Integration homework

  • #26
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I haven't done the calculation myself, but you should, according to Euler's Equation, take the real part of the sum of exponential functions.
 
  • #28
vela
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I know, you are not allowed to solve the question form me but if the student is stuck shouldn't you give some hints? :(
We've given you hints, but for some reason, you've completely ignored them.
 
  • #29
I ignored some because Euler's formula isn't covered by the syllabus. Maybe we should use cis2x+(cis2x)^2+...+(cis2x)^n as a suitable geometric series.

I was able to do the integral without using the series but that isn't acceptable for this question.

Let [tex]I_n=\int_0^{\pi/2}\frac{\sin ((2n+1)x)}{\sin(x)}dx[/tex]
we can use sum to products formula: [tex]\sin((2n+1)x)-\sin((2n-1)x)=2\sin(x)\cos(2nx)[/tex] [tex]\therefore I_n=\int_0^{\pi/2} \frac{\sin((2n+1)x)}{\sin(x)}dx[/tex]
[tex]=\int_0^{\pi/2} \frac{\sin((2n-1)x)}{\sin(x)}dx+2\int_0^{\pi/2}\cos(2nx)dx[/tex]
The cosine integral is zero, so we have that [tex]I_n=I_{n-1}[/tex]
By induction this implies that [tex]I_n=I_0=\pi/2[/tex]
 

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