Even though I had done this question previously and knew the exact steps, I just watched all along, mesmerized by the way you teach. Let me tell you sir that you are indeed, very cool.
I am a guy who studies math in french, we use something called "linéarisation" which means you use the complexe définition of the sinx which is (e^ix-e^-ix)/2 and when you finish you get the answer
In Electrical Engineering "State Equations" or differential equations Calculus also! 👍 Integrating by exp()s eases having to integrate sines and cosines in Engineering courses.
This might be a bit beside the point: I am a lecturer at at technical college, and I write quite a bit on blackboards, myself. I must compliment you on your neat and elegant writing style: It is very nearly perfect (and close to infinitely better than mine).
Your videos are informative and entertaining. At age 78 I have enjoyed having my math memories refreshed. I don't know if you have this book in your library but my favorite reference math book is "Advanced Engineering Mathematics" by Erwin Kreyszig. He was a professor of Mathematics at Ohio State University. Now, your videos are my favorite math refreshers. Keep up the excellent work.
hello love your channel. I would like to ask if we could write sin^2(x) as (1-cos2x) / 2 at the second line cause we know cos2x= 1- 2sin^(x). And then go on
Nice solution! Does seem strange that this integral has non-trigonometric parts in final answer but nice that we don't have any powers of trigonometric functions at the end.
Yes, it seems a bit strange until you realize that sin⁴𝑥 is always non-negative, which means that the primitive function must be growing and therefore can't consist of only trig functions, because trig functions don't grow, they only oscillate.
I would have factored out another 1/2 from the integral again, to make the numbers nicer: 1/8 * integral of (3 - 4cos2x + cos4x) dx You end up with 1/8 [3x - 2sin2x + sin4x/4] + c
It can be done with reduction formula Int(sin^n(x),x)=Int(sin(x)sin^(n-1)(x),x) Int(sin^n(x),x)=-cos(x)sin^(n-1)(x) - Int((-cos(x))((n-1)sin^(n-2)(x)cos(x)),x) Int(sin^n(x),x)=-cos(x)sin^(n-1)(x) + (n - 1)Int(sin^(n-2)(x)cos^2(x),x) Int(sin^n(x),x)=-cos(x)sin^(n-1)(x) + (n - 1)Int(sin^(n-2)(x)(1-sin^2(x)),x) Int(sin^n(x),x)=-cos(x)sin^(n-1)(x) + (n - 1)Int(sin^(n-2)(x),x) - (n - 1)Int(sin^n(x),x) (1-(-(n-1)))Int(sin^n(x),x)=-cos(x)sin^(n-1)(x) + (n - 1)Int(sin^(n-2)(x),x) nInt(sin^n(x),x)=-cos(x)sin^(n-1)(x) + (n - 1)Int(sin^(n-2)(x),x) Int(sin^n(x),x)=-1/n*cos(x)sin^(n-1)(x) + (n - 1)/n*Int(sin^(n-2)(x),x) With this reduction formula we can easily calculate it in mind -1/4cos(x)sin^3(x)-3/8*cos(x)sin(x)+3/8x + C Problems for you 1. Express Int(sin^n(x),x) in terms of sum (with sigma notation) 2. Calclulate Int(cos^n(t),t=0..Pi) Integral from second problem may be useful if you want to get Chebyshov polynomials via orthogonalization To calculate Int(sin^n(x),x) with approach presented in this video Substitute t = Pi/2-x to get (-1)^(n+1)Int(cos^n(t),t) Get coefficients of Chebyshov polynomial via recurrence relation or ordinary differential equation Put coefficients ofChebyshov polynomial into the matrix and invert this matrix
Its problems like this that bug me. I kept running into road blocks. I would use what is essentially the power reducing formula and square it but i didnt use it a second time, hence the road block.
No porque debería estar multiplicando a f(g(x)) la expresión g'(x) y así la integral da f(g(x))+C pero como no está la expresión g'(x) multiplicando no hay forma general a menos que g(x) sea ax+b para algunos a y b números reales.
![Limit of absolute value functions [ lim |x^3 - x|/(x^3 - |x|) as x goes to 0 ]](http://i.ytimg.com/vi/nJ9lEolfVpU/mqdefault.jpg)
If you never stop teaching, I will never stop learning
True
Even though I had done this question previously and knew the exact steps, I just watched all along, mesmerized by the way you teach. Let me tell you sir that you are indeed, very cool.
Thank you!
i struggled with this integral within Signal Processing for 2 years, now my heart finds peace within ❤
You brighten my day. That is all.
... also, that really is a quality hat.
Thank you. Good to hear from you.
thank you guy , i am from Ethiopia thank you again
man you're the best I swear. Even though I'm new to calculus you just made it look so simple and engaging keep teaching please!!
