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Chapter 7: Problem 47

Evaluate the integrals. \begin{equation}\int_{0}^{\pi / 4}\left(1+e^{\tan \theta}\right) \sec ^{2} \theta d \theta\end{equation}

Short Answer

Expert verified
The integral evaluates to \( e \).

Step by step solution

01

Identify the Substitution

To evaluate the integral, we observe that the integrand includes \( \sec^2 \theta \), which suggests the substitution \( u = \tan \theta \). This substitution leads to \( du = \sec^2 \theta \, d\theta \).
02

Change the Limits of Integration

We need to adjust the limits of integration based on the substitution. When \( \theta = 0 \), \( u = \tan(0) = 0 \). When \( \theta = \frac{\pi}{4} \), \( u = \tan(\frac{\pi}{4}) = 1 \). Thus, the new limits of integration for \( u \) are from 0 to 1.
03

Rewrite the Integral in Terms of u

Substitute \( u = \tan \theta \) and \( du = \sec^2 \theta \, d\theta \) into the integral. The integral becomes \( \int_{0}^{1} \left(1 + e^u\right) \ du \).
04

Evaluate the Integral

Separate the integral into two parts: \( \int_{0}^{1} 1 \, du + \int_{0}^{1} e^u \, du \). The first integral, \( \int 1 \, du = u + C \), evaluated from 0 to 1, is \( 1 - 0 = 1 \). The second integral, \( \int e^u \, du = e^u + C \), evaluated from 0 to 1, is \( e^1 - e^0 = e - 1 \).
05

Add the Results

Add the results from the integrals. The total is \( 1 + (e - 1) = e \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Substitution Method
The substitution method is a powerful technique for solving integrals, especially when dealing with composite functions. Essentially, it involves changing the variable of integration to simplify the integral. This is helpful when you spot a function and its derivative within the integrand.
  • Step 1 is to identify the beneficial substitution. By doing this, we aim to transform a complicated integral into a more manageable one.
  • In the given problem, we use the substitution \( u = \tan\theta \). Therefore, \( du = \sec^2\theta\, d\theta \), which matches a portion of the original integrand.
The substitution helps in simplifying the integral into terms of \( u \), allowing us to switch to the variable \( u \) for easier evaluation.
Definite Integrals
Definite integrals are a way to calculate the area under a curve for a specific interval on the x-axis. Unlike indefinite integrals, they come with upper and lower limits.
  • After substitution, it's crucial to change the limits of integration accordingly.
  • In our example, we switched from \( \theta \) limits (0 to \( \frac{\pi}{4} \)) to \( u \) limits (0 to 1) by evaluating \( u = \tan(\theta) \).
This adjustment is necessary because we are changing variables and need to ensure that our new integral still accurately represents the original problem.
Trigonometric Functions
Trigonometric functions are essential in integrals, as they often appear in a variety of settings due to their cyclical nature. Key functions include sine, cosine, and tangent.
  • In the exercise, \( \tan(\theta) \) and \( \sec^2(\theta) \) are pivotal for substitution.
  • Since \( \sec^2(\theta) \) simplifies directly to \( du \) when \( u = \tan(\theta) \), it makes integration much simpler.
Trigonometric identities can offer direct paths to simplification, reducing complex integrals to a form that is easier to integrate.
Exponential Function
Exponential functions arise frequently in integrals due to their unique properties and the ease of differentiating and integrating them. The function \( e^u \), where \( e \) is the base of the natural logarithm, is especially common.
  • In this problem, the integral \( \int_{0}^{1} e^u \, du \) is straightforward owing to the nature of the exponential function.
  • Essentially, the integral of \( e^u \) with respect to \( u \) is \( e^u + C \), a major advantage in calculus.
The unique property of the exponential function, where the derivative of \( e^u \) is itself, means that calculations remain simple throughout the integration process.

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Most popular questions from this chapter

Find the limits in Exercises \(97-104\) $$ \lim _{x \rightarrow \infty} x \tan ^{-1} \frac{2}{x} $$

Evaluate the integrals in Exercises \(85-94\) $$ \int_{-\sqrt{3}}^{1 / \sqrt{3}} \frac{\cos \left(\tan ^{-1} 3 x\right)}{1+9 x^{2}} d x $$

For problems \(43-46\) use implicit differentiation to find \(\frac{d y}{d x}\) at the given point \(P .\) $$ 3 \tan ^{-1} x+\sin ^{-1} y=\frac{\pi}{4} ; \quad P(1,-1) $$

Solve the initial value problems in Exercises \(109-114\) $$ \frac{d y}{d x}=\frac{1}{1+x^{2}}-\frac{2}{\sqrt{1-x^{2}}}, \quad y(0)=2 $$

In Exercises \(21-42,\) find the derivative of \(y\) with respect to the appropriate variable. $$ y=\ln \left(\tan ^{-1} x\right) $$
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