# Math Help - itnegral of sin x using the definition

1. ## 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

2. ## Re: integral of sin x using the definition

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...

3. ## Re: integral of sin x using the definition

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:

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

4. ## Re: integral of sin x using the definition

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.

5. ## Re: integral of sin x using the definition

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' ...

Kind regards

$\chi$ $\sigma$

6. ## 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,

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

7. ## Re: integral of sin x using the definition

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?

8. ## Re: integral of sin x using the definition

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.