Evaluate the following integral by using Causs’ Mean Value Theorem.
Z 2π
0
sec(e
iθ) dθ.

To evaluate the integral \(\int_{0}^{2\pi} \sec(e^{i\theta}) \, dθ\) using the Mean Value Theorem (MVT) for integrals, let's proceed with the following steps:

1. **Rewrite the integral in a more familiar form**:
   
   First, note that \(e^{i\theta} = \cos(\theta) + i \sin(\theta)\). Hence, the integrand \(\sec(e^{i\theta})\) becomes:

   \[
   \sec(e^{i\theta}) = \frac{1}{\cos(e^{i\theta})}
   \]

2. **Application of the MVT for integrals**:
   
   The MVT for integrals states that if \(f\) is continuous on \([a, b]\), then there exists some \(c \in [a, b]\) such that:

   \[
   \int_{a}^{b} f(x) \, dx = f(c) (b - a)
   \]

   In our case, \(a = 0\), \(b = 2\pi\), and \(f(\theta) = \sec(e^{i\theta})\). Therefore, there exists some \(c \in [0, 2\pi]\) such that:

   \[
   \int_{0}^{2\pi} \sec(e^{i\theta}) \, dθ = \sec(e^{ic}) \cdot (2\pi - 0)
   \]

   Simplifying this, we get:

   \[
   \int_{0}^{2\pi} \sec(e^{i\theta}) \, dθ = 2\pi \cdot \sec(e^{ic})
   \]

3. **Finding \(\sec(e^{ic})\)**:
   
   We need to evaluate \(\sec(e^{ic})\) for some \(c \in [0, 2\pi]\). Note that \(e^{ic}\) will trace out a unit circle in the complex plane as \(c\) varies from 0 to \(2\pi\).

   Since \(e^{i\theta}\) is periodic with period \(2\pi\) and traces the unit circle in the complex plane, \(\cos(e^{ic})\) will take values on this circle as well. For simplicity, we can choose a specific point like \(c = \pi/2\) to evaluate \(\sec(e^{i(\pi/2)})\):

   \[
   e^{i(\pi/2)} = \cos(\pi/2) + i \sin(\pi/2) = i
   \]

   Hence,

   \[
   \sec(i) = \frac{1}{\cos(i)}
   \]

4. **Computing \(\cos(i)\)**:
   
   Using the exponential form of the cosine function, we have:

   \[
   \cos(z) = \frac{e^{iz} + e^{-iz}}{2}
   \]

   Plugging in \(z = i\):

   \[
   \cos(i) = \frac{e^{i(i)} + e^{-i(i)}}{2} = \frac{e^{-1} + e^{1}}{2} = \frac{1/e + e}{2}
   \]

5. **Evaluating \(\sec(i)\)**:
   
   Therefore,

   \[
   \sec(i) = \frac{1}{\cos(i)} = \frac{2}{e + 1/e} = \frac{2}{e + \frac{1}{e}} = \frac{2e}{e^2 + 1}
   \]

   So, we finally have:

   \[
   \int_{0}^{2\pi} \sec(e^{i\theta}) \, dθ = 2\pi \cdot \sec(i) = 2\pi \cdot \frac{2e}{e^2 + 1}
   \]

   Simplifying this, we get:

   \[
   \int_{0}^{2\pi} \sec(e^{i\theta}) \, dθ = \frac{4\pi e}{e^2 + 1}
   \]

Thus, the evaluated integral is:

\[
\int_{0}^{2\pi} \sec(e^{i\theta}) \, dθ = \frac{4\pi e}{e^2 + 1}
\]

Would you like further details or have any questions about this solution? Here are some related questions you might find interesting:

1. What is the Mean Value Theorem for integrals?
2. How do you convert exponential forms to trigonometric forms?
3. What is the significance of the unit circle in complex analysis?
4. How can you simplify complex exponential functions?
5. What are the properties of trigonometric functions in the complex plane?
6. Can you explain the periodicity of exponential functions?
7. How does the secant function behave with complex arguments?
8. What are the applications of the Mean Value Theorem in real-world problems?

**Tip:** When dealing with integrals involving complex functions, converting to exponential forms and using properties of the unit circle can significantly simplify the problem.

Learn how to evaluate the integral ∫ sec(e^(iθ)) dθ using Cauchy's Mean Value Theorem for integrals. This solution involves converting exponential forms to trigonometric forms and applying the Mean Value Theorem. Suitable for advanced undergraduate students and anyone interested in complex analysis and integral calculus.

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