# Proving identity of this integral: $\dfrac{1}{\pi}\int_0^\pi \cos(n\theta-z\sin\theta)d\theta$

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?

• How did you obtain the easily obtained result? – copper.hat Jun 6 at 22:44
• @copper.hat $cos(-x)=cos(x)$ – Thom Jun 6 at 22:46
• That would leave you with $-n\theta+z \sin \phi$. – copper.hat Jun 6 at 22:47
• Oops, excuse me, I missed the $-n$. – copper.hat Jun 6 at 22:50
• No problem, Glad to be able to help. – copper.hat Jun 6 at 22:54

As was commented, you forgot to change the bounds with your sub $$\theta=\pi-\varphi$$: $$A_{-n}=\frac1\pi\int_\pi^0\cos(n\pi -n\varphi + z\sin\varphi)(-1)d\varphi \\=-\frac1\pi\int^\pi_0\cos(n\pi -n\varphi + z\sin\varphi)(-1)d\varphi\\ =(-1)^nA_n$$