Chapter 7: Problem 17
Evaluate the integral. $$ \int \frac{d x}{\left(4 x^{2}-9\right)^{3 / 2}} $$
Short Answer
Expert verified
The integral evaluates to \(-\frac{1}{6\sqrt{4x^2 - 9}} + C\).
Step by step solution
01
Recognize Integral Form
The integral \( \int \frac{d x}{(4x^2 - 9)^{3/2}} \) involves a radical and a polynomial raised to a power. This suggests that we might utilize a trigonometric substitution due to the form \( a^2 - u^2 \).
02
Identify Substitution
In the expression \(4x^2 - 9\), recognize it as a difference of squares: \((2x)^2 - 3^2\). This resembles the \(a^2 - u^2\) form, suggesting the substitution \(2x = \frac{3}{\cos(\theta)}\). We can rewrite it as \(x = \frac{3}{2}\sec(\theta)\) and find \(dx = \frac{3}{2}\sec(\theta)\tan(\theta)\,d\theta\).
03
Substitute and Simplify the Integral
Substituting \(x = \frac{3}{2}\sec(\theta)\) into \(4x^2 - 9\) gives:\((2x)^2 - 9 = (3\sec(\theta))^2 - 9 = 9\sec^2(\theta) - 9 = 9\tan^2(\theta)\).So, \((4x^2 - 9)^{3/2} = (9\tan^2(\theta))^{3/2} = 27\tan^3(\theta)\).The integral becomes:\[ \int \frac{ \frac{3}{2}\sec(\theta)\tan(\theta)\, d\theta }{ 27\tan^3(\theta) } = \int \frac{ 1 }{ 18\tan^2(\theta) } \sec(\theta)\, d\theta. \]
04
Simplify Further and Integrate
Rewrite \(\frac{\sec(\theta)}{\tan^2(\theta)}\) as \(\int \frac{1}{18}\cot(\theta)\csc(\theta)\,d\theta\). Use the identity \(\cot(x)\csc(x) = -\frac{d}{d\theta}(\csc(x))\), yielding:\[-\frac{1}{18}\int d(\csc(\theta)) = -\frac{1}{18}\csc(\theta) + C.\]
05
Reverse the Substitution
Returning to the original variable, remember \(\sec(\theta) = \frac{2x}{3}\), so:\[ \csc(\theta) = \frac{1}{\sin(\theta)} = \frac{ \sqrt{(2x)^2 - 9} }{3}\,\text{implying:} \\sin(\theta) = \frac{ 3 }{\sqrt{(2x)^2 - 9}}. \]Therefore, the solution to the integral is:\(-\frac{1}{18}\cdot \frac{ 3 }{ \sqrt{(4x^2 - 9) }} + C = -\frac{1}{6\sqrt{4x^2 - 9}} + C.\)
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.
Trigonometric substitution
Trigonometric substitution is a technique used in calculus to simplify integrals containing radical expressions, especially those of the form \( a^2 - x^2 \), \( a^2 + x^2 \), or \( x^2 - a^2 \). This method involves substituting a trigonometric function for the variable in the integrand, transforming the integral into one that is easier to evaluate.
- For expressions similar to \( \sqrt{a^2 - x^2} \), we use the substitution \( x = a \sin(\theta) \).
- If the form is \( \sqrt{a^2 + x^2} \), we substitute \( x = a \tan(\theta) \).
- In the case of \( \sqrt{x^2 - a^2} \), \( x = a \sec(\theta) \) is appropriate.
Definite integrals
A definite integral evaluates the integral of a function over a specific interval \([a, b]\). It provides the area under the curve of the function within these limits, yielding a real number as a result.
- The expression \( \int_{a}^{b} f(x)dx \) calculates the net area, considering the regions above the x-axis as positive and those below as negative.
- Definite integrals can be computed using the Fundamental Theorem of Calculus: by finding an antiderivative \( F(x) \) of the function \( f(x) \) and evaluating \( F(b) - F(a) \).
- They are widely used in various applications, including physics for calculating displacement, work, and accumulated quantities.
Power of a polynomial
The power of a polynomial involves raising a polynomial expression to a given power, which is a common scenario in calculus. Understanding how to manipulate and simplify these expressions is critical for effective integration and differentiation.
- Polyomial expressions like \((4x^2 - 9)^{3/2}\) can be daunting, but they can often be associated with known algebraic forms, making them easier to handle with techniques like trigonometric substitution.
- Manipulating powers of polynomials often requires familiarity with algebraic identities, such as difference of squares, which can simplify the expression before integration.
- In calculus, these powers add layers of complexity, often requiring creative substitutions or transformations to reduce the integral to a solvable form.