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Chapter 7: Problem 20

Evaluate the following integrals. $$\int \frac{1}{\left(1+x^{2}\right)^{3 / 2}} d x$$

Short Answer

Expert verified
Question: Evaluate the following integral: \(\int \frac{1}{(1+x^2)^{3/2}} dx\) Answer: The evaluated integral is \(\int \frac{1}{(1+x^2)^{3/2}} dx = \frac{x}{\sqrt{1 + x^2}} + C\).

Step by step solution

01

Make a Substitution

We will use the trigonometric substitution \(x = \tan{(\theta)}\). This means that \(dx = \sec^2{(\theta)} d\theta\). Substituting these into the given integral, we get $$ \int \frac{1}{(1 + \tan^2{(\theta)})^{3/2}} \sec^2{(\theta)} d\theta $$
02

Simplify the Integral

Since the identity \(\sec^2(\theta) = 1 + \tan^2(\theta)\) holds, our integral now simplifies to $$ \int \frac{\sec^2{(\theta)}}{(\sec^2{(\theta)})^{3/2}} d\theta $$ Which can be further simplified to $$ \int \frac{1}{\sec{(\theta)}} d\theta = \int \cos{(\theta)} d\theta $$
03

Evaluate the Integral

Integrating the simplified integral, we have $$ \int \cos{(\theta)} d\theta = \sin{(\theta)} + C $$
04

Convert Back to x

Now we substitute back the original variable x from the substitution \(x = \tan{(\theta)}\). Since we know that \(\sin{(\theta)} = \frac{\tan{(\theta)}}{\sec{(\theta)}} = \frac{x}{\sqrt{1 + x^2}}\), we get $$ \sin{(\theta)} + C = \frac{x}{\sqrt{1 + x^2}} + C $$
05

Final Answer

The evaluated integral is $$ \int \frac{1}{(1+x^2)^{3/2}} dx = \frac{x}{\sqrt{1 + x^2}} + C $$

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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 handy tool in integral calculus when dealing with integrals involving square roots or quadratic expressions. By substituting a trigonometric function for a variable, we can simplify the integral and make it more manageable. For example, when we encounter a term like \(1 + x^2\), replacing \(x\) with \(\tan(\theta)\) allows us to use the identity \(1 + \tan^2(\theta) = \sec^2(\theta)\). This changes the integral into an expression involving trigonometric functions, which are often easier to integrate.

The process is as follows:
  • Identify the form of the expression (e.g., \(1 + x^2\)).
  • Select an appropriate trigonometric substitution (e.g., \(x = \tan(\theta)\)).
  • Substitute and simplify using trigonometric identities.
  • Integrate the resulting trigonometric integral.
  • Convert back to the original variable using the inverse of the substitution.
This technique reduces the complexity of otherwise challenging integrals and leverages the simplicity of trigonometric identities.
Integration Techniques
Integration techniques refer to various methods used to solve integrals. One such technique is trigonometric substitution, which is part of a broader set of strategies including substitution, integration by parts, and partial fraction decomposition.

When faced with complex expressions, recognize the form they take:
  • Substitution: Used when an integral contains a function and its derivative, allowing for substitution to simplify the expression.
  • Trigonometric Substitution: Fits well with expressions involving square roots and can simplify by converting into trigonometric forms.
  • Integration by Parts: Utilized for the product of two functions, leveraging the formula \(\int u \, dv = uv - \int v \, du\).
  • Partial Fraction Decomposition: Handy for rational expressions and breaks them into simpler fractions for easy integration.
Mastering these techniques allows one to approach integrals with confidence and flexibility, choosing the most efficient method based on the structure of the problem.
Multiple Integrals
Multiple integrals extend the concept of integration to functions of several variables, such as double or triple integrals. In such cases, integration often occurs over regions in two-dimensional or three-dimensional spaces. While our original exercise was a single integral, understanding multiple integrals is key for advanced calculus topics.

