It is mentioned here that,
\begin{equation} \int_0^{2\pi} exp\left(z\cos(\theta) \right)d\theta = 2\pi I_0(z) \end{equation}
whereas, in equation 30 of this reference, following integral was transformed to the same result, \begin{equation} \int_0^{2\pi} exp\left(z\cos(\theta - \phi) \right)d\theta = \int_{-\phi}^{2\pi-\phi} exp\left(z\cos(\theta') \right)d\theta' =2\pi I_0(z), \end{equation}
where $\phi$ is a constant.
I need to understand that why does the shifting operation impose no change in the result.
Following the operations above, can I say that 1,
\begin{equation} \int_0^{2\pi} \cos(\theta)exp\left(z\cos(\theta - \phi) \right)d\theta = 2\pi I_1(z) \end{equation}
Edit 1:
$I_n(z)$ is the n-th order modified Bessel function of the first kind.
First question about $I_0(z)$: For the first equal sign, by substitution $\theta' = \theta-\phi$, $d\theta' = d\theta$, the upper bound $\theta=2\pi$ becomes $\theta' = 2\pi-\phi$, and the lower bound $\theta=0$ becomes $\theta' = -\phi$.
Then the second equal sign, you may split the LHS into two integrals to match your first equation:
$$\begin{align*} \int_{-\phi}^{2\pi-\phi}\exp(z\cos\theta)\ d\theta &= \int_{-\phi}^0\exp(z\cos\theta)\ d\theta + \int_0^{2\pi-\phi}\exp(z\cos\theta)\ d\theta\\ &= \int_{2\pi-\phi}^{2\pi}\exp(z\cos\theta)\ d\theta + \int_0^{2\pi-\phi}\exp(z\cos\theta)\ d\theta\\ &= \int_0^{2\pi}\exp(z\cos\theta)\ d\theta \end{align*}$$
Second question about $I_1(z)$: After substitution $\theta_1 = \theta-\phi$, the LHS should become
$$\int_{-\phi}^{2\pi-\phi}\cos(\theta_1+\phi)\exp(z\cos\theta_1)\ d\theta_1$$
Even after shifting integration bounds, you still have to handle that $\cos(\theta_1+\phi)$.