How to show this integral is $0$? I want to calculate the integral
$$\int_0^{2\pi} \sin(a\sin \theta - b \theta)\,d\theta$$
where $a \in \mathbb{R}$ and $b \geq 0$ is an integer.
I was wondering how to proceed with this integral. One thought I had is to use Cauchy's Integral Formula, since $e^{ix} = \cos x + i \sin x$, which seems similar, but not directly applicable (and also since this integral is a function of $\theta$ and not $z$).
 A: If $f$ is odd function and integrable, then $\int_{-\pi}^{\pi}f(x)\, {\rm d}x=0$.  Define the change of variables $y:=x-\pi$, then $$\int_{0}^{2\pi}\sin(a\sin(x)-bx)\, {\rm d}x=\int_{-\pi}^{\pi}\sin(a\sin(y+\pi)-b(y+\pi))\, \, {\rm d}y.$$
Define the mapping  $f: y\mapsto \sin(a\sin(y+\pi)-b(y+\pi))$ over $[-\pi,\pi]$ and notice that $f(-y)=-f(y)$ for all $y\in [-\pi,\pi]$, then $f$ is odd over $[-\pi,\pi]$ and then  $$\int_{0}^{2\pi}\sin(a\sin(x)-bx)\, {\rm d}x=0.$$
A: HINT: To this end, let us write $f(\theta) = a \sin(\theta)+\theta$.  Then we are evaluating the integral
$$\int_0^{2\pi} \sin(f(\theta))\,d\theta.$$ We make the following claim:
Claim 1: The equation $$\sin(f(\theta)) \ = \ -\sin(f(\pi+\theta))$$ is satisfied for all $\theta \in [0,\pi]$.
Proof: The  function $f(\theta) = a \sin(\theta)+\theta$ satisfies
$$f(\pi+\theta) = -b\pi-a \sin(\theta)$$ $$=-2b\pi + f(-\theta).$$
Furthermore, $$f(\theta) = -f(-\theta).$$ So for each $\theta \in [0,\pi]$:
$$\sin(f(\theta+\pi)) = \sin(-2b\pi+f(-\theta))$$
$$=\sin(f(-\theta)) = \sin(-f(\theta)) =-\sin(f(\theta)).$$
[Note that we used $\sin(y)$ $=$ $\sin(2n\pi +y)$ $=$ $-\sin(-y)$ for all integers $n$ and all real numbers $y$.] So in particular, for each $\theta \in [0,\pi]$:
$$\sin(f(\pi+\theta)) = -\sin(f(\theta)).$$
Claim 2: The equation $$\int_0^{2\pi} \sin(f(\theta))\,d\theta \ = \ 0$$ is satisfied.
First note: $$\int_0^{2\pi} \sin(\theta)d\theta \ = \ \int_0^{\pi} \sin(f(\theta)) d \theta \ + \ \int_{\pi}^{2\pi} \sin(f(\theta)) d \theta$$
$$= \ \int_0^{\pi} \sin(f(\theta)) d \theta \ + \ \int_{0}^{\pi} \sin(f(\theta'+\pi)) d \theta'.$$
Thus from Claim 1:
$$  \ \int_0^{\pi} \sin(f(\theta)) d \theta \ + \ -\int_{0}^{\pi} \sin(f(\theta')) d \theta'$$ $$ = \ 0.$$
A: \begin{equation*}
\int_{0}^{2\pi }\sin (a\sin \theta -b\theta )d\theta 
\end{equation*}
Applying a change of variable, i.e.,
\begin{equation*}
u=\theta -\pi \text{,}
\end{equation*}
we obtain
\begin{equation*}
\int_{0}^{2\pi }\sin (a\sin \theta -b\theta )d\theta =\int_{-\pi }^{\pi
}\sin (a\sin (u+\pi )-b(u+\pi ))du\text{.}
\end{equation*}
Recall the well-known trigonometric formula:
\begin{equation*}
\sin (x+y)=\sin x\times \cos y+\sin y\times \cos x\text{.}
\end{equation*}
Invoking this formula
\begin{equation*}
\sin (u+\pi )=\sin u\times \cos \pi +\sin \pi \times \cos u=-\sin u\text{,}
\end{equation*}
which, when plugged into the integral, turns it into
\begin{equation*}
\int_{-\pi }^{\pi }\sin (-a\sin u-bu-b\pi )du\text{.}
\end{equation*}
Now, invoking the well-known formula once more, we get%
\begin{equation*}
\sin (-a\sin u-bu-b\pi )=\sin (-a\sin u-bu)\cos (-b\pi )+\sin (-b\pi )\cos
(-a\sin u-bu)\text{,}
\end{equation*}
which, along with $\sin (-b\pi )=0$ and $\cos (-b\pi )=\mp 1$ ($+1$ if $b$
is even and $-1$ if $b$ is odd) implies that
\begin{equation*}
\sin (-a\sin u-bu-b\pi )=\mp \sin (-a\sin u-bu)\text{.}
\end{equation*}
Thus, the integral simplifies to
\begin{equation*}
\mp \int_{-\pi }^{\pi }\sin (-a\sin u-bu)du\text{,}
\end{equation*}
where it is easy to see that $\sin (-a\sin u-bu)$ is an odd function, i.e.,
\begin{equation*}
\sin (-a\sin (-u)-b(-u))=\sin (a\sin (u)+bu)=\sin (-a\sin u-bu)\text{.}
\end{equation*}
Since we are integrating an odd function over a symmetric interval around
zero,
\begin{equation*}
\mp \int_{-\pi }^{\pi }\sin (-a\sin u-bu)du=0\text{,}
\end{equation*}
as claimed.
