Problem 6 - Advanced Engineering Mathematics

Question

Evaluate the given trigonometric integral. \(\int_{0}^{\pi} \frac{1}{1+\sin ^{2} \theta} d \theta\)

Step-by-Step Solution

Answer

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

Step 1: Understand Problem Requirements

We need to evaluate the trigonometric integral \(\int_{0}^{\pi} \frac{1}{1+\sin ^{2} \theta} d \theta\). Our task is to find the definite integral over the interval from 0 to \(\pi\).

Step 2: Simplify the Integrand

Notice that the denominator \(1 + \sin^2 \theta\) does not simplify neatly with basic trigonometric identities. Thus, consider using trigonometric substitution or a known trigonometric identity if possible.

Step 3: Use Trigonometric Identities

Recall the identity \(\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}\). Substitute this into the integrand to get \(1 + \frac{1 - \cos(2\theta)}{2}\), which simplifies to \(\frac{3}{2} - \frac{1}{2}\cos(2\theta)\).

Step 4: Modify the Integrand

The integrand can now be expressed as a single fraction: \(\int_{0}^{\pi} \frac{2}{3 - \cos(2\theta)} d\theta\). This modification allows us to consider further simplification or substitution.

Step 5: Consider Appropriate Substitution

Since the problem involves periodic functions, we might need to involve a substitution that simplifies the cosine, such as \(t = \tan(\theta/2)\), leading to the Weierstrass substitution. Here, \(d\theta = \frac{2}{1+t^2} dt\).

Step 6: Apply Weierstrass Substitution

Substituting \(\cos(\theta) = \frac{1 - t^2}{1 + t^2}\), we can write \(\cos(2\theta) = \frac{1-2t^2+t^4}{1+2t^2+t^4}\). Simplifying further, the integral can be evaluated in terms of \(t\) from \(t=0\) to \(t=\infty\).

Step 7: Evaluate the New Integral

The integral becomes \(\int \frac{4}{(3+t^2)(1+t^2)} dt\), which splits into partial fractions: \(\int \left( \frac{A}{3+t^2} + \frac{B}{1+t^2} \right) dt\), where \(A\) and \(B\) are determined by equating coefficients.

Step 8: Solve Partial Fractions

Equate and solve for \(A\) and \(B\). You will find \(A = 2/3\) and \(B = -2/3\). The integral then splits into two simpler integrals involving the arctangent function: \(\frac{2}{3} \text{tan}^{-1}(\frac{\sqrt{3}}{3}t) - \frac{2}{3} \text{tan}^{-1}(t)\) evaluated from 0 to \(\infty\).

Step 9: Determine Limits and Evaluate

Compute the arctan values as the limits approach from 0 to \(\infty\). Using properties that \(\text{tan}^{-1}(\infty) = \frac{\pi}{2}\), determine the subtraction helps resolve the final value.

Step 10: Write the Final Result

The definite integral evaluates to \(\frac{2\pi}{3} - \frac{2\pi}{3} = \frac{\pi}{3}\), after carefully observing symmetry and cancellation over the interval.

Key Concepts

Definite Integral

When dealing with the problem of evaluating \(\int_{0}^{\pi} \frac{1}{1+\sin ^{2} \theta} d \theta\), the key task is finding the definite integral over a specified interval, in this case from 0 to \(\pi\). The definite integral calculates the area under a curve, bounded by the given limits. Here, we evaluate the integral not just as a general form but specifically from 0 to \(\pi\). This requires a precise approach to handle trigonometric functions within this interval.

• The definite integral provides a numerical result, reflecting the accumulation of areas, accounting for both the shapes involved and periodic properties.
• In this instance, since the function isn’t simplified by elementary identities, strategic substitutions are considered crucial.
• After performing appropriate substitutions, evaluate by integrating from the lower to the upper limit, ensuring careful application of the fundamental theorem of calculus.

Trigonometric Substitution

Trigonometric substitution is a technique used to simplify the integrand to a more familiar form. In this exercise, the expression \(1 + \sin^2 \theta\) doesn’t transform neatly with basic identities, leading us to explore substitutions or ways to reorganize this term.
The idea is to use trigonometric identities or substitutions to convert complex expressions into simpler ones.

• Recall an identity such as \(\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}\), which aids in converting higher power trigonometric terms into integrable forms.
• With this identity, the integrand simplifies to \(\frac{2}{3 - \cos(2\theta)}\), setting the stage for further substitution.

Substitution facilitates oversimplifying the problem by diving deeper into trigonometric relations. This technique often requires intuition about how different trigonometric identities relate and can dismantle more complex expressions into fractions or regular integrals.

Partial Fractions

When integrating a complex rational function, partial fractions become a valuable tool for simplifying the process. In the case of the transformed integral \(\int \frac{4}{(3+t^2)(1+t^2)} dt\), partial fraction decomposition is used. This decomposition allows us to split the integrand into simpler, separate parts which can be independently integrated.

• Express the integrand as a sum of simpler fractions. Here, evaluate as \(\int \left( \frac{A}{3+t^2} + \frac{B}{1+t^2} \right) dt\).
• Determining constants \(A\) and \(B\) involves equating the terms and resolving the system of equations, finding that \(A = \frac{2}{3}\) and \(B = -\frac{2}{3}\).

This decomposition is crucial as it breaks down complex blends of polynomials into tractable pieces, simplifying integration to simple recognized forms such as arctangent, which can be managed directly.

Weierstrass Substitution

Weierstrass substitution is a clever technique often applied in integrals with trigonometric functions. Known also as the tangent half-angle substitution, it simplifies integrals by changing variables to make them linear. In essence, set \(t = \tan(\frac{\theta}{2})\), then derive Weierstrass relations to substitute and simplify functions of \(\theta\).

• With Weierstrass substitution, the trigonometric integrand translate into rational expressions of \(t\). Specifically, handling cosine terms becomes \(\cos(\theta) = \frac{1-t^2}{1+t^2}\).
• This method allows the integral to transform into one involving simpler polynomials over \(t\), which are generally easier to integrate.

Apply the substitution judiciously, observing bounds transform from \(t = 0\) to \(t = \infty\), ensuring to check symmetry and limits carefully. This process deeply aligns with partial fraction decomposition, making the entire integration more straightforward by reducing complexity.