Problem 6 - A First Course in Complex Analysis with Applications

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: Identify the Integral and Setup Substitution

The integral to evaluate is \( \int_{0}^{\pi} \frac{1}{1+\sin^2 \theta} \, d\theta \). For this, we'll use the Weierstrass substitution \( \theta = 2 \tan^{-1}(t) \). This substitution transforms the trigonometric functions into rational functions, making the integral more manageable. The corresponding derivatives are derived as \( d\theta = \frac{2}{1+t^2} \, dt \) and \( \sin^2 \theta = \frac{4t^2}{(1+t^2)^2} \).

Step 2: Change Limits of Integration

After substitution, the limits change as well. When \( \theta = 0 \), \( t = \tan(0) = 0 \). When \( \theta = \pi \), \( t = \tan(\frac{\pi}{2}) \), which tends towards \( \infty \). Thus, the limits of integration change from \( 0 \to \pi \) to \( 0 \to \infty \).

Step 3: Substitute and Simplify

Substitute \( \sin^2 \theta = \frac{4t^2}{(1+t^2)^2} \) and \( d\theta = \frac{2}{1+t^2} \, dt \) into the integral. The integral becomes \[ \int_{0}^{\infty} \frac{1}{1+\frac{4t^2}{(1+t^2)^2}} \cdot \frac{2}{1+t^2} \, dt. \] Simplifying gives \[ \int_{0}^{\infty} \frac{2(1+t^2)}{(1+4t^2)} \, dt \].

Step 4: Solve the Integral

The integral \( \int_{0}^{\infty} \frac{2(1+t^2)}{1+4t^2} \, dt \) can be approached by rewriting it as \[ 2 \int_{0}^{\infty} \left( \frac{1}{1+4t^2} + \frac{1}{4t^2} \right) \, dt. \] Recognizing standard integral forms, compute \( \int_{0}^{\infty} \frac{1}{1+4t^2} \, dt = \frac{\pi}{4} \) and that \( \int_{0}^{\infty} \frac{1}{4t^2} \, dt = 0 \) (the second term doesn't contribute). Hence, the original integral resolves to \( \pi/2 \).

Step 5: Final Result

Combining all parts, the original integral \( \int_{0}^{\pi} \frac{1}{1+\sin^2 \theta} \, d\theta \) evaluates to \( \pi/2 \).

Key Concepts

Weierstrass Substitution

The Weierstrass substitution is a clever technique used in calculus to simplify trigonometric integrals by transforming them into rational functions. This is particularly useful because rational functions are generally easier to integrate.
The basic idea is to replace variables involving trigonometric functions like sine or cosine with a new variable that uses a tangent function. For example, in our exercise, we used the substitution \( \theta = 2 \tan^{-1}(t) \).
This substitution changes the original variable \( \theta \) into a new variable \( t \), which can make solving the integral more straightforward. As part of this, we also adjust the differential \( d\theta \) to match the new variable by using the derivative \( \frac{d\theta}{dt} = \frac{2}{1+t^2} \).
This transformation simplifies even complex trigonometric functions like \( \sin^2 \theta \) into expressions involving \( t \), making the calculus more manageable.

Definite Integral

A definite integral, like the one in this exercise, involves finding the area under a curve within a specified range—here from 0 to \( \pi \). The limits of integration determine the starting and ending points of this interval.
When solving a definite integral, it’s essential to carefully account for these limits, as they influence the final result of your computation.

• The process begins by calculating an antiderivative of the function you’re working with.
• Then you evaluate this antiderivative at both the upper and lower limits of integration.
• Finally, the value of the lower limit antiderivative is subtracted from that of the upper limit to give the result.

In this exercise, after substitution, both the function and its limits are transformed, which we will discuss further under the limit of integration section.

Integration Techniques

Various integration techniques exist to help tackle a wide range of integrals, each suited to different functions. Among the most common are substitution, integration by parts, and recognizing standard form integrals.
In this exercise, we saw the use of substitution, particularly the Weierstrass substitution. Through substitution, the complex trigonometric integral transforms into a more straightforward rational function integral.
Moreover, simplifying the integrand to a combination of standard integral forms is another powerful technique. By breaking down the original problem, we utilized known integral results:

• \( \int \frac{1}{1+4t^2} \, dt \), which is a standard arctangent form.
• \( \int \frac{1}{4t^2} \, dt \), a form which in this context does not contribute to the result due to its limits.

Using these techniques makes complex problems more accessible and solutions more elegant.

Limit of Integration

The limit of integration specifies the boundaries for evaluating a definite integral. When using substitution methods, these limits can evolve, as they need to match the new variable.
In this exercise, as \( \theta \) transitions from 0 to \( \pi \), we evaluate the corresponding values of \( t \):

• When \( \theta = 0 \), \( t = \tan(0) = 0 \).
• When \( \theta = \pi \), \( t = \tan\left(\frac{\pi}{2}\right) \), approaching \( \infty \).

This alteration in limits, from \( 0 \) to \( \pi \) for \( \theta \) to \( 0 \) to \( \infty \) for \( t \), is crucial to computing the definite integral correctly with the new substitution.
Understanding these limits ensures that the evaluation of the integral is accurate and aligns with the new variable and integrand.