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

Question

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?

Answer 1

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