Chapter 7: Problem 31
The integral $$ \int \frac{x}{x^{2}+4} d x $$ can be evaluated either by a trigonometric substitution or by the substitution \(u=x^{2}+4 .\) Do it both ways and show that the results are equivalent.
Short Answer
Expert verified
Both methods yield \(\frac{1}{2}\ln(x^2+4) + C\).
Step by step solution
01
Substitution Method
Let's use the substitution \(u = x^2 + 4\). This means \(du = 2x\,dx\) or \(\frac{du}{2} = x\,dx\). Substituting into the integral, we have:\[\int \frac{x}{x^2 + 4} \, dx = \int \frac{1}{u} \cdot \frac{du}{2} = \frac{1}{2} \int \frac{1}{u} \, du\]The integral becomes:\[ \frac{1}{2} \ln|u| + C\]Substituting back for \(u\), this gives us:\[ \frac{1}{2} \ln(x^2+4) + C\]
02
Trigonometric Substitution
Now, we will use trigonometric substitution. For the integral using trigonometric substitution, use \(x = 2\tan\theta\) which makes \(dx = 2 \sec^2\theta \, d\theta\). The identity \(x^2 + 4 = 4\sec^2\theta\) allows us to substitute:\[\int \frac{2\tan\theta}{4\sec^2\theta} \cdot 2\sec^2\theta \, d\theta = \int \tan\theta \, d\theta = \int \frac{\sin\theta}{\cos\theta} \, d\theta\]This integral becomes:\[-\ln|\cos\theta| + C\]From \(x = 2\tan\theta\), we have \(\cos\theta = \frac{2}{\sqrt{x^2+4}}\) leading to:\[-\ln\left|\frac{2}{\sqrt{x^2+4}}\right| + C = \ln|\sqrt{x^2+4}| + C = \frac{1}{2}\ln(x^2+4) + C\]
03
Equivalence Verification
Both methods yield the result:\[\frac{1}{2}\ln(x^2+4) + C\]Therefore, the two approaches are consistent and give the same result.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The substitution method is a powerful technique for solving integrals where you replace a complex part of the integrand with a single variable, to simplify the integration process. For the given exercise, we use substitution to tackle the integral \(\int \frac{x}{x^{2}+4} dx\).To start, identify a substitution that can simplify the expression. Here, we set \(u = x^2 + 4\), transforming the integral into a simpler form. Correspondingly, the differential \(du\) is derived as \(du = 2x\,dx\), or rearranged to \(\frac{du}{2} = x\,dx\). This allows transformation of the integrand to \(\frac{1}{u}\cdot\frac{du}{2}\).
- Simplify by integrating \(\frac{1}{u}\) to obtain \(\frac{1}{2}\ln|u| + C\).
- Finally, reverse the substitution by replacing \(u\) with \(x^2 + 4\), yielding \(\frac{1}{2}\ln(x^2+4) + C\).
Trigonometric Substitution
Trigonometric substitution is another fascinating method used for solving integrals that involve expressions of the form \(\sqrt{a^2+x^2}\), \(\sqrt{x^2-a^2}\), or \(\sqrt{a^2-x^2}\).In the exercise, using trigonometric substitution involves replacing \(x\) with \(2\tan\theta\). Consequently, \(dx\) becomes \(2\sec^2\theta\,d\theta\) and \(x^2 + 4\) matches \(4\sec^2\theta\). This substitution turns the integrand \(\frac{x}{x^2+4}\) into \(\tan\theta\). Now, the integral becomes \(\int \tan\theta\,d\theta\).
- The integration of \(\tan\theta\) yields \(-\ln|\cos\theta| + C\).
- To revert back to the original variable \(x\), use the trigonometric identity \(\cos\theta = \frac{2}{\sqrt{x^2+4}}\).
- The resulting expression is \(\ln|\sqrt{x^2+4}| + C\), simplifying further to \(\frac{1}{2}\ln(x^2+4) + C\).
Definite Integrals
While the exercise mainly addresses indefinite integrals, understanding the concept of definite integrals is crucial for comprehensive learning.Definite integrals differ as they evaluate the area under a curve between two specified points, \(a\) and \(b\). For definiteness in our integral, you would calculate \[\int_a^b \frac{x}{x^2+4} \, dx\] over a particular interval rather than simply finding the antiderivative.
- Begin by following similar substitution techniques as for indefinite integrals.
- After integration, replace the variable back and evaluate the limits \(a\) and \(b\) in the expression.
- This gives you a numerical value representing the area under the curve defined by the integrand between \(x = a\) and \(x = b\).