I am a guy who studies math in french, we use something called "linéarisation" which means you use the complexe définition of the sinx which is (e^ix-e^-ix)/2 and when you finish you get the answer
In Electrical Engineering "State Equations" or differential equations Calculus also! 👍 Integrating by exp()s eases having to integrate sines and cosines in Engineering courses.
This might be a bit beside the point: I am a lecturer at at technical college, and I write quite a bit on blackboards, myself.
I must compliment you on your neat and elegant writing style: It is very nearly perfect (and close to infinitely better than mine).
Thanks to square master I can now enjoy my cup of coffee.
Love your enthusiasm sir!
A wonderfully engaging manner!
Your videos are informative and entertaining. At age 78 I have enjoyed having my math memories refreshed. I don't know if you have this book in your library but my favorite reference math book is "Advanced Engineering Mathematics" by Erwin Kreyszig. He was a professor of Mathematics at Ohio State University. Now, your videos are my favorite math refreshers. Keep up the excellent work.
bro you teach so well
great to watch while eating dinner
hey Prime Newtons! Love your teaching. i would just walli's formula for this integral as it is the easiest approach
Superb my great sir..
Perfect
hello love your channel. I would like to ask if we could write sin^2(x) as (1-cos2x) / 2 at the second line cause we know cos2x= 1- 2sin^(x). And then go on
Love❤
Nice solution! Does seem strange that this integral has non-trigonometric parts in final answer but nice that we don't have any powers of trigonometric functions at the end.
Yes, it seems a bit strange until you realize that sin⁴𝑥 is always non-negative, which means that the primitive function must be growing and therefore can't consist of only trig functions, because trig functions don't grow, they only oscillate.
I would have factored out another 1/2 from the integral again, to make the numbers nicer:
1/8 * integral of (3 - 4cos2x + cos4x) dx
You end up with
1/8 [3x - 2sin2x + sin4x/4] + c
Wouldn't it be easy if you rewrote sin(x) as (e^ix-e^-ix)/2i?
And 7:12 instantly made me hear the song "Zombie" in my head
If you where my Calculus professor I would be so fucking Happy
It can be done with reduction formula
Int(sin^n(x),x)=Int(sin(x)sin^(n-1)(x),x)
Int(sin^n(x),x)=-cos(x)sin^(n-1)(x) - Int((-cos(x))((n-1)sin^(n-2)(x)cos(x)),x)
Int(sin^n(x),x)=-cos(x)sin^(n-1)(x) + (n - 1)Int(sin^(n-2)(x)cos^2(x),x)
Int(sin^n(x),x)=-cos(x)sin^(n-1)(x) + (n - 1)Int(sin^(n-2)(x)(1-sin^2(x)),x)
Int(sin^n(x),x)=-cos(x)sin^(n-1)(x) + (n - 1)Int(sin^(n-2)(x),x) - (n - 1)Int(sin^n(x),x)
(1-(-(n-1)))Int(sin^n(x),x)=-cos(x)sin^(n-1)(x) + (n - 1)Int(sin^(n-2)(x),x)
nInt(sin^n(x),x)=-cos(x)sin^(n-1)(x) + (n - 1)Int(sin^(n-2)(x),x)
Int(sin^n(x),x)=-1/n*cos(x)sin^(n-1)(x) + (n - 1)/n*Int(sin^(n-2)(x),x)
With this reduction formula we can easily calculate it in mind
-1/4cos(x)sin^3(x)-3/8*cos(x)sin(x)+3/8x + C
Problems for you
1. Express Int(sin^n(x),x) in terms of sum (with sigma notation)
2. Calclulate Int(cos^n(t),t=0..Pi)
Integral from second problem may be useful if you want to get Chebyshov polynomials via orthogonalization
To calculate Int(sin^n(x),x) with approach presented in this video
Substitute t = Pi/2-x to get (-1)^(n+1)Int(cos^n(t),t)
Get coefficients of Chebyshov polynomial
via recurrence relation or ordinary differential equation
Put coefficients ofChebyshov polynomial into the matrix and invert this matrix
But we have a formula for §sin^n(x)dx=-1/n×sin^(n-1)(x)cos(x)+(n-1)/nקsin^(n-2)(x)dx
Its problems like this that bug me. I kept running into road blocks. I would use what is essentially the power reducing formula and square it but i didnt use it a second time, hence the road block.
good
Reduction formula? 😁
What you not use reduction formula?
So, there is no general form for «integral of f(g(x)) dx»?
No porque debería estar multiplicando a f(g(x)) la expresión g'(x) y así la integral da f(g(x))+C pero como no está la expresión g'(x) multiplicando no hay forma general a menos que g(x) sea ax+b para algunos a y b números reales.
yoooo nice haha
D.I. trick
I hope you know your work is appreciated, needed and loved sir🫶🏽