulation is commonly used in Algebra and Number Theory
@@SyberMath yea but I don't see why or how anyone would.come to it naturally?
This method can be used to solve any integeral in the form (Asinx+Bcosx)/(C sinx + D cosx)
By finding k1 and k2
Where
(Asinx+Bcosx)/(C sinx + D cosx)
= k1 (Csinx+Dcosx)/(C sinx + D cosx) + k2 (Ccosx-D sinx)/(C sinx + D cosx)
Nice!
It is some how similar to partial fraction decomposition. You can find k1 and k2 by equating the coefficient of sinx and cos x on both sides
Multiply by the Conj. Of dmntr. (sin x - cos x) both of nmntr. And dmntr.
No it will be useless 😐 it will make the integral more complicated
Very clear explanation
Very nice ,thank you sir for new method of solving this type of integral
Thanks and welcome
I love your techniques how you just logically solve the diophantine equations I am inspired by you after watching your videos I feel more lovely towards mathematics Thanks!!☺☺☺👌👌👌👏👏👏🙏🙏🙏
great job Syber, thanks for sharing bro
Thanks for watching! 💖
Very elegant.
Thank you! 😊
I used Weierstrass substitution. It was a mess.
Really?
@@SyberMath yes sometimes it is not going well.you should use u=cotx or tgx.
Find a,b such that num = a(den)+ d/dx(den) , where num=numerator and den=denominator,
then integrate. So easy..!
Well, it's new. But, I'll say that a simple substitution of sin(pi/4 +x) should have done the trick.
Just as LouLoutroll said.
The definite integral from 0 to π/2 of this function is easy.
But I didn't know how to solve the indefinite integral of this function.
oki, got it!! haha
*Integrate[Sin[x]/(Sin[x] + Cos[x]), x]*
Bruh , is computer is allowed in your exam , i bet your computer can't do this problem
@@pardeepgarg2640 you can bet i caht do the problem . but cell phone is with me all the time ( wolframcloudDOTcom via app or web browser )
@@leecherlarry ok let's do this
Find
Sum from n = 1 to infinity
(nπ Si(2nπ) + (1 - nπ^2)/2)
If your compi says it's divergent than your compi is breaking maths as It's simple to prove this
@@pardeepgarg2640 i have input
*Sum[n \[Pi] Sin[2 n \[Pi]] + (1 - n \[Pi]^2)/2, {n, 1, \[Infinity]}]*
and compi says "Sum::div: Sum does not converge."
So which result should be the answer of the sum?
@@leecherlarry the real answer is (π^2 -9)/24
And you can prove this by simple integration
Very nice solution sir
Thank you! 💖
Multiply numerator by two
And write 2 sinx =(sin +cos) +(sin -cos)
Nice!
How I solved it:
Multiply top and bot by (cos(x) - sin(x))
Write in terms of cos(2x) and sin(2x), and pull out constant term to integrate separately. This part gives 1/2 x
Set u = sin(2x). Remaining integral simplifies as 1/4 int((u-1)/(1-u^2)du), which evaluates to -1/4 ln|sin(2x) +1| = -1/4 ln|2sin(x)cos(x) + sin^2(x) + cos^2(x)| = -1/2 ln|sin(x) + cos(x)|. Add a constant of integration and you get the full answer.
Nice!
I rewrote the equation as 1/(cotx+1) then substituted u=cotx+1
Made it waaay more complicated than it had to be but at least I got more practice out of it
Awesome way of solving problems .thanks sir
Happy to help
Hi, love your videos! Could you please make a video showing the proof of integrating dx/x? It seems a bit weird and I was wondering if you could show a clean proof, as you always do
Good tips. But i prefer all what is about integers. By the way, i will see as more vidéos of your Channel about integers as i can. Thank you for all.
Καλές οι λύσεις αυτές, αλλά δέστε μια απίθανη άλλη λύση:
1. Διαιρούμε πάνω και κάτω με συνχ
2. Στον αριθμητή την μονάδα τη γράφουμε στεμ^2χ-σφ^2χ
3. Χωρίζω το κλάσμα σε 2
4. Το 1ο είναι το ln|1+σφχ|
5 Στο 2ο αφαιρώ και προσθέτω 1
6. Το χωρίζω σε άλλα 2 κ.ο.κ.
Very inventive, never seen anything like this before
Nice question!
Great explanations as always
Thank you! 💖
Interesting method 😳 and this problem may be solve multiplying and dividing by 2.
So I gues int of cosx/(sinx+cosx) is the same as the other, just with a plus sign when you solve for I2.
i want to try something like that next year in my studies .
Its very triggy I think.Also cool content lol.
Edit 1:Nevermind its really easy,but I wanna try It with harmonic addition theorem.
Edit 2:I solved it and got the same answer,guess we didnt need algebra here,but I respect ur method anyways.
Edit 3:I checked my answer and I got the negative of It,but the harmonic addition theorem still applies to other cases.
Thanks a Ton..keep it up & upload more & more videos on Integration 🖤😘😘
I will try my best
thanks, my graduation exam is next week, this helps me out a lot.
Great to hear!
@@SyberMath hey I just done my graduation exam few days ago, it went well, some of my problems are similar to others of yours, I love to have crossed this channel.
Slick, Syber. Very slick. :D
😁
Calculus integral I = sinx/(sinx - cosx) and I = sinx / ( sinx - cosx ) in [ 0 , pi/4 ]..
