The integral of sin (dx)

[argile5]

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

 

[BigBeachBanana]

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

You'd have to explain the notation as it's not standard.

 

[mario99]

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

[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]

 

[Steven G]

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

Maybe if you posted your work that someone can review it.

 

[Steven G]

I know of sin(x) but am not familiar with sin. As BBB already requested, can you give us your definition of sin.

 

[BigBeachBanana]

[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]

This has no meaning. "dx" isn't a variable so sin(dx) is not a function.

 

[Harry_the_cat]

You could say that the integral of \(\displaystyle \sin(dx)\) w. r .t. \(\displaystyle x\) is \(\displaystyle -\frac{1}{d} \cos (dx)\).

 

[argile5]

I know of sin(x) but am not familiar with sin. As BBB already requested, can you give us your definition of sin.

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...

 

[Dr.Peterson]

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...

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.

 

[argile5]

See attached photo below:

 

[argile5]

[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]

Nice idea. The weird thing about my integral is sin(dx) with no dx outside of parentheses. See my derivation below

 

[mario99]

Nice idea. The weird thing about my integral is sin(dx) with no dx outside of parentheses. See my derivation below

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]?

 

[argile5]

Not sure I just followed the steps. Need someone to analyze it

 

[argile5]

You could say that the integral of \(\displaystyle \sin(dx)\) w. r .t. \(\displaystyle x\) is \(\displaystyle -\frac{1}{d} \cos (dx)\).

Haha funny

 

[argile5]

I know of sin(x) but am not familiar with sin. As BBB already requested, can you give us your definition of sin.

I posted a photo of it below. It might need more development. You will know better than me...

 

[argile5]

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]?

I made a minor mistake on the last line when I divided by 2. So disregard last line

 

[mario99]

I posted a photo of it below. It might need more development. You will know better than me...

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]

 

[Dr.Peterson]

Nice idea. The weird thing about my integral is sin(dx) with no dx outside of parentheses. See my derivation below

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]

 

[argile5]

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]

Thanks, Dr. Peterson. I see my error in jumping to an integral from a summation.

 

[argile5]

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]

Thanks Mario this is much cleaner