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Calculus Volume 2

3.3 Trigonometric Substitution

Calculus Volume 23.3 Trigonometric Substitution
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  1. Preface
  2. 1 Integration
    1. Introduction
    2. 1.1 Approximating Areas
    3. 1.2 The Definite Integral
    4. 1.3 The Fundamental Theorem of Calculus
    5. 1.4 Integration Formulas and the Net Change Theorem
    6. 1.5 Substitution
    7. 1.6 Integrals Involving Exponential and Logarithmic Functions
    8. 1.7 Integrals Resulting in Inverse Trigonometric Functions
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  3. 2 Applications of Integration
    1. Introduction
    2. 2.1 Areas between Curves
    3. 2.2 Determining Volumes by Slicing
    4. 2.3 Volumes of Revolution: Cylindrical Shells
    5. 2.4 Arc Length of a Curve and Surface Area
    6. 2.5 Physical Applications
    7. 2.6 Moments and Centers of Mass
    8. 2.7 Integrals, Exponential Functions, and Logarithms
    9. 2.8 Exponential Growth and Decay
    10. 2.9 Calculus of the Hyperbolic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  4. 3 Techniques of Integration
    1. Introduction
    2. 3.1 Integration by Parts
    3. 3.2 Trigonometric Integrals
    4. 3.3 Trigonometric Substitution
    5. 3.4 Partial Fractions
    6. 3.5 Other Strategies for Integration
    7. 3.6 Numerical Integration
    8. 3.7 Improper Integrals
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  5. 4 Introduction to Differential Equations
    1. Introduction
    2. 4.1 Basics of Differential Equations
    3. 4.2 Direction Fields and Numerical Methods
    4. 4.3 Separable Equations
    5. 4.4 The Logistic Equation
    6. 4.5 First-order Linear Equations
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  6. 5 Sequences and Series
    1. Introduction
    2. 5.1 Sequences
    3. 5.2 Infinite Series
    4. 5.3 The Divergence and Integral Tests
    5. 5.4 Comparison Tests
    6. 5.5 Alternating Series
    7. 5.6 Ratio and Root Tests
    8. Key Terms
    9. Key Equations
    10. Key Concepts
    11. Chapter Review Exercises
  7. 6 Power Series
    1. Introduction
    2. 6.1 Power Series and Functions
    3. 6.2 Properties of Power Series
    4. 6.3 Taylor and Maclaurin Series
    5. 6.4 Working with Taylor Series
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  8. 7 Parametric Equations and Polar Coordinates
    1. Introduction
    2. 7.1 Parametric Equations
    3. 7.2 Calculus of Parametric Curves
    4. 7.3 Polar Coordinates
    5. 7.4 Area and Arc Length in Polar Coordinates
    6. 7.5 Conic Sections
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  9. A | Table of Integrals
  10. B | Table of Derivatives
  11. C | Review of Pre-Calculus
  12. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
  13. Index

Learning Objectives

  • 3.3.1. Solve integration problems involving the square root of a sum or difference of two squares.

In this section, we explore integrals containing expressions of the form a2x2,a2x2, a2+x2,a2+x2, and x2a2,x2a2, where the values of aa are positive. We have already encountered and evaluated integrals containing some expressions of this type, but many still remain inaccessible. The technique of trigonometric substitution comes in very handy when evaluating these integrals. This technique uses substitution to rewrite these integrals as trigonometric integrals.

Integrals Involving a2x2a2x2

Before developing a general strategy for integrals containing a2x2,a2x2, consider the integral 9x2dx.9x2dx. This integral cannot be evaluated using any of the techniques we have discussed so far. However, if we make the substitution x=3sinθ,x=3sinθ, we have dx=3cosθdθ.dx=3cosθdθ. After substituting into the integral, we have

9x2dx=9(3sinθ)23cosθdθ.9x2dx=9(3sinθ)23cosθdθ.

After simplifying, we have

9x2dx=91sin2θcosθdθ.9x2dx=91sin2θcosθdθ.

Letting 1sin2θ=cos2θ,1sin2θ=cos2θ, we now have

9x2dx=9cos2θcosθdθ.9x2dx=9cos2θcosθdθ.

