Problem 65 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(47-70\) $$ \int_{-\pi / 2}^{\pi / 2} \frac{2 \cos \theta d \theta}{1+(\sin \theta)^{2}} $$

Step-by-Step Solution

Answer

The integral evaluates to \( \pi \).

Step 1: Identify the Integral and Substitution

The given integral is \( \int_{-\pi / 2}^{\pi / 2} \frac{2 \cos \theta \, d\theta}{1+(\sin \theta)^{2}} \). We can simplify the problem by using a trigonometric identity or substitution. Let's use the substitution \( u = \sin \theta \), which implies \( du = \cos \theta \, d\theta \).

Step 2: Change the Limits of Integration

When \( \theta = -\pi/2 \), \( u = \sin(-\pi/2) = -1 \). When \( \theta = \pi/2 \), \( u = \sin(\pi/2) = 1 \). Therefore, the limits of integration with respect to \( u \) are from \(-1\) to \(1\).

Step 3: Substitute and Transform the Integral

Substituting into the integral, we have \( \int_{-1}^{1} \frac{2 \, du}{1 + u^{2}} \). This transformation relies on the identity \( \cos \theta \, d\theta = du \).

Step 4: Recognize the Integral Form

The integral \( \int \frac{du}{1 + u^{2}} \) is a standard form that is equal to \( \arctan(u) \). Thus, our integral becomes \( 2 \int_{-1}^{1} \frac{du}{1 + u^{2}} = 2[\arctan(u)]_{-1}^{1} \).

Step 5: Evaluate the Integral at the Limits

Substitute the limits of integration: \( 2 [\arctan(1) - \arctan(-1)] \). Since \( \arctan(1) = \frac{\pi}{4} \) and \( \arctan(-1) = -\frac{\pi}{4} \), it follows that \( \arctan(1) - \arctan(-1) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2} \).

Step 6: Final Calculation

Therefore, the integral evaluates to \( 2 \times \frac{\pi}{2} = \pi \).

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a powerful method to simplify integrals involving expressions such as \( a^2 - x^2 \), \( a^2 + x^2 \), or \( x^2 - a^2 \). In this context, it helps convert complex trigonometric functions into simpler, more manageable forms.
For the integral \( \int_{-\pi / 2}^{\pi / 2} \frac{2 \cos \theta d \theta}{1+(\sin \theta)^{2}} \), using a substitution can transform the original variables to make integration easier.
An effective substitution here is \( u = \sin \theta \). This means \( du = \cos \theta \, d\theta \), allowing conversion of the integral of a trigonometric function to a function of \( u \) only.

• This changes the integral limits, translating \( \theta = -\pi/2 \) and \( \pi/2 \) to \( u = -1 \) and \( 1 \), respectively.
• It simplifies the integral to \( \int_{-1}^{1} \frac{2 \, du}{1 + u^{2}} \).

Choosing the right substitution is essential. It not only simplifies the integral but also ensures that solving it is straightforward by using standard integral forms.

Integral Calculus

Integral calculus is centered around the concept of integrals, which are used to find areas, volumes, and central points. It plays a crucial role in understanding and solving problems involving continuous functions.
This problem involves a definite integral, \( \int_{-\pi / 2}^{\pi / 2} \frac{2 \cos \theta d \theta}{1+ (\sin \theta)^{2}} \), which requires evaluation from one point to another.
Integrals can be classified as definite or indefinite, where a definite integral provides a numerical value representing an area under a curve between specified limits.
The substitution reduces the integral to a simpler form, \( \int_{-1}^{1} \frac{2 \ du}{1 + u^{2}} \), involving only basic calculus skills to evaluate the new limits.

• Recognizing and transforming the integral into a standard form is key in solving it efficiently.
• Definite integrals, like this one, have direct applications in computing physical quantities and are foundational in the analysis of mathematical models.

Integral calculus is indispensable for understanding the behavior of functions over intervals, aiding in both theoretical studies and practical applications.

Standard Integral Forms

Standard integral forms serve as a toolkit in calculus, simplifying the process of evaluating integrals that match certain common patterns.
Once an integral is reduced and transformed, like \( \int_{-1}^{1} \frac{2 \, du}{1 + u^{2}} \), recognizing it as a standard form is crucial.
In this instance, \( \int \frac{du}{1 + u^{2}} \) matches the well-known form that evaluates to \( \arctan(u) \). This allows for a straightforward computation.

• By applying the result \( \arctan(u) \), the definite integral becomes \( 2[\arctan(u)]_{-1}^{1} \).
• Each integral form has associated rules and results, familiarizing with them accelerates problem-solving.

Evaluating using these forms simplifies the calculation: \( 2 (\arctan(1) - \arctan(-1)) = 2(\frac{\pi}{4} + \frac{\pi}{4}) = \pi \).
Recognizing and using standard integral forms ensures calculations are quick and accurate, allowing you to focus on more complex aspects of a problem.