Evaluate the real integral ∫ 𝑑𝜃 17−8 sin 𝜃 2𝜋 0 using Cauchy’s Residue theorem

To evaluate the real integral \(\int_{0}^{2\pi} \frac{d\theta}{17 - 8\sin\theta}\) using Cauchy's Residue Theorem, we follow these steps:

### Step 1: Convert the integral into a contour integral

First, express the integral in terms of the complex variable \(z = e^{i\theta}\), where \(\theta \) ranges from 0 to \(2\pi\). Note that \(d\theta = \frac{dz}{iz}\), \( \sin\theta = \frac{z - z^{-1}}{2i} \), and \( \cos\theta = \frac{z + z^{-1}}{2} \).

The integral becomes:
\[
\int_{0}^{2\pi} \frac{d\theta}{17 - 8\sin\theta} = \oint_{|z|=1} \frac{1}{17 - 8\frac{z - z^{-1}}{2i}} \cdot \frac{dz}{iz}
\]

Simplifying further:
\[
\oint_{|z|=1} \frac{1}{17 - 4i(z - z^{-1})} \cdot \frac{dz}{iz}
\]

Multiply numerator and denominator by \(z\) to get rid of \(z^{-1}\):
\[
\oint_{|z|=1} \frac{z^{-1}}{17z - 4i(z^2 - 1)} \cdot \frac{dz}{iz}
\]

Simplify the expression:
\[
\oint_{|z|=1} \frac{dz}{4z^2 + 17iz - 4}
\]

### Step 2: Identify the poles inside the unit circle

The integrand is a rational function, and the poles are the roots of the quadratic equation in the denominator:
\[
4z^2 + 17iz - 4 = 0
\]

Solve for \(z\):
\[
z = \frac{-17i \pm \sqrt{(-17i)^2 - 4(4)(-4)}}{2(4)}
\]
\[
z = \frac{-17i \pm \sqrt{-289 - 64}}{8}
\]
\[
z = \frac{-17i \pm \sqrt{-353}}{8}
\]
\[
z = \frac{-17i \pm i\sqrt{353}}{8}
\]
\[
z = \frac{-17 \pm \sqrt{353}}{8} i
\]

Thus, the roots (poles) are:
\[
z_1 = \frac{-17 + \sqrt{353}}{8} i, \quad z_2 = \frac{-17 - \sqrt{353}}{8} i
\]

We need to determine which of these roots lies inside the unit circle \(|z|=1\). Consider the magnitudes:
\[
|z_1| = \left|\frac{-17 + \sqrt{353}}{8}\right|, \quad |z_2| = \left|\frac{-17 - \sqrt{353}}{8}\right|
\]

Given that \(z_2\) has a negative real part and a larger magnitude, it lies outside the unit circle. Thus, only \(z_1\) lies inside the unit circle.

### Step 3: Apply the Residue Theorem

The integral over the unit circle is \(2\pi i\) times the residue at \(z_1\):
\[
\oint_{|z|=1} \frac{dz}{4z^2 + 17iz - 4} = 2\pi i \cdot \text{Res}\left(\frac{1}{4z^2 + 17iz - 4}, z_1\right)
\]

To find the residue at \(z_1\), differentiate the denominator \(f(z) = 4z^2 + 17iz - 4\) and evaluate it at \(z_1\):
\[
f'(z) = 8z + 17i
\]
\[
f'(z_1) = 8z_1 + 17i
\]

Thus, the residue is:
\[
\text{Res}\left(\frac{1}{4z^2 + 17iz - 4}, z_1\right) = \frac{1}{f'(z_1)} = \frac{1}{8z_1 + 17i}
\]

### Step 4: Evaluate the final integral

Substitute the residue back into the integral:
\[
\int_{0}^{2\pi} \frac{d\theta}{17 - 8\sin\theta} = 2\pi i \cdot \frac{1}{8z_1 + 17i}
\]

Substitute \(z_1\) and simplify to get the final result.

### Conclusion

The final integral is:
\[
\int_{0}^{2\pi} \frac{d\theta}{17 - 8\sin\theta} = \frac{2\pi}{\sqrt{17^2 + 8^2}} = \frac{2\pi}{19}
\]

Would you like further clarification on any part of the process?

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### Related Questions:
1. What is the significance of the Residue Theorem in complex analysis?
2. How do you determine which poles lie inside the unit circle?
3. Can the method be applied to integrals with different trigonometric functions?
4. How do you interpret the result geometrically?
5. What are some other applications of Cauchy's Residue Theorem?

### Tip:
When dealing with complex integrals, always ensure to correctly identify the poles within the contour you are integrating over.

Learn how to evaluate the real integral ∫ 𝑑𝜃 17−8 sin 𝜃 2𝜋 0 using Cauchy's Residue Theorem with a detailed step-by-step solution. This problem covers complex analysis, the Residue Theorem, and trigonometric integrals, suitable for college-level students.

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