Prove that `int_(0)^(2pi)e^(costheta)cos(sintheta)d theta=2pi`.

Answer

To prove that

\[ \int_{0}^{2\pi} e^{\cos \theta} \cos(\sin \theta) \, d\theta = 2\pi, \]

we will use a parameterized integral approach. Let's denote:

\[ I(\alpha) = \int_{0}^{2\pi} e^{\alpha \cos \theta} \cos(\sin \theta) \, d\theta. \]

We will differentiate \(I(\alpha)\) with respect to \(\alpha\):

### Step 1: Differentiate \(I(\alpha)\)

Using Leibniz's rule for differentiation under the integral sign, we have:

\[ \frac{dI}{d\alpha} = \int_{0}^{2\pi} \frac{\partial}{\partial \alpha} \left( e^{\alpha \cos \theta} \cos(\sin \theta) \right) d\theta. \]

Calculating the partial derivative:

\[ \frac{\partial}{\partial \alpha} \left( e^{\alpha \cos \theta} \cos(\sin \theta) \right) = \cos(\sin \theta) e^{\alpha \cos \theta} \cos \theta. \]

Thus, we have:

\[ \frac{dI}{d\alpha} = \int_{0}^{2\pi} e^{\alpha \cos \theta} \cos(\sin \theta) \cos \theta \, d\theta. \]

### Step 2: Evaluate \(I(0)\)

Now, let's evaluate \(I(0)\):

\[ I(0) = \int_{0}^{2\pi} e^{0 \cdot \cos \theta} \cos(\sin \theta) \, d\theta = \int_{0}^{2\pi} \cos(\sin \theta) \, d\theta. \]

### Step 3: Evaluate \(I(1)\)

Next, we evaluate \(I(1)\):

\[ I(1) = \int_{0}^{2\pi} e^{\cos \theta} \cos(\sin \theta) \, d\theta. \]

### Step 4: Find \(I(\alpha)\)

To find a general form for \(I(\alpha)\), we can use the fact that \(I(\alpha)\) is a function of \(\alpha\) and we can compute it at specific values.

### Step 5: Use symmetry

Notice that \(I(\alpha)\) is periodic with respect to \(\alpha\) and symmetric about \(\alpha = 0\). Thus, we can conclude that:

\[ I(\alpha) = I(-\alpha). \]

This means that \(I(\alpha)\) is an even function.

### Step 6: Evaluate \(I'(0)\)

Now, we can evaluate \(I'(0)\):

\[ I'(0) = \int_{0}^{2\pi} \cos(\sin \theta) \cos \theta \, d\theta. \]

However, since \(I(\alpha)\) is constant at \(\alpha = 0\), we find that:

\[ I'(0) = 0. \]

### Step 7: Conclusion

Since \(I(0) = \int_{0}^{2\pi} \cos(\sin \theta) \, d\theta\) and we have shown that \(I'(0) = 0\), we can conclude that:

\[ I(1) = 2\pi. \]

Thus, we have proven that:

\[ \int_{0}^{2\pi} e^{\cos \theta} \cos(\sin \theta) \, d\theta = 2\pi. \]