Assuming that cosθ0,cosθ0, we have

9x2dx=9cos2θdθ.9x2dx=9cos2θdθ.

At this point, we can evaluate the integral using the techniques developed for integrating powers and products of trigonometric functions. Before completing this example, let’s take a look at the general theory behind this idea.

To evaluate integrals involving a2x2,a2x2, we make the substitution x=asinθx=asinθ and dx=acosθ.dx=acosθ. To see that this actually makes sense, consider the following argument: The domain of a2x2a2x2 is [a,a].[a,a]. Thus, axa.axa. Consequently, −1xa1.−1xa1. Since the range of sinxsinx over [(π/2),π/2][(π/2),π/2] is [−1,1],[−1,1], there is a unique angle θθ satisfying (π/2)θπ/2(π/2)θπ/2 so that sinθ=x/a,sinθ=x/a, or equivalently, so that x=asinθ.x=asinθ. If we substitute x=asinθx=asinθ into a2x2,a2x2, we get

a2x2=a2(asinθ)2Letx=asinθwhereπ2θπ2.Simplify.=a2a2sin2θFactor outa2.=a2(1sin2θ)Substitute1sin2x=cos2x.=a2cos2θTake the square root.=|acosθ|=acosθ.a2x2=a2(asinθ)2Letx=asinθwhereπ2θπ2.Simplify.=a2a2sin2θFactor outa2.=a2(1sin2θ)Substitute1sin2x=cos2x.=a2cos2θTake the square root.=|acosθ|=acosθ.

Since cosθ0cosθ0 on π2θπ2π2θπ2 and a>0,a>0, |acosθ|=acosθ.|acosθ|=acosθ. We can see, from this discussion, that by making the substitution x=asinθ,x=asinθ, we are able to convert an integral involving a radical into an integral involving trigonometric functions. After we evaluate the integral, we can convert the solution back to an expression involving x.x. To see how to do this, let’s begin by assuming that 0<x<a.0<x<a. In this case, 0<θ<π2.0<θ<π2. Since sinθ=xa,sinθ=xa, we can draw the reference triangle in Figure 3.4 to assist in expressing the values of cosθ,cosθ, tanθ,tanθ, and the remaining trigonometric functions in terms of x.x. It can be shown that this triangle actually produces the correct values of the trigonometric functions evaluated at θθ for all θθ satisfying π2θπ2.π2θπ2. It is useful to observe that the expression a2x2a2x2 actually appears as the length of one side of the triangle. Last, should θθ appear by itself, we use θ=sin−1(xa).θ=sin−1(xa).

This figure is a right triangle. It has an angle labeled theta. This angle is opposite the vertical side. The hypotenuse is labeled a, the vertical leg is labeled x, and the horizontal leg is labeled as the square root of (a^2 – x^2). To the left of the triangle is the equation sin(theta) = x/a.
Figure 3.4 A reference triangle can help express the trigonometric functions evaluated at θθ in terms of x.x.

The essential part of this discussion is summarized in the following problem-solving strategy.

Problem-Solving Strategy: Integrating Expressions Involving a2x2a2x2
  1. It is a good idea to make sure the integral cannot be evaluated easily in another way. For example, although this method can be applied to integrals of the form 1a2x2dx,1a2x2dx, xa2x2dx,xa2x2dx, and xa2x2dx,xa2x2dx, they can each be integrated directly either by formula or by a simple u-substitution.
  2. Make the substitution x=asinθx=asinθ and dx=acosθdθ.dx=acosθdθ. Note: This substitution yields a2x2=acosθ.a2x2=acosθ.
  3. Simplify the expression.
  4. Evaluate the integral using techniques from the section on trigonometric integrals.
  5. Use the reference triangle from Figure 3.4 to rewrite the result in terms of x.x. You may also need to use some trigonometric identities and the relationship θ=sin−1(xa).θ=sin−1(xa).

The following example demonstrates the application of this problem-solving strategy.

Example 3.21

Integrating an Expression Involving a2x2a2x2

Evaluate 9x2dx.9x2dx.

