integral of sin x using the definition

Hi,

I am trying to calculate:

$\displaystyle \int_0^{\pi}\sin x \,dx $

using the definition only, that is, that $\displaystyle \int_0^{\pi}\sin x\,dx = \lim_{n \to \infty}\sum_{i = 1}^{\infty}\sin(x_{i}^{*})\Delta x$, where $\displaystyle \Delta x = \frac{\pi}{n}$. If we take take $\displaystyle x_i^* = i\Delta x = \frac{\pi}{n}$ we obtain:

$\displaystyle \int_0^{\pi}\sin x\,dx = \lim_{n \to \infty}\left(\frac{\pi}{n}\right)\sum_{i=1}^n \sin\left(\frac{i\pi}{n}\right)$.

I am stuck at this point. Is there any kind of identity, perhaps trigonometric or power series related, to help me evaluate this limit by hand only?

Thank in advanced or any help,

James

Re: integral of sin x using the definition

Quote:

Originally Posted by

**james121515** Hi,

I am trying to calculate:

$\displaystyle \int_0^{\pi}\sin x \,dx $

using the definition only, that is, that $\displaystyle \int_0^{\pi}\sin x\,dx = \lim_{n \to \infty}\sum_{i = 1}^{\infty}\sin(x_{i}^{*})\Delta x$, where $\displaystyle \Delta x = \frac{\pi}{n}$. If we take take $\displaystyle x_i^* = i\Delta x = \frac{\pi}{n}$ we obtain:

$\displaystyle \int_0^{\pi}\sin x\,dx = \lim_{n \to \infty}\left(\frac{\pi}{n}\right)\sum_{i=1}^n \sin\left(\frac{i\pi}{n}\right)$.

I am stuck at this point. Is there any kind of identity, perhaps trigonometric or power series related, to help me evaluate this limit by hand only?

Thank in advanced or any help,

James

Surely you should know that $\displaystyle \displaystyle \int{\sin{x}\,dx} = -\cos{x} + C$ by definition...

Re: integral of sin x using the definition

Quote:

Originally Posted by

**james121515** Hi,

I am trying to calculate:

$\displaystyle \int_0^{\pi}\sin x \,dx $

using the definition only, that is, that $\displaystyle \int_0^{\pi}\sin x\,dx = \lim_{n \to \infty}\sum_{i = 1}^{\infty}\sin(x_{i}^{*})\Delta x$, where $\displaystyle \Delta x = \frac{\pi}{n}$. If we take take $\displaystyle x_i^* = i\Delta x = \frac{\pi}{n}$ we obtain:

$\displaystyle \int_0^{\pi}\sin x\,dx = \lim_{n \to \infty}\left(\frac{\pi}{n}\right)\sum_{i=1}^n \sin\left(\frac{i\pi}{n}\right)$.

I am stuck at this point. Is there any kind of identity, perhaps trigonometric or power series related, to help me evaluate this limit by hand only?

Thank in advanced or any help,

James

Hint:

Use Euler's formula - Wikipedia, the free encyclopedia.

Edit:

An alternative:

$\displaystyle \sum_{i=1}^n \sin\left(\frac{i\pi}{n}\right)=S$

Now multiply both sides by $\displaystyle \frac{\cos{\frac{\pi}{2n}}}{\sin\frac{\pi}{2n}}$.

And third alternative:

Proving by induction:

$\displaystyle \sum_{i=1}^n{\sin(i\alpha)}=\frac{\sin\frac{n}{2}{ \alpha}\cdot\sin\frac{n+1}{2}\alpha}{\sin\frac{ \alpha }{2}}$

Re: integral of sin x using the definition

Quote:

Originally Posted by

**Prove It** Surely you should know that $\displaystyle \displaystyle \int{\sin{x}\,dx} = -\cos{x} + C$ by definition...

I'm not sure it's a definition. I suspect the OP knows the rule and wants to derive the rule from first principles.

