/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 Evaluate the integral and check ... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the integral and check your answer by differentiating. $$\int \sec x(\sec x+\tan x) d x$$

Short Answer

Expert verified
The integral evaluates to \( \tan x + \sec x + C \).

Step by step solution

01

Distribute Terms Inside the Integral

The given integral is \( \int \sec x (\sec x + \tan x) \, dx \). Distribute the \( \sec x \) to both terms inside the parentheses. This turns the integral into \( \int \sec^2 x \, dx + \int \sec x \tan x \, dx \).
02

Evaluate Each Integral Separately

Now we have two separate integrals: \( \int \sec^2 x \, dx \) and \( \int \sec x \tan x \, dx \). We know that \( \int \sec^2 x \, dx = \tan x + C_1 \) because the derivative of \( \tan x \) is \( \sec^2 x \). Similarly, \( \int \sec x \tan x \, dx = \sec x + C_2 \) because the derivative of \( \sec x \) is \( \sec x \tan x \). Thus, the combined integral result is \( \tan x + \sec x + C \).
03

Combine the Results

Combine the results from each part: \( \tan x + \sec x + C \), where \( C \) is the constant of integration that absorbs both \( C_1 \) and \( C_2 \). This is the evaluated integral.
04

Check by Differentiation

Differentiate \( \tan x + \sec x + C \) with respect to \( x \). The derivative of \( \tan x \) is \( \sec^2 x \), and the derivative of \( \sec x \) is \( \sec x \tan x \). The constant \( C \) vanishes during differentiation. Thus, the derivative is \( \sec^2 x + \sec x \tan x \), which matches the integrand of the original integral, confirming the solution is correct.

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Key Concepts

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

Integration Techniques
Integration is a key concept in calculus, and involves finding the integral or antiderivative of a function. One effective integration technique is to break down a complex integral into simpler parts. In the given exercise, the integral is split into two simpler integrals after distributing terms. By distributing \( \sec x \) over \( \sec x + \tan x \), we break it down into \( \int \sec^2 x \, dx \) and \( \int \sec x \tan x \, dx \). This makes it easier to solve since each part has a known integral:
- The integral of \( \sec^2 x \) is straightforward and known to be \( \tan x \).
- The integral of \( \sec x \tan x \) is \( \sec x \), which stems from its derivative characteristics.
Using these techniques reduces the complexity of integration tasks by transforming them into formulas that are easier to manage and understand.
Differentiation
Differentiation is the process of finding the derivative of a function, essentially reversing integration. The exercise checks the integrated result by differentiating the solution \( \tan x + \sec x + C \). Differentiation of each part is as follows:
  • The derivative of \( \tan x \) is \( \sec^2 x \).
  • The derivative of \( \sec x \) is \( \sec x \tan x \).
  • The constant \( C \), when differentiated, becomes zero.
Combining these results gives \( \sec^2 x + \sec x \tan x \) which is the original integrand, demonstrating that differentiation verifies the integration solution. This check illustrates how integration and differentiation are interconnected, allowing one to verify results and better understand the relationship between a function and its rate of change.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a pivotal link between differentiation and integration, illustrating they are inverse processes. It states that the integral of a function can be reversed by differentiation, and vice versa. In the exercise, once the integral \( \int \sec x(\sec x+\tan x) \, dx \) is evaluated, the solution is checked through differentiation.
The theorem consists of two main parts:
  • The first part establishes that if a function is continuous over an interval, and \( F \) is its indefinite integral, then \( F \) is differentiable on the interval and its derivative is the original function.
  • The second part states that if \( f \) is continuous over an interval \([a, b]\), and \( F \) is an antiderivative of \( f \) over that interval, then \( \int_a^b f(x) \, dx = F(b) - F(a) \).
This theorem underlines that solving an integral and then checking by differentiation ensures consistency between these fundamental concepts, reinforcing the principles of calculus.

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