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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 us perform a substitution by letting \( u = x^2 + 4 \). Then, the derivative is \( du = 2x \, dx \). We can rearrange this to get \( x \, dx = \frac{1}{2} du \). Substitute this into the integral:\[\int \frac{x}{x^{2}+4} \, dx = \int \frac{1}{u} \cdot \frac{1}{2} \, du = \frac{1}{2} \int \frac{1}{u} \, du.\]This integral becomes:\[\frac{1}{2} \ln |u| + C = \frac{1}{2} \ln |x^2 + 4| + C.\]So, the result of this method is \( \frac{1}{2} \ln |x^2 + 4| + C \).
02

Trigonometric Substitution Method

For trigonometric substitution, recognize that the denominator suggests using a tangent substitution. Let \( x = 2 \tan \theta \), which gives \( dx = 2 \sec^2 \theta \, d\theta \). Then:\[x^2 + 4 = 4\tan^2\theta + 4 = 4(\tan^2\theta + 1) = 4\sec^2\theta.\]Thus, our integral becomes:\[\int \frac{2 \tan \theta}{4 \sec^2 \theta} \, 2 \sec^2 \theta \, d\theta = \int \frac{2 \tan \theta}{4} \, d\theta = \frac{1}{2} \int \tan \theta \, d\theta.\]The integral of \( \tan \theta \) is\[\int \tan \theta \, d\theta = \ln |\sec \theta| + C.\]Thus, our integral becomes:\[\frac{1}{2} \ln |\sec \theta| + C.\]
03

Returning from Trigonometric Substitution

Now we need to express \( \ln |\sec \theta| \) in terms of \( x \). Recall \( \sec \theta = \frac{1}{\cos \theta} = \sqrt{1 + \tan^2\theta} \), so we have:\[\sec \theta = \sqrt{1 + \left(\frac{x}{2}\right)^2} = \frac{\sqrt{x^2 + 4}}{2}.\]Thus, \( \ln |\sec \theta| = \ln \left|\frac{\sqrt{x^2 + 4}}{2}\right| \), which can be split using properties of logarithms into:\[\ln |\sec \theta| = \ln |\sqrt{x^2 + 4}| - \ln 2.\]Therefore, \( \frac{1}{2} \ln |\sec \theta| = \frac{1}{2} (\ln |\sqrt{x^2 + 4}| - \ln 2) = \frac{1}{2} \ln |x^2 + 4| - \frac{1}{2} \ln 2 \).After adding the constant, our result matches previous step \( \frac{1}{2} \ln |x^2 + 4| + C \) after incorporating the constant of integration.

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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 powerful technique used in integral calculus when the integrand contains expressions like \(a^2 + x^2\), \(x^2 - a^2\), or \(a^2 - x^2\). These expressions suggest a relation to trigonometric identities.
When you see \(x^2 + a^2\), it's useful to let \(x = a \tan \theta\). This substitution simplifies the integral, often utilizing the Pythagorean identity \(1 + \tan^2 \theta = \sec^2 \theta\).

Here's why this method works beautifully:
  • Converts complex algebraic integrands into simpler trigonometric forms.
  • Uses known derivatives of trigonometric functions like \(\tan\), \(\sin\), or \(\cos\).
  • Allows conversion back to the original variable once integrated, relating \(\theta\) back to \(x\).
In our original exercise, by letting \( x = 2 \tan \theta \), we transformed the integral \(\int \frac{x}{x^2+4}\) into one involving \(\tan \theta\), which we know integrates to \(\ln |\sec \theta|\). This ties back to the original variable using trigonometric identities, giving us an easy path back into reality with algebraic expressions.
u-substitution
U-substitution, or integration by substitution, is an integral calculus technique often taught as a natural extension of the chain rule. It's about making integrals less daunting by substituting a complicated portion of the integral with a single variable \(u\).

Here's how it generally proceeds:
  • Identify a portion of the integral to substitute, usually the part that complicates the differentiation.
  • Let \(u\) represent this part, differentiating to find \(du\). Adjust your integral accordingly.
  • Recast the integral in terms of \(u\), which should result in a much simpler form.
  • Integrate and, finally, revert \(u\) back to the original variable.
In our example, substituting \(u = x^2 + 4\) transformed this exercise by changing \(dx\) to \(\frac{1}{2} du\). This leads to the integral becoming \(\int \frac{1}{u} \, du\), a classic derivative we recognize as \(\ln |u|\). The solution ends by back-substituting \(x\) in place of \(u\).
U-substitution is like untangling a knot in calculus, simplifying what can appear to be a daunting integral.
definite integrals
Definite integrals represent the accumulation of quantities, often thought of as the "net area" under a curve between two points. Unlike indefinite integrals that include a constant of integration, definite integrals produce a specific numerical value.

To solve a definite integral, we:
  • Evaluate the indefinite integral.
  • Apply the Fundamental Theorem of Calculus, using the limits of integration (the two endpoints).
  • Calculate the difference between the integral at these limits, \(F(b) - F(a)\).
While our original exercise didn't explicitly deal with the bounded limits of definite integrals, understanding their evaluation provides context to when substitutions, like our \(u\)-substitution, are readily applied to bounded integrals.
Integrals across specified ranges benefit greatly from these techniques, easing the process of computing exact areas or accumulated quantities.

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Most popular questions from this chapter

The exact value of the given integral is \(\pi\) (verify). Approximate the integral using (a) the midpoint approximation \(M_{10},(\mathrm{~b})\) the trapezoidal approximation \(T_{10}\), and (c) Simpson's rule approximation \(S_{20}\) using Formula (7). Approximate the absolute error and express your answers to at least four decimal places. $$ \int_{0}^{2} \frac{8}{x^{2}+4} d x $$

Evaluate the integral. $$ \int_{0}^{1 / 4} \sec \pi x \tan \pi x d x $$

(a) Make \(u\) -substitution (5) to convert the integrand to a rational function of \(u\), and then evaluate the integral. (b) If you have a CAS, use it to evaluate the integral (no substitution), and then confirm that the result is equivalent to that in part (a). $$ \int \frac{d \theta}{1-\cos \theta} $$

The trigonometric identity $$ \sin \alpha \cos \beta=\frac{1}{2}[\sin (\alpha-\beta)+\sin (\alpha+\beta)] $$ is often useful for evaluating integrals of the form \(\int \sin ^{m} x \cos ^{n} x d x\)

(a) Confirm graphically and algebraically that $$ \frac{1}{2 x+1} \leq \frac{e^{x}}{2 x+1} \quad(x \geq 0) $$ (b) Evaluate the integral $$ \int_{0}^{+\infty} \frac{d x}{2 x+1} $$ (c) What does the result obtained in part (b) tell you about the integral $$ \int_{0}^{+\infty} \frac{e^{x}}{2 x+1} d x ? $$
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