Can the integration of sine be proven using Euler's formula?

[mnada]

Is there any proof for the integration of sine function beside the "fundamental theorem of calculus"

I know that the derivative of sine function is driven by the limit of sin(x)/x , is there a way to prove the integration of sine like this.

Thanks

 

[Char. Limit]

Er, what exactly are you looking for here? Are you looking for a proof that the integral of sin(x) is -cos(x)?

 

[mnada]

Yes, exactly, this is what I am looking for

 

[olivermsun]

The power series for sin and cos are one way to look at it. You can also derive it from the definition of derivative and the angle addition formulas.

 

[chiro]

In case you're unaware of what power series are (and taylor series) see:

http://en.wikipedia.org/wiki/Taylor_series

 

[Char. Limit]

I'm wondering if there would be some way to prove it using Riemann sums, which I think is what the OP is going for. But I'm not sure how to do that... set $$\Delta x$$ to 1/n, I guess, and then what for sin(x_i)?

 

[chiro]

I'm wondering if there would be some way to prove it using Riemann sums, which I think is what the OP is going for. But I'm not sure how to do that... set $$\Delta x$$ to 1/n, I guess, and then what for sin(x_i)?

I guess you could do something like that, but the differentiation/integration are readily available (and easy to derive) for your standard a . x^n, which is a lot easier to do on a term by term basis.

I remember a very very long time ago (when I was in high school), we used some results of certain limits (like lim sin(x)/x as x -> 0) and the definition of a derivative to get sin(x) from first principles. That may be what you're asking but to verify its along these lines:

http://www.math.com/tables/derivatives/more/trig.htm

 

[mnada]

I'm wondering if there would be some way to prove it using Riemann sums, which I think is what the OP is going for. But I'm not sure how to do that... set $$\Delta x$$ to 1/n, I guess, and then what for sin(x_i)?

Actually this is exactly what I hope to find. a way to prove it using Riemann sums.
I didn't ask this question before until I noticed that the integration of sine looks this

How can i tell from the figure that the integration is -cos(x)

 

[gb7nash]

It would probably be easier to show that the derivative of -cos(x) is sin(x), using the difference quotient definition. After that you could use the fundamental theorem of calculus (assuming you can use that?)

 

[mnada]

In case you're unaware of what power series are (and taylor series) see:

http://en.wikipedia.org/wiki/Taylor_series

I got your point but if I integrate sin(x) as in the form of Taylor series :

I end up missing 1

it should be -cos(x)+1

 

[uart]

You're forgetting about the constant of integration. This is exactly the reason that it's important, the integral is not complete without it!

 

[mnada]

In this case can you help me to show how to calculate C. it should be -1 in order to have the integration of sin(x) as -cos(x)

 

[Mute]

I'm not sure what that plot is showing, but it's not the integration of sin(x) to get cos(x). At first glance the curve doesn't have the correct period or height to be the integral of the sine curve, but even if the two curves were plotted on different scales, the large yellow curve is not just a cosine. It looks more like a sin^2(x) curve - look at the way the curve bends as it reaches the x axis, and compare it to the sine curve beneath it. EDIT: Okay, it's a cos(x)+const, I guess I missed that.

 

[uart]

In this case can you help me to show how to calculate C. it should be -1 in order to have the integration of sin(x) as -cos(x)

Yes its an arbitrary constant. If you want it to be -1 then you make it -1. That's what arbitrary means.

Why is it arbitrary? Because the derivative of any constant is zero. It doesn't have to be any specific number in order to have this property.

 

[mnada]

Thank you all for your help. I really appreciate it.
First of all I understood the idea of C and how to get the integration of sin by Taylor series.
I also found another mathematical proof which is a kind that I was looking for "by using Euler’s formula

http://www.mathslogic.com/integration-of-trigonometric-function-sine/
For me now the mathematical proof is set.

For the above graph the thick curve is the integration of the sin as the area under the curve (note that the integration of sin from 0-pi is 2 and this is what you have here).

you can look at this graph as 1-cos(x). My last goal is how to bring together the mathematical solution with the geometrical solution

Thanks