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- Thread starter VertexOperator
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- #26

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- #27

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Isn't the real part the cos2nx series?

- #28

vela

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We've given you hints, but for some reason, you've completely ignored them.I know, you are not allowed to solve the question form me but if the student is stuck shouldn't you give some hints? :(

- #29

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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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