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Definitions: We have an integral: $$ A_n= \dfrac{1}{\pi}\int_0^\pi \cos(n\theta-z\sin\theta)d\theta, n=0,\pm 1,... $$ Question: How to prove this identity: $$ A_{-n}=(-1)^n A_n$$ My attempt: I obtatin this easily: $$A_{-n} = \dfrac{1}{\pi}\int_0^\pi \cos(n\theta+z\sin\theta)d\theta,$$ then I use substitution: $\theta = \pi - \varphi$: $$A_{-n} = \dfrac{1}{\pi}\int_0^\pi \cos(n\pi -n\varphi + z\sin\varphi)(-1)d\varphi ,$$ By using formula for cosine of sum I obtain: $$A_{-n} = \dfrac{1}{\pi}\int_0^\pi \cos(n\pi)\cos(n\varphi - z\sin\varphi)(-1)d\varphi,$$ $$A_{-n} =(-1)^{n+1} \dfrac{1}{\pi}\int_0^\pi \cos(n\varphi - z\sin\varphi)d\varphi$$ Where did I make a mistake?

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  • $\begingroup$ How did you obtain the easily obtained result? $\endgroup$ – copper.hat Jun 6 at 22:44
  • $\begingroup$ @copper.hat $cos(-x)=cos(x)$ $\endgroup$ – Thom Jun 6 at 22:46
  • $\begingroup$ That would leave you with $-n\theta+z \sin \phi$. $\endgroup$ – copper.hat Jun 6 at 22:47
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    $\begingroup$ Oops, excuse me, I missed the $-n$. $\endgroup$ – copper.hat Jun 6 at 22:50
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    $\begingroup$ No problem, Glad to be able to help. $\endgroup$ – copper.hat Jun 6 at 22:54
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As was commented, you forgot to change the bounds with your sub $\theta=\pi-\varphi$: $$A_{-n}=\frac1\pi\int_\pi^0\cos(n\pi -n\varphi + z\sin\varphi)(-1)d\varphi \\=-\frac1\pi\int^\pi_0\cos(n\pi -n\varphi + z\sin\varphi)(-1)d\varphi\\ =(-1)^nA_n$$

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