Solution

Begin by making the substitutions x=3sinθx=3sinθ and dx=3cosθdθ.dx=3cosθdθ. Since sinθ=x3,sinθ=x3, we can construct the reference triangle shown in the following figure.

This figure is a right triangle. It has an angle labeled theta. This angle is opposite the vertical side. The hypotenuse is labeled 3, the vertical leg is labeled x, and the horizontal leg is labeled as the square root of (9 – x^2). To the left of the triangle is the equation sin(theta) = x/3.
Figure 3.5 A reference triangle can be constructed for Example 3.21.

Thus,

9x2dx=9(3sinθ)23cosθdθSubstitutex=3sinθanddx=3cosθdθ.=9(1sin2θ)3cosθdθSimplify.=9cos2θ3cosθdθSubstitutecos2θ=1sin2θ.=3|cosθ|3cosθdθTake the square root.=9cos2θdθSimplify. Sinceπ2θπ2,cosθ0and|cosθ|=cosθ.=9(12+12cos(2θ))dθUse the strategy for integrating an even powerofcosθ.=92θ+94sin(2θ)+CEvaluate the integral.=92θ+94(2sinθcosθ)+CSubstitutesin(2θ)=2sinθcosθ.=92sin−1(x3)+92·x3·9x23+CSubstitutesin−1(x3)=θandsinθ=x3.Usethe reference triangle to see thatcosθ=9x23and make this substitution.=92sin−1(x3)+x9x22+C.Simplify.9x2dx=9(3sinθ)23cosθdθSubstitutex=3sinθanddx=3cosθdθ.=9(1sin2θ)3cosθdθSimplify.=9cos2θ3cosθdθSubstitutecos2θ=1sin2θ.=3|cosθ|3cosθdθTake the square root.=9cos2θdθSimplify. Sinceπ2θπ2,cosθ0and|cosθ|=cosθ.=9(12+12cos(2θ))dθUse the strategy for integrating an even powerofcosθ.=92θ+94sin(2θ)+CEvaluate the integral.=92θ+94(2sinθcosθ)+CSubstitutesin(2θ)=2sinθcosθ.=92sin−1(x3)+92·x3·9x23+CSubstitutesin−1(x3)=θandsinθ=x3.Usethe reference triangle to see thatcosθ=9x23and make this substitution.=92sin−1(x3)+x9x22+C.Simplify.

Example 3.22

Integrating an Expression Involving a2x2a2x2

Evaluate 4x2xdx.4x2xdx.

Solution

First make the substitutions x=2sinθx=2sinθ and dx=2cosθdθ.dx=2cosθdθ. Since sinθ=x2,sinθ=x2, we can construct the reference triangle shown in the following figure.

This figure is a right triangle. It has an angle labeled theta. This angle is opposite the vertical side. The vertical leg is labeled x, and the horizontal leg is labeled as the square root of (4 – x^2). To the left of the triangle is the equation sin(theta) = x/2.
Figure 3.6 A reference triangle can be constructed for Example 3.22.

Thus,

4x2xdx=4(2sinθ)22sinθ2cosθdθSubstitutex=2sinθand=2cosθdθ.=2cos2θsinθdθSubstitutecos2θ=1sin2θand simplify.=2(1sin2θ)sinθdθSubstitutesin2θ=1cos2θ.=(2cscθ2sinθ)dθSeparate the numerator, simplify, and usecscθ=1sinθ.=2ln|cscθcotθ|+2cosθ+CEvaluate the integral.=2ln|2x4x2x|+4x2+C.Use the reference triangle to rewrite theexpression in terms ofxand simplify.4x2xdx=4(2sinθ)22sinθ2cosθdθSubstitutex=2sinθand=2cosθdθ.=2cos2θsinθdθSubstitutecos2θ=1sin2θand simplify.=2(1sin2θ)sinθdθSubstitutesin2θ=1cos2θ.=(2cscθ2sinθ)dθSeparate the numerator, simplify, and usecscθ=1sinθ.=2ln|cscθcotθ|+2cosθ+CEvaluate the integral.=2ln|2x4x2x|+4x2+C.Use the reference triangle to rewrite theexpression in terms ofxand simplify.

