The above result is complete.
I think,it can be done in the following way.
I consider that $a \in \mathbb{R}$ We know, from the representation of Bessel function that $\exp(ia\cos \phi)=\sum_{m=-\infty}^{m=\infty}i^m J_m(a) \exp(im\phi)$ such that $\cos(a \cos \phi)=J_0(a)+2\sum_{m=1}^{\infty}(-1)^m J_{2m}(a)\cos(2m\phi)$ and $\sin(a \cos \phi)=2\sum_{m=0}^{\infty}(-1)^m J_{2m+1}(a)\cos((2m+1)\phi)$, where $J_m$ represents Bessel function of first kind with order $m$.
As the integrand is the even function, for simplicity,we can write it as
$$\frac{1}{2}\int_0^{2\pi}\bigg(J_0(a)+2\sum_{m=1}^{\infty}(-1)^m J_{2m}(a)\cos 2m\phi+ 2 i\sum_{m=0}^{\infty}(-1)^m J_{2m+1}(a)\cos(2m+1)\phi\bigg)\frac{(1-\cos 2\phi)}{2}\,d\phi$$ where, the imaginary part will be $+$ve or $-$ve depending on $a$ is $+$ve or $-$ve. Using the fact that $\int_0^{2\pi}\cos nx \,dx=\int_0^{2\pi}\sin nx \, dx=0$ and $\int_0^{2\pi}\cos nx \sin mx \, dx=\pi \delta_{nm}$, it becomes,
$$\frac{1}{4}\int_0^{2\pi}\bigg(J_0(a)+2 J_2(a)\cos^2 (2\phi)\bigg)\,d\phi $$
Now, using $\cos^2 2\phi= \frac{1+\cos 4\phi}{2}$, the final result will be,
$$\frac{1}{2}(\pi J_0(a)+\pi J_2(a))=\frac{\pi J_1(a)}{a}$$ using the recurrence relation $J_{n-1}(x)+J_{n+1}(x)=\frac{2nJ_n(x)}{x}$Where, $a$ can be taken forboth the $+ve$ and $-ve$ values, and accordingly the answer will be finally $$\frac{\pi J_1(|a|)}{|a|}$$ as shown above.
I also request you to check whether there is any misconception or calculation error.
