Problem 9
Question
Evaluate the integrals in Exercises \(1-28\). $$ \int \frac{d x}{\sqrt{4 x^{2}-49}}, \quad x>\frac{7}{2} $$
Step-by-Step Solution
Verified Answer
\( \ln \left| \frac{2x}{7} + \sqrt{\left(\frac{2x}{7}\right)^2 - 1} \right| + C \).
1Step 1: Identify the integral form
The integral is of the form \( \int \frac{dx}{\sqrt{4x^2 - 49}} \). This can be recognized as resembling a standard form for trigonometric substitution, specifically involving \( \sqrt{a^2x^2 - a^2} \) which suggests the use of hyperbolic functions.
2Step 2: Factor the expression inside the square root
Rewrite \( 4x^2 - 49 \) as \( 4(x^2 - \frac{49}{4}) \) to make the expression inside the square root easier to handle for substitution. This simplifies to \( 2\sqrt{x^2 - \frac{49}{4}} \).
3Step 3: Choose the appropriate substitution
Since the expression now is \( \sqrt{x^2 - \left(\frac{7}{2}\right)^2} \), we can use the trigonometric substitution \( x = \frac{7}{2} \sec(\theta) \). This gives a derivative of \( dx = \frac{7}{2}\sec(\theta)\tan(\theta) \,d\theta \).
4Step 4: Substitute and simplify
Substitute \( x = \frac{7}{2} \sec(\theta) \) into the integral. The expression \( \sqrt{4x^2 - 49} \) becomes \( \sqrt{\left(\frac{7}{2}\right)^2 \sec^2(\theta) - \left(\frac{7}{2}\right)^2} = \frac{7}{2}\tan(\theta) \). Thus, the integral becomes: \[ \int \frac{\frac{7}{2}\sec(\theta)\tan(\theta)\, d\theta}{\frac{7}{2}\tan(\theta)} = \int \sec(\theta) \, d\theta \].
5Step 5: Integrate the simplified expression
The integral \( \int \sec(\theta) \, d\theta \) is known to be \( \ln |\sec(\theta) + \tan(\theta)| + C \), where \( C \) is the constant of integration.
6Step 6: Convert back to variable x
Since \( x = \frac{7}{2} \sec(\theta) \), then \( \sec(\theta) = \frac{2x}{7} \) and \( \tan(\theta) = \sqrt{\sec^2(\theta) - 1} = \sqrt{\left(\frac{2x}{7}\right)^2 - 1} \). Substitute these back to get: \[ \ln \left| \frac{2x}{7} + \sqrt{\left(\frac{2x}{7}\right)^2 - 1} \right| + C \] as the final expression in terms of \( x \).
Key Concepts
Hyperbolic FunctionsIntegral CalculusTrigonometric Identities
Hyperbolic Functions
Hyperbolic functions, such as the hyperbolic sine (\(\sinh\)) and hyperbolic cosine (\(\cosh\)), are analogues of the trigonometric functions but for hyperbolas rather than circles. These functions often arise in problems of calculus due to their properties that mimic certain trigonometric identities, making them suitable for particular types of integrations, especially those involving quadratic expressions.
A key characteristic of hyperbolic functions is their definition through exponential functions:
Using hyperbolic identities such as \(\cosh^2(x) - \sinh^2(x) = 1\) can be a powerful method in evaluations where traditional trigonometric substitutions are cumbersome.
A key characteristic of hyperbolic functions is their definition through exponential functions:
- \(\sinh(x) = \frac{e^x - e^{-x}}{2}\)
- \(\cosh(x) = \frac{e^x + e^{-x}}{2}\)
Using hyperbolic identities such as \(\cosh^2(x) - \sinh^2(x) = 1\) can be a powerful method in evaluations where traditional trigonometric substitutions are cumbersome.
Integral Calculus
Integral calculus is the branch of mathematics concerned with finding the total or accumulation of quantities. It often deals with the calculation of areas under curves or the total size accumulation when given rates of change.
At the core of integral calculus is the concept of an "integral," which can be either definite or indefinite:
At the core of integral calculus is the concept of an "integral," which can be either definite or indefinite:
- Indefinite integrals, denoted \(\int f(x) \, dx\), represent a family of functions whose derivative is \(f(x)\).
- Definite integrals, written as \(\int_a^b f(x) \, dx\), calculate the net area under \(f(x)\) from \(a\) to \(b\).
Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that hold true for all values of the variable where both sides of the identity are defined. Basic identities stem from the definitions of sine, cosine, and tangent, and include:
In trigonometric substitution, one often uses identities to convert square roots into trigonometric functions that are easier to integrate. For example, \(\sec^2(\theta) - 1 = \tan^2(\theta)\) is used in the process of resolving complications within expressions involving square roots.
Understanding these identities and how to apply them facilitates problem-solving in calculus, especially when traditional methods of integration fall short.
- \(\sin^2(x) + \cos^2(x) = 1\)
- \(1 + \tan^2(x) = \sec^2(x)\)
- \(1 + \cot^2(x) = \csc^2(x)\)
In trigonometric substitution, one often uses identities to convert square roots into trigonometric functions that are easier to integrate. For example, \(\sec^2(\theta) - 1 = \tan^2(\theta)\) is used in the process of resolving complications within expressions involving square roots.
Understanding these identities and how to apply them facilitates problem-solving in calculus, especially when traditional methods of integration fall short.
Other exercises in this chapter
Problem 8
Evaluate the integrals. \(\int \sin ^{-1} y d y\)
View solution Problem 8
Evaluate each integral in Exercises \(1-36\) by using a substitution to reduce it to standard form. $$ \int \frac{d x}{x-\sqrt{x}} $$
View solution Problem 9
Use the table of integrals at the back of the book to evaluate the integrals. \(\int x \sqrt{4 x-x^{2}} d x\)
View solution Problem 9
Evaluate the integrals in Exercises \(1-34\) without using tables. $$ \int_{-\infty}^{-2} \frac{2 d x}{x^{2}-1} $$
View solution