Differentiate under the integral sign: $\int_0^{2π} e^{\cos (x)}\cos(\sin x) \, \mathrm{d}x$

I have again a doubt regarding an exercise of differentiation under the integral sign. In this case, it concerns the integral:

$$\int_0^{2π} e^{\cos(x)}\cos(\sin x) \, \mathrm{d}x$$

I tried the substitution:

$$f(a,b) = \int_0^{2π} e^{acos(x)}\cos(b\sin x) \, \mathrm{d}x$$

So then:

$$\frac {\partial f} {\partial a} = \int_0^{2π} e^{a\cos(x)}\cos(b\sin x)\cos x \, \mathrm{d}x$$

and:

$$\frac {\partial f} {\partial b} = -\int_0^{2π} e^{a\cos(x)}\sin(b\sin x)\sin x \, \mathrm{d}x$$

Then what i did was try to integrate $$\frac{\partial f}{\partial b}$$ by parts, making:

$$dv = -e^{a\cos x}\sin x\, \mathrm{d}x$$

so:

$$v = \frac{1}{a}e^{a\cos x}$$

and:

$$u = \sin(b\sin x)$$

so:

$$du = \cos(b\sin x)b\cos x$$

Then we got that:

$$\frac{\partial f}{\partial b} = - \frac{b}{a} \int_0^{2π} e^{a\cos(x)}\cos(b\sin x)\cos x \, \mathrm{d}x = -\frac{b}{a} \frac{\partial f}{\partial a}$$

(Because evaluating $$uv$$ from $$0$$ to $$2π$$ gives us $$0$$)

So we get the partial differential equation:

$$\frac{1}{b} \frac{\partial f}{\partial b} = - \frac{1}{a} \frac{\partial f}{\partial a}$$

I solved this by separation of variables, making $$f(a,b) = A(a)B(b)$$. The end result was:

$$f(a,b) = Ce^{\frac{h^2}{2}(a^2 - b^2)}$$

Where $$C$$ is an integration constant and $$h$$ comes from solving the ODE associated with each $$A$$ and $$B$$.

So, my problem begins with this, because I am not sure of my result, this is mostly because of the $$h$$, should I pick an specific value? Or my result is just wrong? If this is the case, where did I made the mistake? Because evaluating $$f(0,0)$$ we get: $$f(0,0) = C = 2π$$, and so: $$f(1,1) = 2π$$, that is the correct result evaluating in wolfram alfa. I just want to know if there is a way in wich I should get the value.

Let $$I(a)=\int_{0}^{2\pi}e^{a\cos x}\cos(a\sin x)dx$$ Then \begin{align} I’(a) &=\int_0^{2\pi} e^{a\cos x}(\cos(a\sin x)\cos x- \sin(a\sin x)\sin x)dx\\ &=\frac1a\int_{0}^{2\pi}d(e^{a\cos x}\sin(a\sin x))=0\\ \end{align} which leads to $$\int_{0}^{2\pi}e^{\cos x}\cos(\sin x)dx = I(1) = I(0)=2\pi$$