Problem 43 - Thomas Calculus in SI Units

Question

Evaluate the integrals in Exercises \(29-50.\) $$\int_{0}^{\pi / 4}\left(1+e^{\tan \theta}\right) \sec ^{2} \theta d \theta$$

Step-by-Step Solution

Answer

The integral evaluates to \( e \).

Step 1: Recognize the Substitution

Notice that the integrand contains a \( \sec^2 \theta \) term, which is the derivative of \( \tan \theta \). This suggests that a substitution using \( u = \tan \theta \) could be useful.

Step 2: Perform the Substitution

Let \( u = \tan \theta \). Then, \( du = \sec^2 \theta \, d\theta \). The limits of integration change accordingly: when \( \theta = 0, \, u = \tan 0 = 0 \); when \( \theta = \pi/4, \, u = \tan(\pi/4) = 1 \).

Step 3: Rewrite the Integral

Substitute \( u \) for \( \tan \theta \) and \( du \) for \( \sec^2 \theta \, d\theta \) into the integral, transforming it into: \[ \int_{0}^{1} (1 + e^u) \, du \]

Step 4: Integrate Each Term

Break the integral \( \int_{0}^{1} (1 + e^u) \, du \) into two separate integrals: \( \int_{0}^{1} 1 \, du + \int_{0}^{1} e^u \, du \).1. The integral \( \int_{0}^{1} 1 \, du = [u]_{0}^{1} = 1 - 0 = 1 \).2. The integral \( \int_{0}^{1} e^u \, du = [e^u]_{0}^{1} = e^1 - e^0 = e - 1 \).

Step 5: Combine the Results

Add the results of the two integrals: \[ 1 + (e - 1) = e \] Thus, the evaluated integral is \( e \).

Key Concepts

Substitution Method

The Substitution Method is a powerful technique in calculus used to transform complex integrals into simpler forms, making them easier to evaluate. Recognizing when to use substitution is key. In the exercise, the presence of \( \sec^2 \theta \), the derivative of \( \tan \theta \), signaled that substitution might simplify the integral. Here's how it works:

• Select a substitution that will simplify the integral. For instance: \( u = \tan \theta \).
• Express the differential in terms of \( u \) through differentiation: \( du = \sec^2 \theta \, d\theta \).
• Change the limits of integration if dealing with definite integrals, adjusting them from the original variable \( \theta \) to the new variable \( u \).

These steps convert the integral into a new form, often simplifying the process of finding antiderivatives.

Definite Integrals

Definite Integrals provide a way to calculate the area under a curve between two points on the x-axis. They are written as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration.

• The Fundamental Theorem of Calculus connects derivatives and integrals, allowing us to evaluate definite integrals by finding an antiderivative.
• For substitution in definite integrals, adjust the limits to match the substituted variable. In the exercise, \( \theta = 0 \) became \( u = 0 \) and \( \theta = \pi/4 \) became \( u = 1 \).
• Once the antiderivative is found, plug in the limits to obtain the final result.

By applying these principles, we can solve complex integrals like the one in the exercise efficiently.

Exponential Functions

Exponential functions, identified by the form \( e^x \), play a crucial role in calculus due to their unique properties. In the exercise, the function \( e^{\tan \theta} \) was part of the integrand, adding to its complexity.

• Exponential functions are particularly straightforward to integrate, as the antiderivative of \( e^x \) is \( e^x \) itself.
• When such functions are involved in definite integrals, after substitution, they maintain their basic property, simplifying the integration step.
• In the given exercise, after transforming the integral, \( \int_{0}^{1} e^u \, du \) directly translates to \( [e^u]_{0}^{1} = e - 1 \).

These characteristics make exponential functions highly significant and often easier to handle within integrals than many other functions.