Problem 34 - Advanced Engineering Mathematics

Question

$$\begin{aligned} &\int_{0}^{2 \pi} \frac{\sin ^{2} \theta}{a+b \cos \theta} d \theta=\frac{i}{2 b} \oint_{C} \frac{z^{2}-1}{z^{2}\left(z-r_{1}\right)\left(z-r_{2}\right)} d z(C \text { is }|z|=1) \text { where } r_{1}=(-a+\sqrt{a^{2}-b^{2}}) / b\\\ &r_{2}=(-a-\sqrt{a^{2}-b^{2}}) / b . \text { Now } \end{aligned}$$ $$\oint_{C} \frac{z^{2}-1}{z^{2}\left(z-r_{1}\right)\left(z-r_{2}\right)} d z=2 \pi i\left[\operatorname{Res}(f(z), 0)+\operatorname{Res}\left(f(z), r_{1}\right)\right]=2 \pi i\left[-\frac{2 a}{b}+\frac{2 \sqrt{a^{2}-b^{2}}}{b}\right]$$ Thus, \int_{0}^{2 \pi} \frac{\sin ^{2} \theta}{a+b \cos \theta} d \theta=\frac{2 \pi}{b^{2}}(a-\sqrt{a^{2}-b^{2}}), \quad a > b > 0

Step-by-Step Solution

Answer

The integral evaluates to \[ \frac{2\pi}{b^2}(a - \sqrt{a^2 - b^2}) \] with \( a > b > 0 \).

Step 1: Convert the Integral to Complex Form

The given real integral \( \int_{0}^{2 \pi} \frac{\sin^{2}\theta}{a+b\cos\theta} \, d\theta \) can be expressed in terms of a complex contour integral. By using the substitution \( z = e^{i\theta} \), \( d\theta = \frac{dz}{iz} \), and \( \cos\theta = \frac{1}{2}(z + z^{-1}) \), we convert the trigonometric integral into a complex contour integral around the unit circle \( |z|=1 \).

Step 2: Setup the Complex Integral

Substitute \( \sin^{2}\theta = \frac{1}{2}(1 - \cos{2\theta}) \) into the integral and after applying the substitution, the integral becomes \[ \int_{0}^{2 \pi} \frac{\sin^{2}\theta}{a+b\cos\theta} \, d\theta = \frac{i}{2b} \oint_{C} \frac{z^{2} - 1}{z^{2} (z - r_1)(z - r_2)} \, dz \] where \( r_1 = \frac{-a + \sqrt{a^2 - b^2}}{b} \) and \( r_2 = \frac{-a - \sqrt{a^2 - b^2}}{b} \).

Step 3: Evaluate the Residues

We need to calculate the residues of the function inside the contour at its poles. The poles within the unit circle \( |z|=1 \) are at \( z = 0 \) and potentially at \( z = r_1 \). We compute the residues: \( \operatorname{Res}(f(z), 0) = -\frac{2a}{b} \) and \( \operatorname{Res}(f(z), r_1) = \frac{2\sqrt{a^2 - b^2}}{b} \).

Step 4: Apply the Residue Theorem

According to the Residue Theorem, the integral is \[ \oint_{C} \frac{z^{2} - 1}{z^{2} (z - r_1)(z - r_2)} \, dz = 2\pi i \left( \operatorname{Res}(f(z), 0) + \operatorname{Res}(f(z), r_1) \right) \] which evaluates to \( 2\pi i \left( -\frac{2a}{b} + \frac{2\sqrt{a^2 - b^2}}{b} \right) \).

Step 5: Relate Back to Real Integral

From the relation obtained from the complex contour integral, we have that \[ \int_{0}^{2 \pi} \frac{\sin^{2}\theta}{a+b\cos\theta} \, d\theta = \frac{2\pi}{b^2}(a - \sqrt{a^2 - b^2}) \] This completes the transformation back to the original real integral.

Key Concepts

Contour Integration

In complex analysis, contour integration is a method of evaluating certain integrals along paths or contours in the complex plane. It is particularly powerful for integrating functions that are difficult using standard real analysis techniques.
Contour integrals often involve integrating over circular paths. For the given integral, we use a unit circle in the complex plane, represented as \(|z|=1\). This circle represents our contour, denoted by \(C\).
This conversion from a real integral to a complex one allows us to harness the complex plane's properties, making difficult problems more approachable.

Residue Theorem

The Residue Theorem is a crucial concept in complex analysis for evaluating integrals over closed curves. It relates the contour integral of a function to the sum of residues of its poles inside the contour.
Residues are values calculated at singularities, which are points where the function becomes undefined or infinitely large. For a given function \(f(z)\), its residue at a pole \(z_0\) is a measure of its behavior around that point.
In this exercise, the poles are determined using the roots derived from the transformation. By calculating the residues at these poles, we use the Residue Theorem to evaluate the entire contour integral around \(|z| = 1\).

• Residue at \(z=0\)
• Residue at \(z=r_1\)

Carrying this out simplifies our task to calculating the sum of these residues, turned into a product involving \(2\pi i\).

Trigonometric Substitutions

Trigonometric substitutions involve substituting trigonometric functions to simplify integrals. They are particularly useful when dealing with periodic functions like sine and cosine.
In the given problem, substituting \(z=e^{i\theta}\) helps convert the trigonometric functions into polynomial terms in \(z\).
Some other helpful substitutions are:

• \(\sin^{2}\theta = \frac{1}{2}(1 - \cos{2\theta})\)
• \(\cos\theta = \frac{1}{2}(z + z^{-1})\)

Such substitutions assist in bridging the gap between trigonometric functions and complex analysis, making complex integrals more manageable.

Complex Substitution

Complex substitution involves expressing a real integral as a complex one, using certain transformations. This technique is particularly useful for simplifying complex expressions and taking advantage of the powerful tools available in complex analysis.
In this problem, by letting \(z = e^{i\theta}\) and therefore transforming \(d\theta\) to \(\frac{dz}{iz}\), we move from a real setting to a complex domain.
This approach aids in converting a trigonometric integral into a format where applying the Residue Theorem or law of residues becomes feasible. This method enables you to take an otherwise cumbersome integral and leverage complex function theory to break it down into simpler, solvable parts.

• Converts the range of integration from a real interval to a closed path.
• Simplifies handling of trigonometric expressions.