You obviously do not know Bioche rules:
1. If the integrand f(cos x,sin x) dx is invariant by -x, t=cos x
2 If it is invariant by pi-x, t=sin x
3. If it is invariant by pi+x, t=tan x
4. Otherwise t = tan (x/2)
Your integrand is in case 3; so t=tan x will give you the result with no effort.
Good solution
Thanks!
Very, very cool!
Thanks!
Do a Q&A
Substitute u = x + π/4. Then, sin(u = x+π/4)=(sin x + cos x)/√2 and sin(x = u - π/4) = (sin u - cos u)/√2 and du = dx. Then integrate (sin u - cos u) / (√2 √2 sin u) du = 1/2 [ u - ln sin u] + c = 1/2 [ x + π/4 - ln | sin x + cos x | + ln √2 ] + c = x - ln | sin x + cos x | + C [ where C = c + π/8 + 1/2 ln 2 ].
Wow amazing please show me more you are underrated
💖
Very nice you explain so nicely 😊😊😊😊😊😊😊 thanks for these joyful moments 😊☺️😊😊😊😊
Thank you so much 😀
Very smart method !!
Glad you think so!
very nice so awesome you are the best pro
Άλλη λύση έξυπνη :
1. Πολαπλασιάζω πάνω και κάτω με συνχ - ημχ. Άρα έχω ολοκλ.
ημχ . συνχ - ημ^2 χ / συν2χ
2. Το χωρίζω σε 2 ολοκληρώματα:
Το πρώτο 1/2 .ημ2χ / συν2χ και το δεύτερο 1/2 . (1 -συν2χ) / συν2χ.
3. Το πρώτο ολοκλ. 1/2 . εφ2χ και το δεύτερο το χωρίζω πάλι σε 1/2 τεμ2χ
- 1/2.
4. Άρα έχω -1/4.ln|συν2χ| +1/4 .ln| τεμ2χ+εφ2χ| - 1/2. χ +c.
There are other ways to solve this but this way is very nice and unusual!
Διορθώνω την λύση που έστειλα:
1. Διαιρώ πάνω και κάτω με ημχ
Έχω ολοκλ. 1/ 1+σφχ
2. Την μονάδα την γράφω στεμ^2 χ _ σφ^2 χ
Έχω ολοκλ. στεμ^2 χ _σφ^2 χ/ 1+σφχ
3. Το χωρίζω σε 2 ολοκλ.
Το πρώτο -ln|1+σφχ| και στο δεύτερο αφαιρώ 1 και προσθέτω 1
4. Τι σφ^2 χ -1=(σφχ-1)(σφχ+1) κ.ο.κ.
That’s a great way to solve it
Thank you! 😊
great idea sir thank u
Most welcome
Amazing method !!!
Glad you think so!
This is brilliant
Thank you! 😊
Fantastic
Thank you so much 😀
“u” are cool!
U substitution at play again!! 😁
Thank u! 😊
Integrale (0)=k, k is Realnumber
Please Solve JEE Advanced tough questions,that will help me for my exam
They are tough! 😁
@@SyberMath I agree 😂😂😂
useful video. clever.
Great
My teacher told me that in calculus log is taken as base e by default
Not really, the log without a subscript means common log, or log base 10. log base e is the same as ln. I think it's clear just to write ln instead of log. Otherwise, it would be marked wrong all time.
On my channel log(x)=log_10(x) and lnx=log_e(x)
@@SyberMath I know that but in algebra if no base is given then its base 10 and in calculus it is base e
Excelent
Thank you! 💖
This is lenghthy, there is another easy way...
Show me. @reddy
I’m wondering will it work if I use this sub x=(π/2-θ)? And thx for the great explanation :)
It should! Thanks
The easiest function to integrate is ∫ 0dx 😁
No this is the toughest one....because you it will give you only a constant which value can never be found because it will not have any variable term so that we can put the variable value and get constant value
@@paragswarnkar6858
What if you are given C = 1?
@@davidbrisbane7206 🧐 if they given then why we need to find... I am telling that we can't apply any boundry conditions in practical problems...
@@paragswarnkar6858
Do you see any limits on the integral proposed? No. So the constant is C. If the integral had limits its evaluation would be zero, and why do mathematical constructs need to be practical anyway?
@@davidbrisbane7206 I am not talking about limit 🙄 I am talking about boundry conditions and maths is for practical use that's why I am taking about practical condition.....and these constant values either found by boundry conditions or by experiments...but in case when some entity is constant then it can only be found by experiments
4x^2 - 4x + 1 > (x^2 -1)/(x^2 + 1)
i dont have any idea how to do this
Sub x=Tan theta and remember 1-tan^2 theta/1+tan^2 theta=cos(2theta)
I would suggest changing the numerator of the right hand side as follows:
x^2-1 = (x^2+1)-2, and then separate those two terms into separate fractions. You should be left with 1-(2/(x^2+1))
I’ll let you do the rest. If you need help though, do ask
@@fredthelegend7673 pls tell the rest
You can transform this to the form:
(2x-1)(2x-1) > (x-1)(x+1) / (x^2+1)
and test 4 intervals using points -1, 1/2, 1
@@advaykumar9726 Other people’s methods are better, ignore mine
2nd! like you so much
Yay! Thank you! 🤩
How to integrate the expression x.tan(x)?
www.wolframalpha.com/input/?i=integrate+xtanx
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جميل جداً جداً
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