Problem 37 - University Calculus: Early Transcendentals

Question

Use the Substitution Formula in Theorem 7 to evaluate the integrals in Exercises \(1-46\). $$\int_{-\pi / 2}^{\pi / 2} \frac{2 \cos \theta d \theta}{1+(\sin \theta)^{2}}$$

Step-by-Step Solution

Answer

The integral evaluates to \( \pi \).

Step 1: Identify the Integral

The integral we need to evaluate is \( \int_{-\pi / 2}^{\pi / 2} \frac{2 \cos \theta \, d\theta}{1+(\sin \theta)^{2}} \). We need to use substitution to simplify this integral.

Step 2: Choose an Appropriate Substitution

The key to solving this integral is recognizing a substitution that can simplify the expression. Let \( u = \sin \theta \), then \( du = \cos \theta \, d\theta \). This substitution will simplify handling the trigonometric functions.

Step 3: Change the Limits of Integration

Since \( u = \sin \theta \), when \( \theta = -\pi/2 \), \( u = \sin(-\pi/2) = -1 \) and when \( \theta = \pi/2 \), \( u = \sin(\pi/2) = 1 \). So our new limits of integration for \( u \) are from \(-1\) to \(1\).

Step 4: Substitute and Simplify the Integral

Substitute \( u = \sin \theta \) and \( du = \cos \theta \, d\theta \) into the integral: \[ \int_{-1}^{1} \frac{2 \cdot 1 \, du}{1 + u^2} = 2 \int_{-1}^{1} \frac{du}{1 + u^2}. \]

Step 5: Recognize the Integral of Arctangent

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

Step 6: Evaluate the Definite Integral

Compute \( 2 [\arctan(u)]_{-1}^{1} = 2 (\arctan(1) - \arctan(-1)) \). Since \( \arctan(1) = \frac{\pi}{4} \) and \( \arctan(-1) = -\frac{\pi}{4} \), the expression simplifies to \( 2 \left( \frac{\pi}{4} - (-\frac{\pi}{4}) \right) = 2 \cdot \frac{\pi}{2} = \pi. \)

Step 7: Final Step: Conclusion

After evaluating, the integral \( \int_{-\pi / 2}^{\pi / 2} \frac{2 \cos \theta \, d\theta}{1+(\sin \theta)^{2}} \) is equal to \( \pi \).

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a powerful technique used in calculus to simplify integrals involving square roots and trigonometric expressions. In the given problem, we are dealing with an integral that has \( \cos \theta \) and \( \sin \theta \) as part of the expression. The goal of trigonometric substitution is to transform trigonometric components into algebraic forms that are easier to integrate.
To apply this method, you typically choose a substitution based on known trigonometric identities. For example, using \( u = \sin \theta \) simplifies expressions like \( 1-(\sin \theta)^2 \), using the identity \( \cos^2 \theta = 1-(\sin \theta)^2 \). This is particularly effective when our integration limits are symmetric around 0, such as \( -\pi/2 \) to \( \pi/2 \).

• Choose a substitution to simplify the integral.
• Apply trigonometric identities to reduce the complexity.
• Change the limits of the integral to match the substitution.

This technique is a conduit for transforming complex-looking integrals into more manageable forms, enabling smoother calculations.

Definite Integrals

Definite integrals are a fundamental concept in calculus that involves calculating the area under a curve, within certain bounds. In our exercise, we're evaluating the integral from \(-\pi / 2\) to \(\pi / 2\). The result of a definite integral is a specific number, representing the net area between the curve and the x-axis over the interval.

• Provides the net area within specified bounds.
• Integrates a function over an interval to give a fixed value.
• Is essential for applications in physics, engineering, and other fields.

When computed, the definite integral reflects how the function behaves across the chosen interval. In our case, after using substitution and simplifying, the transformed integral computes to \(\pi\), representing the total area under the modified curve for the interval \(-1\) to \(1\), which are the limits after substitution.
Understanding definite integrals allows us to apply calculus concepts to solve real-world problems by quantifying physical quantities like distance, area, and more.

Substitution Method

The substitution method is a clever algebraic technique in integration that simplifies the process of finding integrals. This technique often uncovers a simpler expression for later evaluation. Substitution involves rewriting the integral in terms of a new variable, generally \( u \), which transforms the integration problem into a simpler form.
In our original problem, we used the substitution \( u = \sin \theta \) and consequently, \( du = \cos \theta \, d\theta \). This transforms the integral into a function of \( u \), making it straightforward to integrate.

• Choose a substitution \( u \) to simplify the integral equation.
• Rewrite the integral limits according to the new variable \( u \).
• Simplify the integral for easier computation.

This method effectively handles complexities that would otherwise make direct integration challenging. It's particularly beneficial in converting trigonometric integrals into polynomial forms or recognizing standard integral forms for faster solutions.

Arctangent Function

The arctangent function, denoted as \( \arctan(x) \), is an inverse trigonometric function often encountered in integration, especially in cases involving \( 1+x^2 \). The integral of \( \frac{1}{1+u^2} \) evaluates to the arctangent function.
In our problem, the integral evaluates to \( \int \frac{du}{1+u^2} \), recognized as the integral form of \( \arctan(u) \). Implementing this, we resolve into \( 2[\arctan(u)]_{-1}^{1} \), which is calculated using the known values of the arctangent:

• \( \arctan(1) = \frac{\pi}{4} \)
• \( \arctan(-1) = -\frac{\pi}{4} \)

Evaluating these results in: \( 2(\frac{\pi}{4} - (-\frac{\pi}{4})) = \pi \). The arctangent function translates the algebraic solution back into a trigonometric context, allowing us to conclude the value of the integral.