The integral of sin (dx)

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argile5

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Is it possible to formally solve the integral of sin(dx)?

Not to be mistaken with integral sin( x )dx witch is easy.

I believe I found the answer but I need someone to review my work
 
argile5 said:
Is it possible to formally solve the integral of sin(dx)?

Not to be mistaken with integral sin( x )dx witch is easy.

I believe I found the answer but I need someone to review my work
Click to expand...
[imath]\displaystyle \int\sin x \ dx[/imath] means the area between the curve of the sine function and x-axis.

[imath]\displaystyle \int \sin(dx) \ dx[/imath] means what?

But I have an idea:

[imath]\sin x \approx x[/imath] when [imath]x[/imath] is very small. Therefore,

[imath]\displaystyle \int \sin(dx) \ dx = \int\int dx \ dx = \int x \ dx = \frac{x^2}{2} + C[/imath]
 
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argile5 said:
Is it possible to formally solve the integral of sin(dx)?

Not to be mistaken with integral sin( x )dx witch is easy.

I believe I found the answer but I need someone to review my work
Click to expand...
Maybe if you posted your work that someone can review it.
 
I know of sin(x) but am not familiar with sin. As BBB already requested, can you give us your definition of sin.
 
You could say that the integral of \(\displaystyle \sin(dx)\) w. r .t. \(\displaystyle x\) is \(\displaystyle -\frac{1}{d} \cos (dx)\). ;)
 
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Steven G said:
I know of sin(x) but am not familiar with sin. As BBB already requested, can you give us your definition of sin.
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Same definition as sin(x) except it's sin(dx). So dx would just be an infinitely small angle. I arrived at it by finding the area of a circle in a different way. I can send you my work if you would like to see it...
 
argile5 said:
Same definition as sin(x) except it's sin(dx). So dx would just be an infinitely small angle. I arrived at it by finding the area of a circle in a different way. I can send you my work if you would like to see it...
Click to expand...
Go ahead, since that will implicitly show us what you mean.

It can make some sense to talk about sin(dx), in the sense of a differential approximation to the sine of a small number; but what would you be integrating with respect to?

Once we see what you think the answer is, we can probably talk about what question you are really answering.
 
mario99 said:
[imath]\displaystyle \int\sin x \ dx[/imath] means the area between the curve of the sine function and x-axis.

[imath]\displaystyle \int \sin(dx) \ dx[/imath] means what?

But I have an idea:

[imath]\sin x \approx x[/imath] when [imath]x[/imath] is very small. Therefore,

[imath]\displaystyle \int \sin(dx) \ dx = \int\int dx \ dx = \int x \ dx = \frac{x^2}{2} + C[/imath]
Click to expand...
Nice idea. The weird thing about my integral is sin(dx) with no dx outside of parentheses. See my derivation below
 
argile5 said:
Nice idea. The weird thing about my integral is sin(dx) with no dx outside of parentheses. See my derivation below
Click to expand...
But the area of a circle is [imath]\pi r^2[/imath]. If you have done your analysis on a unit circle the area is [imath]\pi[/imath]. Why is your area [imath]2\pi[/imath]?
 
mario99 said:
But the area of a circle is [imath]\pi r^2[/imath]. If you have done your analysis on a unit circle the area is [imath]\pi[/imath]. Why is your area [imath]2\pi[/imath]?
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I made a minor mistake on the last line when I divided by 2. So disregard last line
 
argile5 said:
I posted a photo of it below. It might need more development. You will know better than me...
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I would do this:

[imath]\displaystyle \Delta A = \frac{1}{2}bh = \frac{1}{2}\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)*2 = \frac{1}{2}\sin \theta[/imath]


[imath]\displaystyle dA = \frac{1}{2}\sin d\theta = \frac{1}{2} \ d\theta[/imath]


[imath]\displaystyle \int dA = \frac{1}{2}\int d\theta[/imath]


[imath]\displaystyle A = \frac{1}{2}\int_{0}^{2\pi} d\theta = \frac{1}{2}\theta\bigg|_0^{2\pi} = \frac{1}{2}(2\pi - 0) = \pi[/imath]
 
argile5 said:
Nice idea. The weird thing about my integral is sin(dx) with no dx outside of parentheses. See my derivation below
Click to expand...
This is the trouble: the definition of the (Riemann) integral includes dx outside the function. What you are doing violates the definition of an integral.

Specifically, when you claim that your summation is an integral, there is no basis for saying that.

What you could do at the end is to use the fact that for small [imath]\theta[/imath], [imath]\sin(\theta)\approx \theta[/imath], so that your [imath]\sin(\Delta\theta)\approx \Delta\theta[/imath], and the integral is actually [math]\frac{1}{2}\int_0^{2\pi}d\theta=\frac{1}{2}\left.\theta\right|_0^{2\pi}=\frac{1}{2}2\pi=\pi[/math]
 
Dr.Peterson said:
This is the trouble: the definition of the (Riemann) integral includes dx outside the function. What you are doing violates the definition of an integral.

Specifically, when you claim that your summation is an integral, there is no basis for saying that.

What you could do at the end is to use the fact that for small [imath]\theta[/imath], [imath]\sin(\theta)\approx \theta[/imath], so that your [imath]\sin(\Delta\theta)\approx \Delta\theta[/imath], and the integral is actually [math]\frac{1}{2}\int_0^{2\pi}d\theta=\frac{1}{2}\left.\theta\right|_0^{2\pi}=\frac{1}{2}2\pi=\pi[/math]
Click to expand...
Thanks, Dr. Peterson. I see my error in jumping to an integral from a summation.
 
mario99 said:
I would do this:

[imath]\displaystyle \Delta A = \frac{1}{2}bh = \frac{1}{2}\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)*2 = \frac{1}{2}\sin \theta[/imath]


[imath]\displaystyle dA = \frac{1}{2}\sin d\theta = \frac{1}{2} \ d\theta[/imath]


[imath]\displaystyle \int dA = \frac{1}{2}\int d\theta[/imath]


[imath]\displaystyle A = \frac{1}{2}\int_{0}^{2\pi} d\theta = \frac{1}{2}\theta\bigg|_0^{2\pi} = \frac{1}{2}(2\pi - 0) = \pi[/imath]
Click to expand...
Thanks Mario this is much cleaner
 
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