Problem 6 - Advanced Engineering Mathematics

Question

$$\begin{aligned} \int_{0}^{\pi} \frac{d \theta}{1+\sin ^{2} \theta} &=\frac{1}{2} \int_{0}^{2 \pi} \frac{d \theta}{1+\sin ^{2} \theta}=-\frac{2}{i} \oint_{C} \frac{z}{z^{4}-6 z^{2}+1} d z \\ &=\left(-\frac{2}{i}\right) 2 \pi i[\operatorname{Res}(f(z), \sqrt{3-2 \sqrt{2}})+\operatorname{Res}(f(z),-\sqrt{3-2 \sqrt{2}})]=\frac{\pi}{\sqrt{2}} \end{aligned}$$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{\pi}{\sqrt{2}} \).

Step 1: Recognize the integral

The given integral \[\int_{0}^{\pi} \frac{d \theta}{1+\sin^{2} \theta}\]is a definite integral over the interval from 0 to \( \pi \). The integrand is a trigonometric function that can potentially be transformed using complex numbers for easier evaluation.

Step 2: Use symmetry to relate integrals

Recognize that the integral from 0 to \( \pi \) is half of an integral from 0 to \( 2\pi \) due to symmetry, stated as\[\int_{0}^{\pi} \frac{d \theta}{1+\sin^{2} \theta} = \frac{1}{2} \int_{0}^{2\pi} \frac{d \theta}{1+\sin^{2} \theta}.\]This allows us to consider the full period of the sine function for transformations.

Step 3: Transform to complex plane

Rewrite the integral in terms of a complex variable \( z \) where \( z = e^{i\theta} \). The relation \( \sin \theta = \frac{1}{2i}(z - z^{-1}) \) allows transforming the integral into a contour integral:\[\int \frac{d\theta}{1 + \sin^2 \theta} \rightarrow -\frac{2}{i} \oint_{C} \frac{z}{z^4 - 6z^2 + 1} \, dz,\]where C is the unit circle in the complex plane.

Step 4: Identify poles and residues

The function to be integrated, \( \frac{z}{z^4 - 6z^2 + 1} \), can be factored and its poles can be found. Determine the residues at the poles inside the unit circle. These residues are at \( \sqrt{3-2\sqrt{2}} \) and \( -\sqrt{3-2\sqrt{2}} \).

Step 5: Apply the residue theorem

Using the residue theorem, compute the integral:\[-\frac{2}{i} \times 2\pi i \left[\operatorname{Res}(f(z), \sqrt{3-2\sqrt{2}}) + \operatorname{Res}(f(z), -\sqrt{3-2\sqrt{2}})\right],\]which simplifies to \( \frac{\pi}{\sqrt{2}} \) by calculating the residues at the poles.

Key Concepts

Residue Theorem

The Residue Theorem is an essential tool in complex analysis used to evaluate complex integrals, especially for functions with singularities. This theorem connects the integral over a closed contour in the complex plane to the sum of residues enclosed by the contour. A residue of a function at a specific point is essentially a coefficient of the \( rac{1}{z-a} \) term in its Laurent series expansion.

When using the Residue Theorem, it simplifies evaluating integrals by focusing on the singularities of the integrand. The theorem states: \[ \oint_{C} f(z) \, dz = 2 \pi i \sum \operatorname{Res}(f, a_i),\] where \( a_i \) are the singular points within the contour \( C \). By calculating the residues at these points and summing them, the value of the integral is determined effectively.

In the exercise, the use of the Residue Theorem allows for the transformation of the initial definite integral \( \int_{0}^{\pi} \frac{d \theta}{1+\sin^2 \theta} \) into a contour integral that is much simpler to evaluate.

Contour Integration

Contour Integration involves integrating a complex function along a specific contour or path in the complex plane. This path, often a closed loop, serves as the domain of integration for the complex integral.

In practice, contour integration simplifies the evaluation of integrals with periodic or complex trigonometric functions. It leverages properties of complex numbers such as analyticity, allowing complex integrals to provide solutions to real problems.

In the given exercise, the original integral is complexified by substituting real variables with complex ones. The path of integration, denoted as \( C \), is typically chosen as the unit circle \( |z| = 1 \). This approach turns a trigonometric integral into one involving the complex variable \( z = e^{i\theta} \), resulting in straightforward evaluation via techniques like poles and residues.

Trigonometric Substitution

Trigonometric Substitution is a technique used in calculus to simplify integrals involving trigonometric functions. By substituting the trigonometric function with an equivalent expression in terms of a complex variable, the integral can often be transformed into a more manageable form.

For example, in the given problem, the substitution \( z = e^{i\theta} \) allows for the expression \( \sin \theta \) to be replaced by \( \frac{1}{2i} (z - z^{-1}) \). This substitution reformulates the trigonometric integral into a rational function of \( z \), suitable for contour integration.

By transforming a typical integral into one involving complex variables, trigonometric substitution streamlines the process of integration, reducing trigonometric complexities to simpler algebraic equivalents.

Definite Integrals

Definite Integrals represent the signed area under a curve within a specified interval on the real number line. They provide a way to calculate total accumulated quantities, like area or volume, based on a function over a certain domain.

In the context of this exercise, the integral \( \int_{0}^{\pi} \frac{d \theta}{1+\sin^2 \theta} \) is a definite integral over the interval \( [0, \pi] \). By using the properties of definite integrals, such as symmetry or periodicity, one can simplify the evaluation as seen where it equates half of the interval over \( [0, 2\pi] \).

Understanding definite integrals is crucial in real-world applications and in situations when integrating bounded functions to find quantities represented by the function's values over a specific range.