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.
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.
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).
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.
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.
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
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
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.
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