Solving S_N Integral with Small Angle Formula

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ghostyc
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Homework Statement



[tex]S_N (x) = \frac{2}{\pi} \int_0^x \frac{\sin (2 N t )}{\sin (t)} \; d{t}[/tex]

use suitable small angle formula to show[tex]S_N \Big( \frac{\pi}{2 N} \Big) = \frac{2}{\pi} \int_0^{\pi} \frac{\sin u}{u} d{u}[/tex]

Homework Equations



i guess the suitable small angle formula is

[tex]\sin (\theta) \sim \theta[/tex]

when [tex](\theta)[/tex] is small...

The Attempt at a Solution

i have tried to do some substations but just can't get both numerator and denominator to the right thingany sugguestions will be appreciated

Thank YOU
 
on Phys.org
It looks like you already know they want you to assume [tex]$\sin(t)=t$[/tex]. Your next step is to find an appropriate "u substitution." Try [tex]$ u = 2Nt$[/tex] so that you have [tex]$ \int_0^{\pi/2N} \frac{\sin{(2Nt)}dt}{t} = \int_{0}^{?} \frac{\sin{u}}{u}du$[/tex]. Use algebra to find [tex]$ ? $[/tex] and [tex]$ du $[/tex].
 
Last edited:
Hi there
I have tried this already (actually 7 days ago)
still using
[tex]u = 2 N t[/tex] i can get the correct limits but
i just can't justify that the bottom
[tex]t[/tex] just goes to [tex]u[/tex]
how do i jusitfy that?

Thank you

++++++++++++++++++++++++++++
holly!

I got it
right after the click "post quick reply"...

THANK YOU ALL

:P
 
[tex]\frac{du}{u} = \frac{2Ndt}{2Nt} = \frac{dt}{t}[/tex]