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Magazine Chapter 7: Problem 43
Evaluate the integrals. $$\int_{0}^{\pi / 4}\left(1+e^{\tan \theta}\right) \sec ^{2} \theta d \theta$$
Short Answer
Expert verified
The value of the integral is \( e \).
Step by step solution
01
Understand the Integral
We are given the integral \(\int_{0}^{\pi / 4}\left(1+e^{\tan \theta}\right) \sec ^{2} \theta \, d\theta\). The goal is to evaluate this definite integral over the interval from \(0\) to \(\pi/4\).
02
Identify Substitution
The integral involves the term \(\sec^2 \theta\), which is the derivative of \(\tan \theta\). Hence, it is useful to perform the substitution \( u = \tan \theta \), so \( \frac{du}{d\theta} = \sec^2 \theta \). This makes \( d\theta = \frac{du}{\sec^2 \theta}\) or equivalently, \( d\theta = du\).
03
Change of Variables
With the substitution \( u = \tan \theta \), the limits of integration will also change. When \( \theta = 0 \), \( u = \tan(0) = 0 \). When \( \theta = \pi/4 \), \( u = \tan(\pi/4) = 1 \). The integral now becomes \(\int_{0}^{1} (1 + e^u) \, du\).
04
Integrate with Respect to u
We now integrate the expression \( 1 + e^u \) with respect to \( u \). The antiderivative of \( 1 \) is \( u \), and the antiderivative of \( e^u \) is \( e^u \). Thus, the integral becomes \( \int (1 + e^u) \, du = u + e^u \).
05
Evaluate the Definite Integral
We need to evaluate the antiderivative at the bounds \( u = 0 \) and \( u = 1 \). This gives us:\[ \left[ u + e^u \right]_{0}^{1} = (1 + e^1) - (0 + e^0) = (1 + e) - (0 + 1) = e. \]
06
Write the Final Answer
Therefore, the value of the definite integral \( \int_{0}^{\pi / 4}\left(1+e^{\tan \theta}\right) \sec ^{2} \theta \, d\theta \) is \( e \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Trigonometric Substitution
Trigonometric substitution is a technique used in calculus to simplify integrals involving trigonometric functions. It transforms the integral into a form that is easier to evaluate. In this problem, we utilize the substitution \( u = \tan \theta \). This choice is strategic because the derivative of \( \tan \theta \) is \( \sec^2 \theta \), which appears in the original integral.
By substituting \( u = \tan \theta \):
By substituting \( u = \tan \theta \):
- \( du = \sec^2 \theta \, d\theta \) makes the integration straightforward.
- The limits change according to the substitution, which in this case from \( \theta = 0 \) to \( \theta = \pi/4 \) transform to \( u = 0 \) to \( u = 1 \).
Exponential Functions
Exponential functions are a fundamental concept in calculus, characterized by the base of the natural logarithm \( e \) raised to the power of a variable. In this problem, we encounter an exponential function within the integral: \( e^{\tan \theta} \).
After trigonometric substitution, this term modifies to \( e^u \). Exponential functions have straightforward antiderivatives, which make them easy to integrate. For example, the antiderivative of \( e^u \) is \( e^u \) itself.
After trigonometric substitution, this term modifies to \( e^u \). Exponential functions have straightforward antiderivatives, which make them easy to integrate. For example, the antiderivative of \( e^u \) is \( e^u \) itself.
- This property of exponential functions significantly simplifies the process of solving integrals.
- Once converted through substitution, dealing with exponential components often requires direct integration, as showcased in this example.
Antiderivatives
The concept of antiderivatives is central to calculus as it involves finding a function whose derivative is the given function. In the context of definite integrals, calculating the antiderivative is a crucial step before applying the limits of the integral.
For this problem, after integrating the expression \( 1 + e^u \), the antiderivative is determined as \( u + e^u \).
For this problem, after integrating the expression \( 1 + e^u \), the antiderivative is determined as \( u + e^u \).
- The polynomial part, represented by \( 1 \), integrates to simply \( u \).
- The exponential component \( e^u \) remains the same after integration due to its unique derivative properties.
