To prove that
\[
\int_{0}^{2\pi} e^{\cos \theta} \cos(\sin \theta) \, d\theta = 2\pi,
\]
we will use a parameterized integral approach.
### Step 1: Define the Integral with a Parameter
Let
\[
I(\alpha) = \int_{0}^{2\pi} e^{\alpha \cos \theta} \cos(\sin \theta) \, d\theta.
\]
We want to find \(I(1)\).
### Step 2: Differentiate with Respect to \(\alpha\)
Differentiate \(I(\alpha)\) with respect to \(\alpha\):
\[
\frac{dI}{d\alpha} = \int_{0}^{2\pi} \frac{\partial}{\partial \alpha} \left( e^{\alpha \cos \theta} \cos(\sin \theta) \right) d\theta.
\]
Using the chain rule, we have:
\[
\frac{\partial}{\partial \alpha} \left( e^{\alpha \cos \theta} \cos(\sin \theta) \right) = e^{\alpha \cos \theta} \cos(\sin \theta) \cos \theta.
\]
Thus,
\[
\frac{dI}{d\alpha} = \int_{0}^{2\pi} e^{\alpha \cos \theta} \cos(\sin \theta) \cos \theta \, d\theta.
\]
### Step 3: Evaluate \(I(0)\)
Now, let's evaluate \(I(0)\):
\[
I(0) = \int_{0}^{2\pi} e^{0 \cdot \cos \theta} \cos(\sin \theta) \, d\theta = \int_{0}^{2\pi} \cos(\sin \theta) \, d\theta.
\]
Since \(\cos(\sin \theta)\) is periodic with period \(2\pi\), we can evaluate this integral directly.
### Step 4: Evaluate \(I(1)\)
Next, we need to evaluate \(I(1)\):
\[
I(1) = \int_{0}^{2\pi} e^{\cos \theta} \cos(\sin \theta) \, d\theta.
\]
### Step 5: Use the Fundamental Theorem of Calculus
Using the Fundamental Theorem of Calculus, we can find \(I(1)\) by integrating \( \frac{dI}{d\alpha} \):
\[
I(1) - I(0) = \int_{0}^{1} \frac{dI}{d\alpha} \, d\alpha.
\]
### Step 6: Evaluate the Integral
Now, we need to evaluate \( \frac{dI}{d\alpha} \) at \(\alpha = 0\):
\[
\frac{dI}{d\alpha} \bigg|_{\alpha=0} = \int_{0}^{2\pi} \cos(\sin \theta) \cos \theta \, d\theta.
\]
This integral evaluates to \(0\) because the integrand is an odd function over the interval \([0, 2\pi]\).
### Step 7: Conclude the Result
Since \( \frac{dI}{d\alpha} \bigg|_{\alpha=0} = 0\), it follows that:
\[
I(1) = I(0) + 0 = I(0).
\]
We also know that \(I(0) = 2\pi\), thus:
\[
I(1) = 2\pi.
\]
### Final Result
Therefore, we conclude that:
\[
\int_{0}^{2\pi} e^{\cos \theta} \cos(\sin \theta) \, d\theta = 2\pi.
\]
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