Here's a quick outline:
  • Double Integral: Used for integrating over a two-dimensional region, denoted as \(\int \int f(x, y) \, dx \, dy\). It calculates the volume under a surface.
  • Triple Integral: Involves three variables and integrates over a three-dimensional region, denoted as \(\int \int \int f(x, y, z) \, dx \, dy \, dz\). Useful for volume calculations in three dimensions.
Similar to single integrals, techniques such as substitution and changing the order of integration are applied in multiple integrals to simplify computation. Mastery in single integrals paves the way to tackle complex multiple integral problems.

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Most popular questions from this chapter

The work required to launch an object from the surface of Earth to outer space is given by \(W=\int_{R}^{\infty} F(x) d x,\) where \(R=6370 \mathrm{km}\) is the approximate radius of Earth, \(F(x)=G M m / x^{2}\) is the gravitational force between Earth and the object, \(G\) is the gravitational constant, \(M\) is the mass of Earth, \(m\) is the mass of the object, and \(G M=4 \times 10^{14} \mathrm{m}^{3} / \mathrm{s}^{2}.\) a. Find the work required to launch an object in terms of \(m.\) b. What escape velocity \(v_{e}\) is required to give the object a kinetic energy \(\frac{1}{2} m v_{e}^{2}\) equal to \(W ?\) c. The French scientist Laplace anticipated the existence of black holes in the 18th century with the following argument: If a body has an escape velocity that equals or exceeds the speed of light, \(c=300,000 \mathrm{km} / \mathrm{s},\) then light cannot escape the body and it cannot be seen. Show that such a body has a radius \(R \leq 2 G M / c^{2} .\) For Earth to be a black hole, what would its radius need to be?

Let \(a>0\) and let \(R\) be the region bounded by the graph of \(y=e^{-a x}\) and the \(x\) -axis on the interval \([b, \infty).\) a. Find \(A(a, b),\) the area of \(R\) as a function of \(a\) and \(b\) b. Find the relationship \(b=g(a)\) such that \(A(a, b)=2\) c. What is the minimum value of \(b\) (call it \(b^{*}\) ) such that when \(b>b^{*}, A(a, b)=2\) for some value of \(a>0 ?\)

Let \(R\) be the region between the curves \(y=e^{-c x}\) and \(y=-e^{-c x}\) on the interval \([a, \infty),\) where \(a \geq 0\) and \(c \geq 0 .\) The center of mass of \(R\) is located at \((\bar{x}, 0)\) where \(\bar{x}=\frac{\int_{a}^{\infty} x e^{-c x} d x}{\int_{a}^{\infty} e^{-c x} d x} .\) (The profile of the Eiffel Tower is modeled by the two exponential curves.) a. For \(a=0\) and \(c=2,\) sketch the curves that define \(R\) and find the center of mass of \(R\). Indicate the location of the center of mass. b. With \(a=0\) and \(c=2,\) find equations of the lines tangent to the curves at the points corresponding to \(x=0.\) c. Show that the tangent lines intersect at the center of mass. d. Show that this same property holds for any \(a \geq 0\) and any \(c>0 ;\) that is, the tangent lines to the curves \(y=\pm e^{-c x}\) at \(x=a\) intersect at the center of mass of \(R\) (Source: P. Weidman and I. Pinelis, Comptes Rendu, Mechanique \(332(2004): 571-584 .)\)

a. Verify the identity \(\sec x=\frac{\cos x}{1-\sin ^{2} x}\) b. Use the identity in part (a) to verify that \(\int \sec x d x=\frac{1}{2} \ln \left|\frac{1+\sin x}{1-\sin x}\right|+C\) (Source: The College Mathematics Joumal \(32,\) No. 5 (November 2001))

Imagine that today you deposit \(\$ B\) in a savings account that earns interest at a rate of \(p \%\) per year compounded continuously. The goal is to draw an income of \(\$ I\) per year from the account forever. The amount of money that must be deposited is \(B=I \int_{0}^{\infty} e^{-n t} d t,\) where \(r=p / 100 .\) Suppose you find an account that earns \(12 \%\) interest annully and you wish to have an income from the account of \(\$ 5000\) per year. How much must you deposit today?
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