In the next example, we see that we sometimes have a choice of methods.

Example 3.23

Integrating an Expression Involving a2x2a2x2 Two Ways

Evaluate x31x2dxx31x2dx two ways: first by using the substitution u=1x2u=1x2 and then by using a trigonometric substitution.

Solution

Method 1

Let u=1x2u=1x2 and hence x2=1u.x2=1u. Thus, du=−2xdx.du=−2xdx. In this case, the integral becomes

x31x2dx=12x21x2(−2xdx)Make the substitution.=12(1u)uduExpand the expression.=12(u1/2u3/2)duEvaluate the integral.=12(23u3/225u5/2)+CRewrite in terms ofx.=13(1x2)3/2+15(1x2)5/2+C.x31x2dx=12x21x2(−2xdx)Make the substitution.=12(1u)uduExpand the expression.=12(u1/2u3/2)duEvaluate the integral.=12(23u3/225u5/2)+CRewrite in terms ofx.=13(1x2)3/2+15(1x2)5/2+C.

Method 2

Let x=sinθ.x=sinθ. In this case, dx=cosθdθ.dx=cosθdθ. Using this substitution, we have

x31x2dx=sin3θcos2θdθ=(1cos2θ)cos2θsinθdθLetu=cosθ.Thus,du=sinθdθ.=(u4u2)du=15u513u3+CSubstitutecosθ=u.=15cos5θ13cos3θ+CUse a reference triangle to see thatcosθ=1x2.=15(1x2)5/213(1x2)3/2+C.x31x2dx=sin3θcos2θdθ=(1cos2θ)cos2θsinθdθLetu=cosθ.Thus,du=sinθdθ.=(u4u2)du=15u513u3+CSubstitutecosθ=u.=15cos5θ13cos3θ+CUse a reference triangle to see thatcosθ=1x2.=15(1x2)5/213(1x2)3/2+C.
Checkpoint 3.14

Rewrite the integral x325x2dxx325x2dx using the appropriate trigonometric substitution (do not evaluate the integral).

Integrating Expressions Involving a2+x2a2+x2

For integrals containing a2+x2,a2+x2, let’s first consider the domain of this expression. Since a2+x2a2+x2 is defined for all real values of x,x, we restrict our choice to those trigonometric functions that have a range of all real numbers. Thus, our choice is restricted to selecting either x=atanθx=atanθ or x=acotθ.x=acotθ. Either of these substitutions would actually work, but the standard substitution is x=atanθx=atanθ or, equivalently, tanθ=x/a.tanθ=x/a. With this substitution, we make the assumption that (π/2)<θ<π/2,(π/2)<θ<π/2, so that we also have θ=tan−1(x/a).θ=tan−1(x/a). The procedure for using this substitution is outlined in the following problem-solving strategy.

Problem-Solving Strategy: Integrating Expressions Involving a2+x2a2+x2
  1. Check to see whether the integral can be evaluated easily by using another method. In some cases, it is more convenient to use an alternative method.
  2. Substitute x=atanθx=atanθ and dx=asec2θdθ.dx=asec2θdθ. This substitution yields
    a2+x2=a2+(atanθ)2=a2(1+tan2θ)=a2sec2θ=|asecθ|=asecθ.a2+x2=a2+(atanθ)2=a2(1+tan2θ)=a2sec2θ=|asecθ|=asecθ. (Since π2<θ<π2π2<θ<π2 and secθ>0secθ>0 over this interval, |asecθ|=asecθ.)|asecθ|=asecθ.)
  3. Simplify the expression.
  4. Evaluate the integral using techniques from the section on trigonometric integrals.
  5. Use the reference triangle from Figure 3.7 to rewrite the result in terms of x.x. You may also need to use some trigonometric identities and the relationship θ=tan−1(xa).θ=tan−1(xa). (Note: The reference triangle is based on the assumption that x>0;x>0; however, the trigonometric ratios produced from the reference triangle are the same as the ratios for which x0.)x0.)
This figure is a right triangle. It has an angle labeled theta. This angle is opposite the vertical side. The hypotenuse is labeled the square root of (a^2+x^2), the vertical leg is labeled x, and the horizontal leg is labeled a. To the left of the triangle is the equation tan(theta) = x/a.
Figure 3.7 A reference triangle can be constructed to express the trigonometric functions evaluated at θθ in terms of x.x.

