itnegral of sin x using the definition

• Jul 10th 2011, 09:32 PM
james121515
integral of sin x using the definition
Hi,

I am trying to calculate:

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

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

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

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
• Jul 10th 2011, 10:39 PM
Prove It
Re: integral of sin x using the definition
Quote:

Originally Posted by james121515
Hi,

I am trying to calculate:

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

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

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

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 $\displaystyle \int{\sin{x}\,dx} = -\cos{x} + C$ by definition...
• Jul 10th 2011, 11:05 PM
Also sprach Zarathustra
Re: integral of sin x using the definition
Quote:

Originally Posted by james121515
Hi,

I am trying to calculate:

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

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

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

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:

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

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

And third alternative:

Proving by induction:

$\sum_{i=1}^n{\sin(i\alpha)}=\frac{\sin\frac{n}{2}{ \alpha}\cdot\sin\frac{n+1}{2}\alpha}{\sin\frac{ \alpha }{2}}$
• Jul 11th 2011, 01:39 AM
mr fantastic
Re: integral of sin x using the definition
Quote:

Originally Posted by Prove It
Surely you should know that $\displaystyle \int{\sin{x}\,dx} = -\cos{x} + C$ 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.
• Jul 11th 2011, 02:53 AM
chisigma
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'...

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

Using the 'identity'...

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

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

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

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

Kind regards

$\chi$ $\sigma$
• Jul 11th 2011, 03:16 AM
FernandoRevilla
Re: integral of sin x using the definition
Following previous outlines and denoting

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

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

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

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

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

Identifying imaginary parts

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

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

Then,

$\int_0^{\pi}\sin x\;dx=\lim_{n\to +\infty} \dfrac{\pi}{n}S_{\pi/n}=2$
• Jul 11th 2011, 04:38 PM
james121515
Re: integral of sin x using the definition
Quote:

Originally Posted by chisigma
Using the 'identity'...

$\sin (k\ \frac{\pi}{n}) = \frac{e^{i\ k\ \frac{\pi}{n}} -e^{- i\ k\ \frac{\pi}{n}}}{2\ i}$ (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 $e^{i\theta} = \cos \theta + i\sin \theta$ or is it more involved?
• Jul 11th 2011, 04:44 PM
mr fantastic
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 $e^{i\theta} = \cos \theta + i\sin \theta$ or is it more involved?

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