Re: integral of sin x using the definition

Quote:

Originally Posted by

**mr fantastic** ... I suspect the OP knows the rule and wants to derive the rule from first principles...

May be that that is not deriving 'from the first principles' but a possibility is to use the 'geometric sum'...

$\displaystyle \sum_{k=1}^{n} x^{n} = x\ \frac{1-x^{n}}{1-x}$ (1)

Using the 'identity'...

$\displaystyle \sin (k\ \frac{\pi}{n}) = \frac{e^{i\ k\ \frac{\pi}{n}} -e^{- i\ k\ \frac{\pi}{n}}}{2\ i}$ (2)

... and (1) You arrive to write...

$\displaystyle \frac{\pi}{n}\ \sum_{k=1}^{n} \sin (k\ \frac{\pi}{n}) = \frac{\pi}{2\ i\ n}\ \{e^{i\ \frac{\pi}{n}}\ \frac{1 - e^{i\ k\ \frac{\pi}{n}} }{1- e^{i\ \frac{\pi}{n}}} - e^{- i\ \frac{\pi}{n}}\ \frac{1 - e^{-i\ k\ \frac{\pi}{n}} }{1- e^{-i\ \frac{\pi}{n}}} \} $ (3)

The effective computation of (3) is however 'just a little hard' for me and is left to 'a young mind' (Itwasntme)...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

Re: integral of sin x using the definition

Following previous outlines and denoting

$\displaystyle C_a=\cos a+\cos 2a+\ldots+\cos na,\; S_a=\sin a+\sin 2a+\ldots+\sin na$

If $\displaystyle e^{ia}\neq 1$ , we have:

$\displaystyle C_a+iS_a=e^{ia}+e^{2ia}+\ldots+e^{nia}=e^{ia}\;\df rac{(e^{nia}-1)}{e^{ia}-1}=$

$\displaystyle e^{ia}\;\dfrac{e^{\frac{nia}{2}}}{{e^{\frac{ia}{2} }}}\;\dfrac{e^{\frac{nia}{2}}-e^{\frac{-nia}{2}}}{e^{\frac{ia}{2}}-e^{\frac{-ia}{2}}}=e^{\frac{i(n+1)a}{2}}\;\dfrac{\sin \frac{na}{2}}{\sin \frac{a}{2}}=$

$\displaystyle \frac{\sin \frac{na}{2}}{\sin \frac{a}{2}}\left(\cos \frac{(n+1)a}{2}+i\sin \frac{(n+1)a}{2} \right)$

Identifying imaginary parts

$\displaystyle S_a=\dfrac{\sin \frac{na}{2}}{\sin \frac{a}{2}}\;\sin\frac{(n+1)a}{2}$

So, $\displaystyle S_{\pi/n}=\dfrac{1}{\sin \frac{\pi}{2n}}\;\sin \frac{(n+1)\pi}{2n}$

Then,

$\displaystyle \int_0^{\pi}\sin x\;dx=\lim_{n\to +\infty} \dfrac{\pi}{n}S_{\pi/n}=2 $

Re: integral of sin x using the definition

Quote:

Originally Posted by

**chisigma** Using the 'identity'...

$\displaystyle \sin (k\ \frac{\pi}{n}) = \frac{e^{i\ k\ \frac{\pi}{n}} -e^{- i\ k\ \frac{\pi}{n}}}{2\ i}$ (2)

Thanks for your reply. I had no idea the solution to this involves complex analysis. This first 'identity' you mention here, is this derived from Euler's formula $\displaystyle e^{i\theta} = \cos \theta + i\sin \theta$ or is it more involved?

Re: integral of sin x using the definition

Quote:

Originally Posted by

**james121515** Thanks for your reply. I had no idea the solution to this involves complex analysis. This first 'identity' you mention here, is this derived from Euler's formula $\displaystyle e^{i\theta} = \cos \theta + i\sin \theta$ or is it more involved?

Yes. And the sum of a geometric series is being used.