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Chapter 7: Problem 2

What change of variables is suggested by an integral containing \(\sqrt{x^{2}+36} ?\)

Short Answer

Expert verified
Answer: The appropriate trigonometric substitution for an integral containing the expression \(\sqrt{x^2+36}\) is \(x = 6\tan{\theta}\). After the substitution, the expression will be replaced with \(6\sec{\theta}\) and \(dx = 6\sec^2{\theta}d\theta\).

Step by step solution

01

Identify possible trigonometric substitutions

The most common trigonometric substitutions are: 1. \(x = a\sin{\theta}\), when an expression is of the form \(\sqrt{a^2-x^2}\). 2. \(x = a\cos{\theta}\), when an expression is of the form \(\sqrt{a^2-x^2}\) (can be used interchangeably with the first one). 3. \(x = a\tan{\theta}\), when an expression is of the form \(\sqrt{x^2+a^2}\). 4. \(x = a\sec{\theta}\), when an expression is of the form \(\sqrt{x^2-a^2}\). 5. \(x = a\csc{\theta}\), when an expression is of the form \(\sqrt{x^2-a^2}\) (can be used interchangeably with the fourth one). In this exercise, we have the expression \(\sqrt{x^2+36}\), which is of the form \(\sqrt{x^2+a^2}\).
02

Choose the appropriate trigonometric substitution

Since our expression is of the form \(\sqrt{x^2+a^2}\), we will choose the substitution: \(x = a\tan{\theta}\) In this case, \(a=6\), so our substitution will be: \(x = 6\tan{\theta}\)
03

Compute dx and the new expression

We need to compute \(\frac{dx}{d\theta}\) and replace the expression \(\sqrt{x^2+36}\) with the new variables. First, let's compute \(\frac{dx}{d\theta}\): \(\frac{dx}{d\theta} = 6\sec^2{\theta}\) Now, let's replace \(x\) in the expression with our substitution: \(\sqrt{x^2+36} = \sqrt{(6\tan{\theta})^2+36} = \sqrt{36\tan^2{\theta} + 36} = \sqrt{36(\tan^2{\theta}+1)}\) From the trigonometric identity, we know that \(\tan^2{\theta}+1 = \sec^2{\theta}\). Thus, we get: \(\sqrt{36(\tan^2{\theta}+1)} = \sqrt{36\sec^2{\theta}} = 6\sec{\theta}\) With this change of variable, the expression \(\sqrt{x^2+36}\) has been replaced with \(6\sec{\theta}\) and \(dx = 6\sec^2{\theta}d\theta\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Calculus
Calculus is the branch of mathematics that focuses on rates of change and the accumulation of quantities. It's an essential tool in understanding how functions behave, especially when analyzing curves and areas under curves. In this context, we deal with integrals, which are a key part of calculus.

Integration is the process of finding a function given its derivative. It鈥檚 the reverse operation of differentiation. Calculus allows us to evaluate areas, volumes, and other concepts in higher dimensions.

When dealing with complex integrals, such as ones containing square roots or other challenging expressions, calculus offers techniques like trigonometric substitution to simplify the process. This results in easier-to-evaluate integrals, aiding in the computation of areas and other problems.
Change of Variables
Change of variables is a powerful integration technique that simplifies complex integrals by transforming them into a more manageable form. It involves substituting a challenging part of the integral with a new variable, making the resulting integral easier to solve.

In our example, the expression \( \sqrt{x^2 + 36} \) necessitates a change of variables to simplify the integration process. By substituting \( x \) with \( 6\tan{\theta} \), the expression becomes more manageable due to the trigonometric identity \( \tan^2{\theta} + 1 = \sec^2{\theta} \).

The change of variables not only simplifies the integral but often reveals underlying relationships between different mathematical quantities, enhancing our understanding of their interactions in calculus problems.
Integration Techniques
Integration techniques are methods used to find the integral of complex mathematical expressions. These techniques aim to simplify the process and make challenging integrals more approachable.

One common technique is trigonometric substitution, which exploits trigonometric identities to replace complex expressions with simpler trigonometric functions. For expressions like \( \sqrt{x^2 + a^2} \), the substitution \( x = a\tan{\theta} \) is chosen because it transforms the square root into a straightforward trigonometric expression.

These techniques are critical in various applications, allowing engineers, physicists, and mathematicians to solve problems across diverse fields. Mastering them can help tackle integrals involving radicals, exponential functions, and even more complex structures with greater ease and understanding.

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Most popular questions from this chapter

Recall that the substitution \(x=a \sec \theta\) implies either \(x \geq a\) (in which case \(0 \leq \theta<\pi / 2\) and \(\tan \theta \geq 0)\) or \(x \leq-a\) (in which case \(\pi / 2<\theta \leq \pi\) and \(\tan \theta \leq 0\) ). $$\begin{aligned} &\text { Show that } \int \frac{d x}{x \sqrt{x^{2}-1}}=\\\ &\left\\{\begin{array}{ll} \sec ^{-1} x+C=\tan ^{-1} \sqrt{x^{2}-1}+C & \text { if } x>1 \\ -\sec ^{-1} x+C=-\tan ^{-1} \sqrt{x^{2}-1}+C & \text { if } x<-1 \end{array}\right. \end{aligned}$$

Use the window \([-2,2] \times[-2,2]\) to sketch a direction field for the following equations. Then sketch the solution curve that corresponds to the given initial condition. $$y^{\prime}(x)=\sin x, y(-2)=2$$

Compare the errors in the Midpoint and Trapezoid Rules with \(n=4,8,16,\) and 32 subintervals when they are applied to the following integrals (with their exact values given). \(\int_{0}^{\pi} \ln (5+3 \cos x) d x=\pi \ln \frac{9}{2}\)

Let \(R\) be the region bounded by the graph of \(f(x)=x^{-p}\) and the \(x\) -axis, for \(x \geq 1\) a. Let \(S\) be the solid generated when \(R\) is revolved about the \(x\) -axis. For what values of \(p\) is the volume of \(S\) finite? b. Let \(S\) be the solid generated when \(R\) is revolved about the \(y\) -axis. For what values of \(p\) is the volume of \(S\) finite?

A differential equation of the form \(y^{\prime}(t)=F(y)\) is said to be autonomous (the function \(F\) depends only on \(y\) ). The constant function \(y=y_{0}\) is an equilibrium solution of the equation provided \(F\left(y_{0}\right)=0\) (because then \(y^{\prime}(t)=0,\) and the solution remains constant for all \(t\) ). Note that equilibrium solutions correspond to horizontal line segments in the direction field. Note also that for autonomous equations, the direction field is independent of \(t\). Consider the following equations. a. Find all equilibrium solutions. b. Sketch the direction field on either side of the equilibrium solutions for \(t \geq 0\). c. Sketch the solution curve that corresponds to the initial condition \(y(0)=1\). $$y^{\prime}(t)=y(2-y)$$
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