Hello, I would like to know if you can help me with this problem the intergal of (cos (6x)+1), I’ve understood it as far as cos (1/2 (ax)) but where is the sqrt of 2 coming from?! Thank you.
integral of (cos (6x)+1) dx ??? from where to where or indefinite?
anyway if you do not have a typo, that is the sum of two integrals
cos 6x dx + 1 dx
(1/6) sin 6x + x + c
Definite
I mean indefinite
from where to where then and what is the a in (1/2)ax ?
Please check what you typed carefully.
Well, I did the indefinite integral and have no idea where your a and your square root come from.
There is something your are not telling us.
I have a photo of it, is there a way that I can post it on here?!
Of course! I'd be happy to help you understand the integral of cos(6x) + 1. However, I'm not sure where you encountered the square root of 2 in the context of this integral. Could you please provide some additional details or context?
In general, the integral of cos(ax) with respect to x can be calculated using the formula:
∫cos(ax) dx = (1/a)sin(ax) + C,
where C is the constant of integration. This formula is derived using basic rules of integration and the chain rule.
For the integral of cos(6x), we can use this formula with a = 6:
∫cos(6x) dx = (1/6)sin(6x) + C.
The integral of 1 with respect to x is simply x, so the integral of cos(6x) + 1 with respect to x would be:
∫(cos(6x) + 1) dx = ∫cos(6x) dx + ∫1 dx = (1/6)sin(6x) + x + C.
If you can provide more information about where the square root of 2 is coming into play, I'll be able to give you an explanation specific to that context.