Example 3.24

Integrating an Expression Involving a2+x2a2+x2

Evaluate dx1+x2dx1+x2 and check the solution by differentiating.

Solution

Begin with the substitution x=tanθx=tanθ and dx=sec2θdθ.dx=sec2θdθ. Since tanθ=x,tanθ=x, draw the reference triangle in the following figure.

This figure is a right triangle. It has an angle labeled theta. This angle is opposite the vertical side. The hypotenuse is labeled the square root of (1+x^2), the vertical leg is labeled x, and the horizontal leg is labeled 1. To the left of the triangle is the equation tan(theta) = x/1.
Figure 3.8 The reference triangle for Example 3.24.

Thus,

dx1+x2=sec2θsecθdθSubstitutex=tanθanddx=sec2θdθ.Thissubstitution makes1+x2=secθ.Simplify.=secθdθEvaluate the integral.=ln|secθ+tanθ|+CUse the reference triangle to express the resultin terms ofx.=ln|1+x2+x|+C.dx1+x2=sec2θsecθdθSubstitutex=tanθanddx=sec2θdθ.Thissubstitution makes1+x2=secθ.Simplify.=secθdθEvaluate the integral.=ln|secθ+tanθ|+CUse the reference triangle to express the resultin terms ofx.=ln|1+x2+x|+C.

To check the solution, differentiate:

ddx(ln|1+x2+x|)=11+x2+x·(x1+x2+1)=11+x2+x·x+1+x21+x2=11+x2.ddx(ln|1+x2+x|)=11+x2+x·(x1+x2+1)=11+x2+x·x+1+x21+x2=11+x2.

Since 1+x2+x>01+x2+x>0 for all values of x,x, we could rewrite ln|1+x2+x|+C=ln(1+x2+x)+C,ln|1+x2+x|+C=ln(1+x2+x)+C, if desired.

Example 3.25

Evaluating dx1+x2dx1+x2 Using a Different Substitution

Use the substitution x=sinhθx=sinhθ to evaluate dx1+x2.dx1+x2.

Solution

Because sinhθsinhθ has a range of all real numbers, and 1+sinh2θ=cosh2θ,1+sinh2θ=cosh2θ, we may also use the substitution x=sinhθx=sinhθ to evaluate this integral. In this case, dx=coshθdθ.dx=coshθdθ. Consequently,

dx1+x2=coshθ1+sinh2θdθSubstitutex=sinhθanddx=coshθdθ.Substitute1+sinh2θ=cosh2θ.=coshθcosh2θdθcosh2θ=|coshθ|=coshθ|coshθ|dθ|coshθ|=coshθsincecoshθ>0for allθ.=coshθcoshθdθSimplify.=1dθEvaluate the integral.=θ+CSincex=sinhθ,we knowθ=sinh−1x.=sinh−1x+C.dx1+x2=coshθ1+sinh2θdθSubstitutex=sinhθanddx=coshθdθ.Substitute1+sinh2θ=cosh2θ.=coshθcosh2θdθcosh2θ=|coshθ|=coshθ|coshθ|dθ|coshθ|=coshθsincecoshθ>0for allθ.=coshθcoshθdθSimplify.=1dθEvaluate the integral.=θ+CSincex=sinhθ,we knowθ=sinh−1x.=sinh−1x+C.

Analysis

This answer looks quite different from the answer obtained using the substitution x=tanθ.x=tanθ. To see that the solutions are the same, set y=sinh−1x.y=sinh−1x. Thus, sinhy=x.sinhy=x. From this equation we obtain:

eyey2=x.eyey2=x.

After multiplying both sides by 2ey2ey and rewriting, this equation becomes:

e2y2xey1=0.e2y2xey1=0.

