@@xibalbam Technically, it is the same function if you consider all the function dominions properly, but as a general approach, it doesn't work (it becomes too complicated), and it's not even the simplest thing to do for sin/cos basic functions.
A good exercise. It was a nice lesson you presented with various methods for approach to the problem. I like students seeing the options they have. You are a good mathematics professor. Cheerful Calculations 🧮
Try the 100 integrals: th-cam.com/video/dgm4-3-Iv3s/w-d-xo.htmlsi=bJQFZjMjHBnkV_5b
Just go from zero to pi and double it.
sin(x) is 2π periodic. |sin(x)| is π periodic. That's an ideal strategy.
I thought the same bro🎉🎉
4:10 He mentioned that was a valid method, too. He just showed how to do it when there is no symmetry.
It is too easy like that
Integration from 0 to nπ, abs(sin(x)) with respect to x is always 2n.
Here n=2 this implies, 4
PS: n is a positive integer.
so logically just do integral of sinx from 0 to pi. then double it.
👍
Similarly
(A) integral of sqrt { (1+cos 2x)/2 }
in [0,π] = 2
(B) integral of । x sin π x। in [ -1,3/2] = (3/π) + (1/ π^2)
I interpret the original question as, "Why does this trigonometric integral evaluate to an integer? That's surprising."
A failure in matrix at 2:54, hah
Oops lol
Why am i always watching the videos where i know how to solve it, but clicking away the ones i dont know how to solve
"There is no shame in not knowing; the shame lies in not finding out."
Can't you replace abs(sin x) by sqrt(sin^2 x) and integrate that?
Why would you do that? It's more difficult to solve that way.
@@antoniousai1989 I just wanted to be sure it was possible because it can be a good way to integrate absolute value of more complex functions
@@xibalbam Technically, it is the same function if you consider all the function dominions properly, but as a general approach, it doesn't work (it becomes too complicated), and it's not even the simplest thing to do for sin/cos basic functions.
@@antoniousai1989 ok thanks !
@@antoniousai1989 small typo *domain*
I integrated from 0 to pi. Then multiplied by 2 since both areas are equal.
i had this type of question in my exam but in the format of first principle of derivative so instead of |sin x| i had |sin(x+t) - sin(x)|/|t|
What exam is this?
@@dannyyeung8237 no formal exam bro just a class test
Wo ist der deutsche Ton?
Easy as π
a piece of cake
The cake is a lie.
😇
A piece of pi
First to comment
Thanks prof
tralalalaaa
A good exercise. It was a nice lesson you presented with various methods for approach to the problem. I like students seeing the options they have. You are a good mathematics professor. Cheerful Calculations 🧮