Chapter 6: Problem 47
Evaluate the integral. \(\int \frac{d x}{\sqrt{x^{2}+16}}\)
Short Answer
Expert verified
\( \int \frac{dx}{\sqrt{x^2+16}} = \ln |\sqrt{x^2+16} + x| + C \).
Step by step solution
01
Identify the Form of the Integral
Recognize that the integral \( \int \frac{dx}{\sqrt{x^2 + 16}} \) resembles the standard form \( \int \frac{dx}{\sqrt{x^2 + a^2}} \), which suggests the use of a trigonometric substitution.
02
Choose the Appropriate Trigonometric Substitution
For the integral \( \int \frac{dx}{\sqrt{x^2 + 16}} \), let \( x = 4\tan(\theta) \). This substitution stems from the identity \( 1 + \tan^2(\theta) = \sec^2(\theta) \). Consequently, \( dx = 4\sec^2(\theta)d\theta \) and \( \sqrt{x^2+16} = 4\sec(\theta) \).
03
Substitute and Simplify the Integral
Substitute \( x = 4\tan(\theta) \), \( dx = 4\sec^2(\theta)d\theta \), and \( \sqrt{x^2+16} = 4\sec(\theta) \) into the integral: \[ \int \frac{4\sec^2(\theta) d\theta}{4\sec(\theta)} = \int \sec(\theta) d\theta \]
04
Integrate with Respect to \(\theta\)
Integrate \( \int \sec(\theta) d\theta \). The integral of \( \sec(\theta) \) is \( \ln |\sec(\theta) + \tan(\theta)| + C \). Therefore, \[ \int \sec(\theta) d\theta = \ln |\sec(\theta) + \tan(\theta)| + C \]
05
Substitute Back to Terms of \(x\)
Using the original substitution \( x = 4\tan(\theta) \), find expressions for \( \tan(\theta) \) and \( \sec(\theta) \). We have \( \tan(\theta) = \frac{x}{4} \) and \( \sec(\theta) = \frac{\sqrt{x^2+16}}{4} \). Substitute these back:\[ \ln \left| \frac{\sqrt{x^2+16}}{4} + \frac{x}{4} \right| + C \]
06
Simplify the Result
Simplify the restored integral expression:\[ \ln \left| \frac{\sqrt{x^2 + 16} + x}{4} \right| + C = \ln \left| \sqrt{x^2 + 16} + x \right| - \ln(4) + C \]
07
Final Step: Combine Constants
Combine constants to get a finalized expression:\[ \ln \left| \sqrt{x^2 + 16} + x \right| + C' \]where \( C' = C - \ln(4) \) is the new integration constant.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integral Calculus
Integral calculus is a fundamental area of calculus focused on the concepts of integration. It provides a way to calculate areas, volumes, and other related quantities. In integral calculus, the integral of a function represents the accumulation of quantities, such as area under a curve.
There are two main types of integrals: definite integrals and indefinite integrals. They serve different purposes and have different characteristics.
This process involves substituting the variable in the integral with a trigonometric expression, capitalizing on trigonometric identities to simplify the square root and apply standard integral formulas effectively.
There are two main types of integrals: definite integrals and indefinite integrals. They serve different purposes and have different characteristics.
- Definite integrals calculate the net area under a curve from one point to another. The result is a number.
- Indefinite integrals, also known as antiderivatives, represent a family of functions. They describe the accumulation of quantities without specific limits.
This process involves substituting the variable in the integral with a trigonometric expression, capitalizing on trigonometric identities to simplify the square root and apply standard integral formulas effectively.
Definite Integral
Definite integrals are a powerful way to quantify areas, volumes, and other measures between two specific bounds. They are represented as \int\_{a}^{b} f(x) \,dx\, where "a" and "b" are the bounds of integration.
The definite integral has several key properties:
The definite integral has several key properties:
- Linearity: You can break down integrals of sums, allowing operations such as *\int_a^b[f(x) + g(x)\, dx]* to equal *\int_a^b f(x)\, dx + \int_a^b g(x)\, dx*
- Zero Width: If "a" equals "b", then the definite integral is zero since there is no interval.
- Reversing Limits: Reversing the order of "a" and "b" multiplies the integral by -1.
Indefinite Integral
An indefinite integral is a general form of integration expressed without limits, denoted as \( \int f(x) \, dx \). It represents a family of functions known as antiderivatives. The indefinite integral of a function describes the accumulation of quantities along an interval without specific bounds.
The result of an indefinite integral is a function plus a constant, denoted by "C." This constant arises because differentiating with respect to "x" eliminates constant terms, and integrates with multiple possible solutions that vary by a constant.
Here's why the constant of integration is important:
The result of an indefinite integral is a function plus a constant, denoted by "C." This constant arises because differentiating with respect to "x" eliminates constant terms, and integrates with multiple possible solutions that vary by a constant.
Here's why the constant of integration is important:
- It ensures the family of antiderivatives includes every possible solution.
- It reflects the indefinite nature of the integral, emphasizing that different constants can shift the graph vertically.