The image contains two integrals with a goal to show their equivalence. The integrals are:

1. $\int_0^{2\pi} \frac{e^{mx} \cos(n\theta)\, d\theta}{5 - 4\cos(\theta - \phi)} = \frac{2\pi}{5} e^{m\phi} \cos(n\phi)$
2. $\int_0^{2\pi} \frac{e^{mx} \sin(n\theta)\, d\theta}{5 - 4\cos(\theta - \phi)} = \frac{2\pi}{5} e^{m\phi} \sin(n\phi)$

Let me break down the steps to solve these integrals.

Step 1: Use the Complex Exponential Representation

The complex exponential representation for cosine and sine can be written as:

• $\cos(n\theta) = \frac{e^{in\theta} + e^{-in\theta}}{2}$
• $\sin(n\theta) = \frac{e^{in\theta} - e^{-in\theta}}{2i}$

Step 2: Consider the Integral's Denominator Structure

The denominator $5 - 4\cos(\theta - \phi)$ can be recognized as part of a trigonometric identity. This structure often relates to Chebyshev polynomials or complex function representations. To simplify, convert the cosine term into its exponential form:

$\cos(\theta - \phi) = \frac{e^{i(\theta - \phi)} + e^{-i(\theta - \phi)}}{2}$

This may transform the integral into a form that can be evaluated using residue calculus or by recognizing it as a known Fourier series expansion.

Step 3: Use the Residue Theorem

The key idea is to transform the integral into a contour integral in the complex plane. By substituting $z = e^{i\theta}$, we convert the integral over $\theta$ to a contour integral over the unit circle in the complex plane:

• $d\theta = \frac{dz}{iz}$
• $\cos(\theta - \phi)$ transforms into a function of $z$.

This will simplify the integral, allowing us to apply the residue theorem to calculate the value.

Step 4: Find the Poles and Residues

The poles of the integrand will occur where the denominator vanishes. Solve for the values of $z$ where:

$5 - 4\cos(\theta - \phi) = 0$

Once the poles are identified, calculate the residues at those points. Summing the residues inside the unit circle will give the result of the contour integral.

Step 5: Compare with the Given Result

The final expression should match the given right-hand side expressions $\frac{2\pi}{5} e^{m\phi} \cos(n\phi)$ and $\frac{2\pi}{5} e^{m\phi} \sin(n\phi)$, confirming the equality.

Next Steps

• Would you like to see detailed workings of the complex analysis method (residue calculus)?
• Do you have any specific questions about transforming trigonometric integrals into complex plane integrals?

Here are 5 related questions to expand your understanding:

1. How is the residue theorem applied to contour integrals over the unit circle?
2. What are Chebyshev polynomials and how do they relate to trigonometric integrals?
3. How do you handle integrals involving exponential and trigonometric functions?
4. What is the significance of Fourier series in solving periodic integrals?
5. How does the complex plane transformation simplify trigonometric integrals?

Tip: When facing integrals involving periodic functions like sine and cosine, converting them to complex exponentials often simplifies the problem, especially when paired with contour integration techniques.