Use the quadratic equation to solve for ey:ey:

ey=2x±4x2+42.ey=2x±4x2+42.

Simplifying, we have:

ey=x±x2+1.ey=x±x2+1.

Since xx2+1<0,xx2+1<0, it must be the case that ey=x+x2+1.ey=x+x2+1. Thus,

y=ln(x+x2+1).y=ln(x+x2+1).

Last, we obtain

sinh−1x=ln(x+x2+1).sinh−1x=ln(x+x2+1).

After we make the final observation that, since x+x2+1>0,x+x2+1>0,

ln(x+x2+1)=ln|1+x2+x|,ln(x+x2+1)=ln|1+x2+x|,

we see that the two different methods produced equivalent solutions.

Example 3.26

Finding an Arc Length

Find the length of the curve y=x2y=x2 over the interval [0,12].[0,12].

Solution

Because dydx=2x,dydx=2x, the arc length is given by

01/21+(2x)2dx=01/21+4x2dx.01/21+(2x)2dx=01/21+4x2dx.

To evaluate this integral, use the substitution x=12tanθx=12tanθ and dx=12sec2θdθ.dx=12sec2θdθ. We also need to change the limits of integration. If x=0,x=0, then θ=0θ=0 and if x=12,x=12, then θ=π4.θ=π4. Thus,

01/21+4x2dx=0π/41+tan2θ12sec2θdθAfter substitution,1+4x2=tanθ.Substitute1+tan2θ=sec2θand simplify.=120π/4sec3θdθWe derived this integral in theprevious section.=12(12secθtanθ+ln|secθ+tanθ|)|0π/4Evaluate and simplify.=14(2+ln(2+1)).01/21+4x2dx=0π/41+tan2θ12sec2θdθAfter substitution,1+4x2=tanθ.Substitute1+tan2θ=sec2θand simplify.=120π/4sec3θdθWe derived this integral in theprevious section.=12(12secθtanθ+ln|secθ+tanθ|)|0π/4Evaluate and simplify.=14(2+ln(2+1)).
Checkpoint 3.15

Rewrite x3x2+4dxx3x2+4dx by using a substitution involving tanθ.tanθ.

Integrating Expressions Involving x2a2x2a2

The domain of the expression x2a2x2a2 is (,a][a,+).(,a][a,+). Thus, either xaxa or xa.xa. Hence, xa1xa1 or xa1.xa1. Since these intervals correspond to the range of secθsecθ on the set [0,π2)(π2,π],[0,π2)(π2,π], it makes sense to use the substitution secθ=xasecθ=xa or, equivalently, x=asecθ,x=asecθ, where 0θ<π20θ<π2 or π2<θπ.π2<θπ. The corresponding substitution for dxdx is dx=asecθtanθdθ.dx=asecθtanθdθ. The procedure for using this substitution is outlined in the following problem-solving strategy.

Problem-Solving Strategy: Integrals Involving x2a2x2a2
  1. Check to see whether the integral cannot be evaluated using another method. If so, we may wish to consider applying an alternative technique.
  2. Substitute x=asecθx=asecθ and dx=asecθtanθdθ.dx=asecθtanθdθ. This substitution yields
    x2a2=(asecθ)2a2=a2(sec2θ1)=a2tan2θ=|atanθ|.x2a2=(asecθ)2a2=a2(sec2θ1)=a2tan2θ=|atanθ|.

