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Magazine Chapter 28: Problem 33
Integrate each of the given functions. $$\int \frac{\sec e^{-x}}{e^{x}} d x$$
Short Answer
Expert verified
\(- \ln |\sec(e^{-x}) + \tan(e^{-x})| + C\), where \( C \) is a constant.
Step by step solution
01
Identify the Integration Formula
The integral given is \( \int \frac{\sec e^{-x}}{e^{x}} \, dx \). To integrate this, we recognize that \( e^{-x} \) is inside the secant function, suggesting a substitution method might be appropriate. In such cases, a common technique is to perform a substitution to simplify the integral.
02
Perform the Substitution
Let \( u = e^{-x} \). Then, differentiate with respect to \( x \): \( \frac{du}{dx} = -e^{-x} \), or equivalently, \( du = -e^{-x} \, dx \). Rearrange to express \( dx \): \( dx = -\frac{du}{e^{-x}} \). Substitute \( u \) into the integral, and replace \( dx \): \[ \int \sec u \frac{1}{e^x} \, dx = \int \sec u \cdot (-du). \]
03
Simplify the Integral
Notice that the \( e^x \) terms have canceled out due to the substitution. The integral simplifies to \( -\int \sec u \, du \). This is a standard integral. The integral of \( \sec u \) is known to be \( \ln |\sec u + \tan u| + C \), where \( C \) is the constant of integration.
04
Integrate and Substitute Back
Perform the integration: \( - \int \sec u \, du = - ( \ln |\sec u + \tan u| + C ) = -\ln |\sec u + \tan u| - C \). Substitute back \( u = e^{-x} \) to express the answer in terms of \( x \): \( - \ln |\sec(e^{-x}) + \tan(e^{-x})| + C \).
05
Present the Final Answer
The integral \( \int \frac{\sec e^{-x}}{e^{x}} \, dx \) is equal to \( - \ln |\sec(e^{-x}) + \tan(e^{-x})| + C \), where \( C \) is a constant of integration.
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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, in the realm of integration, is a powerful technique used to simplify integrals. The idea behind it is to replace a complex expression with a simpler one, using a meaningfully chosen substitution which simplifies the calculation of the integral. Let's see how we can apply this method effectively:
- Start by identifying a part of the integral that can be substituted. Often, this is an expression inside a trigonometric or exponential function.
- Choose a new variable, often denoted as \( u \), and define it in terms of the original variable. For the example given, we chose \( u = e^{-x} \).
- Calculate the differential \( du \), and solve for \( dx \). This step involves differentiating the chosen substitution \( u \) with respect to the original variable \( x \).
- The original integral is then rewritten in terms of \( u \) and \( du \). This transforms the integral into a simpler form, making it easier to evaluate.
Secant Function
The Secant Function, represented by \( \sec(x) \), is a crucial trigonometric function related to the cosine function. It's defined as the reciprocal of the cosine function. This means:\[ \sec(x) = \frac{1}{\cos(x)} \]Understanding the secant function's properties helps when integrating expressions involving it:
- The secant function has a domain wherever the cosine function is not zero, making it important to watch for undefined points.
- It has a range of \([-\infty, -1] \cup [1, \infty]\), as the value of cosine never exceeds these limits.
- When integrated, the secant function often results in a logarithmic function. As seen in the example, the integral \( \int \sec u \, du = \ln |\sec u + \tan u| + C \) is a standard result.
Integral Calculus
Integral Calculus is a branch of calculus focusing on integrals and their properties. The main objective is to find the anti-derivative or the original function given its derivative:
- An integral can be definite or indefinite. An indefinite integral results in a function plus a constant of integration \( C \), indicating the family of all anti-derivatives.
- The process involves finding a function whose derivative matches the given function. Techniques like substitution and integration by parts aid this process.
- The primary applications of integral calculus include calculating areas under curves and solving differential equations, impacting physics, engineering, and other scientific fields.
