Chapter 8: Problem 21
Find the integral. $$ \int \frac{x}{\sqrt{x^{2}+9}} d x $$
Short Answer
Expert verified
The integral \(\int \frac{x}{\sqrt{x^{2}+9}} d x\) is \(3\cos\left(\frac{1}{\arctan\left(\frac{x}{3}\right)}\right) + C\)
Step by step solution
01
Perform the substitution
Replace \(x = 3 \tan(\theta)\). Therefore, \(dx = 3\sec^2(\theta)d \theta\). Substituting these into the original integral, we get:\[\int \frac{3 \tan(\theta)}{\sqrt{(3 \tan(\theta))^{2}+9}} \cdot 3\sec^2(\theta)d \theta\]
02
Simplify the integral
Simplify the above equation to:\[\int \frac{3 \tan(\theta)}{\sqrt{9 \tan^2(\theta)+9}} \cdot 3\sec^2(\theta)d \theta = 9\int \frac{\tan(\theta)\sec^2(\theta)}{3\sqrt{\tan^2(\theta)+1}} d\theta\]Using the trigonometric identity \(1 + \tan^2(\theta) = \sec^2(\theta)\), this simplifies to:\[9\int \frac{\tan(\theta)\sec^2(\theta)}{3 \sec(\theta)}d\theta\]This can be further reduced to:\[3\int \tan(\theta) \sec(\theta) d\theta\]
03
Evaluate the integral
Now, the integral can be solved by recognising it is the derivative of \(\sec(\theta)\). So, the integral of \(\tan(\theta) \sec(\theta)\) is \(\sec(\theta)\). So we get:\[3\sec(\theta) + C\]where \(C\) is the constant of integration.
04
Substitute back original variable
Recall that we made the substitution \(x = 3 \tan(\theta)\) at the beginning. Solving for \(\theta\), we find \(\theta = \arctan\left(\frac{x}{3}\right)\). Substituting this back into our result gives:\[3\sec\left(\arctan\left(\frac{x}{3}\right)\right) + C\]\(\sec(\theta)\) is the reciprocal of \(\cos(\theta)\). Therefore the result is:\[= 3\cos\left(\frac{1}{\arctan\left(\frac{x}{3}\right)}\right) + C\]This is the final result.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Trigonometric Substitution
Trigonometric substitution is a clever technique used in integral calculus to simplify integrals that involve square roots, particularly when the integrand includes forms like \( \sqrt{a^2 + x^2} \), \( \sqrt{a^2 - x^2} \), or \( \sqrt{x^2 - a^2} \). By using trigonometric identities and substitutions, we can transform the radical expression into a more manageable form.
For the given integral \( \int \frac{x}{\sqrt{x^2+9}} \, dx \), we can use the substitution \( x = 3 \tan(\theta) \). Why \( \tan(\theta) \)? Because it takes advantage of the identity \( 1 + \tan^2(\theta) = \sec^2(\theta) \), which will help us deal with the square root. In this substitution:
For the given integral \( \int \frac{x}{\sqrt{x^2+9}} \, dx \), we can use the substitution \( x = 3 \tan(\theta) \). Why \( \tan(\theta) \)? Because it takes advantage of the identity \( 1 + \tan^2(\theta) = \sec^2(\theta) \), which will help us deal with the square root. In this substitution:
- \( dx = 3 \sec^2(\theta) \, d\theta \)
- The integral becomes easier to tackle as the radicals drop out.
Definite Integral
The concept of a definite integral is foundational in calculus, allowing us to calculate the exact area under a curve between two specified points on the x-axis. While the given exercise involves finding an indefinite integral, understanding the process of definite integration is key to more advanced applications.
In definite integrals, represented as \( \int_{a}^{b} f(x) \, dx \), you actually compute the limit of a sum that approximates the area. The endpoints \( a \) and \( b \) define the interval. After integrating the function, you evaluate the antiderivative at each of these bounds and subtract:
In definite integrals, represented as \( \int_{a}^{b} f(x) \, dx \), you actually compute the limit of a sum that approximates the area. The endpoints \( a \) and \( b \) define the interval. After integrating the function, you evaluate the antiderivative at each of these bounds and subtract:
- Evaluate at \( b \) (upper limit)
- Subtract the evaluation at \( a \) (lower limit)
Integration Techniques
Integration techniques are methods used to simplify and solve more complex integrals. Different problems require different techniques; thus, being familiar with a variety of methods is beneficial.
The exercise at hand employs trigonometric substitution—a specific technique targeted at expressions under a square root. Here are some key techniques to know:
Knowing which integration technique to apply when faced with a daunting integral comes with practice and a good grasp of these foundational methods, allowing students to tackle a vast array of problems with confidence.
The exercise at hand employs trigonometric substitution—a specific technique targeted at expressions under a square root. Here are some key techniques to know:
- **Substitution Method:** Useful for integrals that can be simplified using a simple algebraic substitution.
- **Integration by Parts:** Based on the product rule for derivatives, it is usually applied to products of functions.
- **Partial Fractions:** Effective for rational functions where the denominator can be factored.
Knowing which integration technique to apply when faced with a daunting integral comes with practice and a good grasp of these foundational methods, allowing students to tackle a vast array of problems with confidence.