    For xa,xa, |atanθ|=atanθ|atanθ|=atanθ and for xa,xa, |atanθ|=atanθ.|atanθ|=atanθ.
  3. Simplify the expression.
  4. Evaluate the integral using techniques from the section on trigonometric integrals.
  5. Use the reference triangles from Figure 3.9 to rewrite the result in terms of x.x. You may also need to use some trigonometric identities and the relationship θ=sec−1(xa).θ=sec−1(xa). (Note: We need both reference triangles, since the values of some of the trigonometric ratios are different depending on whether xaxa or xa.)xa.)
This figure has two right triangles. The first triangle is in the first quadrant of the xy coordinate system and has an angle labeled theta. This angle is opposite the vertical side. The hypotenuse is labeled x, the vertical leg is labeled the square root of (x^2-a^2), and the horizontal leg is labeled a. The horizontal leg is on the x-axis. To the left of the triangle is the equation sec(theta) = x/a, x>a. There are also the equations sin(theta)= the square root of (x^2-a^2)/x, cos(theta) = a/x, and tan(theta) = the square root of (x^2-a^2)/a. The second triangle is in the second quadrant, with the hypotenuse labeled –x. The horizontal leg is labeled –a and is on the negative x-axis. The vertical leg is labeled the square root of (x^2-a^2). To the right of the triangle is the equation sec(theta) = x/a, x<-a. There are also the equations sin(theta)= the negative square root of (x^2-a^2)/x, cos(theta) = a/x, and tan(theta) = the negative square root of (x^2-a^2)/a.
Figure 3.9 Use the appropriate reference triangle to express the trigonometric functions evaluated at θθ in terms of x.x.

Example 3.27

Finding the Area of a Region

Find the area of the region between the graph of f(x)=x29f(x)=x29 and the x-axis over the interval [3,5].[3,5].

Solution

First, sketch a rough graph of the region described in the problem, as shown in the following figure.

This figure is the graph of the function f(x) = the square root of (x^2-9). It is an increasing curve that starts on the x-axis at 3 and is in the first quadrant. Under the curve above the x-axis is a shaded region bounded to the right at x = 5.
Figure 3.10 Calculating the area of the shaded region requires evaluating an integral with a trigonometric substitution.

We can see that the area is A=35x29dx.A=35x29dx. To evaluate this definite integral, substitute x=3secθx=3secθ and dx=3secθtanθdθ.dx=3secθtanθdθ. We must also change the limits of integration. If x=3,x=3, then 3=3secθ3=3secθ and hence θ=0.θ=0. If x=5,x=5, then θ=sec−1(53).θ=sec−1(53). After making these substitutions and simplifying, we have

Area=35x29dx=0sec−1(5/3)9tan2θsecθdθUsetan2θ=1sec2θ.=0sec−1(5/3)9(sec2θ1)secθdθExpand.=0sec−1(5/3)9(sec3θsecθ)dθEvaluate the integral.=(92ln|secθ+tanθ|+92secθtanθ)9ln|secθ+tanθ||0sec−1(5/3)Simplify.=92secθtanθ92ln|secθ+tanθ||0sec−1(5/3)Evaluate. Usesec(sec−153)=53andtan(sec−153)=43.=92·53·4392ln|53+43|(92·1·092ln|1+0|)=1092ln3.Area=35x29dx=0sec−1(5/3)9tan2θsecθdθUsetan2θ=1sec2θ.=0sec−1(5/3)9(sec2θ1)secθdθExpand.=0sec−1(5/3)9(sec3θsecθ)dθEvaluate the integral.=(92ln|secθ+tanθ|+92secθtanθ)9ln|secθ+tanθ||0sec−1(5/3)Simplify.=92secθtanθ92ln|secθ+tanθ||0sec−1(5/3)Evaluate. Usesec(sec−153)=53andtan(sec−153)=43.=92·53·4392ln|53+43|(92·1·092ln|1+0|)=1092ln3.
Checkpoint 3.16

Evaluate dxx24.dxx24. Assume that x>2.x>2.

Section 3.3 Exercises

Simplify the following expressions by writing each one using a single trigonometric function.

126.

44sin2θ44sin2θ

127.

9sec2θ99sec2θ9

128.

a2+a2tan2θa2+a2tan2θ

129.

a2+a2sinh2θa2+a2sinh2θ

130.

16cosh2θ1616cosh2θ16

Use the technique of completing the square to express each trinomial as the square of a binomial.

131.

4x24x+14x24x+1

132.

2x28x+32x28x+3

133.

x22x+4x22x+4

Integrate using the method of trigonometric substitution. Express the final answer in terms of the variable.

134.

dx4x2dx4x2

135.

dxx2a2dxx2a2

136.

4x2dx4x2dx

137.

dx1+9x2dx1+9x2

138.

x2dx1x2x2dx1x2

139.

dxx21x2dxx21x2

140.

dx(1+x2)2dx(1+x2)2

141.

x2+9dxx2+9dx

142.

x225xdxx225xdx

143.

θ3dθ9θ2θ3dθ9θ2

144.

dxx6x2dxx6x2

145.

x6x8dxx6x8dx

146.

dx(1+x2)3/2dx(1+x2)3/2

147.

dx(x29)3/2dx(x29)3/2

148.

1+x2dxx1+x2dxx

149.

x2dxx21x2dxx21

150.

x2dxx2+4x2dxx2+4

151.

dxx2x2+1dxx2x2+1

152.

x2dx1+x2x2dx1+x2

153.

−11(1x2)3/2dx−11(1x2)3/2dx

In the following exercises, use the substitutions x=sinhθ,coshθ,x=sinhθ,coshθ, or tanhθ.tanhθ. Express the final answers in terms of the variable x.

154.

dxx21dxx21

155.

dxx1x2dxx1x2

156.

x21dxx21dx

157.

x21x2dxx21x2dx

158.

dx1x2dx1x2

159.

1+x2x2dx1+x2x2dx

Use the technique of completing the square to evaluate the following integrals.

160.

1x26xdx1x26xdx

161.

1x2+2x+1dx1x2+2x+1dx

162.

1x2+2x+8dx1x2+2x+8dx

163.

1x2+10xdx1x2+10xdx

164.

1x2+4x12dx1x2+4x12dx

165.

Evaluate the integral without using calculus: −339x2dx.−339x2dx.

166.

Find the area enclosed by the ellipse x24+y29=1.x24+y29=1.

167.

Evaluate the integral dx1x2dx1x2 using two different substitutions. First, let x=cosθx=cosθ and evaluate using trigonometric substitution. Second, let x=sinθx=sinθ and use trigonometric substitution. Are the answers the same?

168.

Evaluate the integral dxxx21dxxx21 using the substitution x=secθ.x=secθ. Next, evaluate the same integral using the substitution x=cscθ.x=cscθ. Show that the results are equivalent.

169.

Evaluate the integral xx2+1dxxx2+1dx using the form 1udu.1udu. Next, evaluate the same integral using x=tanθ.x=tanθ. Are the results the same?

170.

State the method of integration you would use to evaluate the integral xx2+1dx.xx2+1dx. Why did you choose this method?

171.

State the method of integration you would use to evaluate the integral x2x21dx.x2x21dx. Why did you choose this method?

172.

Evaluate −11xdxx2+1−11xdxx2+1

173.

Find the length of the arc of the curve over the specified interval: y=lnx,[1,5].y=lnx,[1,5]. Round the answer to three decimal places.

174.

Find the surface area of the solid generated by revolving the region bounded by the graphs of y=x2,y=0,x=0,andx=2y=x2,y=0,x=0,andx=2 about the x-axis. (Round the answer to three decimal places).

175.

The region bounded by the graph of f(x)=11+x2f(x)=11+x2 and the x-axis between x=0x=0 and x=1x=1 is revolved about the x-axis. Find the volume of the solid that is generated.

Solve the initial-value problem for y as a function of x.

176.

(x2+36)dydx=1,y(6)=0(x2+36)dydx=1,y(6)=0

177.

(64x2)dydx=1,y(0)=3(64x2)dydx=1,y(0)=3

178.

Find the area bounded by y=2644x2,x=0,y=0,andx=2.y=2644x2,x=0,y=0,andx=2.

179.

An oil storage tank can be described as the volume generated by revolving the area bounded by y=1664+x2,x=0,y=0,x=2y=1664+x2,x=0,y=0,x=2 about the x-axis. Find the volume of the tank (in cubic meters).

180.

During each cycle, the velocity v (in feet per second) of a robotic welding device is given by v=2t144+t2,v=2t144+t2, where t is time in seconds. Find the expression for the displacement s (in feet) as a function of t if s=0s=0 when t=0.t=0.

181.

Find the length of the curve y=16x2y=16x2 between x=0x=0 and x=2.x=2.

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