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Calculus Volume 2

3.2 Trigonometric Integrals

Calculus Volume 23.2 Trigonometric Integrals
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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

  • 3.2.1. Solve integration problems involving products and powers of sinxsinx and cosx.cosx.
  • 3.2.2. Solve integration problems involving products and powers of tanxtanx and secx.secx.
  • 3.2.3. Use reduction formulas to solve trigonometric integrals.

In this section we look at how to integrate a variety of products of trigonometric functions. These integrals are called trigonometric integrals. They are an important part of the integration technique called trigonometric substitution, which is featured in Trigonometric Substitution. This technique allows us to convert algebraic expressions that we may not be able to integrate into expressions involving trigonometric functions, which we may be able to integrate using the techniques described in this section. In addition, these types of integrals appear frequently when we study polar, cylindrical, and spherical coordinate systems later. Let’s begin our study with products of sinxsinx and cosx.cosx.

Integrating Products and Powers of sinx and cosx

A key idea behind the strategy used to integrate combinations of products and powers of sinxsinx and cosxcosx involves rewriting these expressions as sums and differences of integrals of the form sinjxcosxdxsinjxcosxdx or cosjxsinxdx.cosjxsinxdx. After rewriting these integrals, we evaluate them using u-substitution. Before describing the general process in detail, let’s take a look at the following examples.

Example 3.8

Integrating cosjxsinxdxcosjxsinxdx

Evaluate cos3xsinxdx.cos3xsinxdx.

Solution

Use uu-substitution and let u=cosx.u=cosx. In this case, du=sinxdx.du=sinxdx. Thus,

cos3xsinxdx=u3du=14u4+C=14cos4x+C.cos3xsinxdx=u3du=14u4+C=14cos4x+C.
Checkpoint 3.5

Evaluate sin4xcosxdx.sin4xcosxdx.

Example 3.9

A Preliminary Example: Integrating cosjxsinkxdxcosjxsinkxdx Where k is Odd

Evaluate cos2xsin3xdx.cos2xsin3xdx.

Solution

To convert this integral to integrals of the form cosjxsinxdx,cosjxsinxdx, rewrite sin3x=sin2xsinxsin3x=sin2xsinx and make the substitution sin2x=1cos2x.sin2x=1cos2x. Thus,

cos2xsin3xdx=cos2x(1cos2x)sinxdxLetu=cosx;thendu=sinxdx.=u2(1u2)du=(u4u2)du=15u513u3+C=15cos5x13cos3x+C.cos2xsin3xdx=cos2x(1cos2x)sinxdxLetu=cosx;thendu=sinxdx.=u2(1u2)du=(u4u2)du=15u513u3+C=15cos5x13cos3x+C.
Checkpoint 3.6

Evaluate cos3xsin2xdx.cos3xsin2xdx.

In the next example, we see the strategy that must be applied when there are only even powers of sinxsinx and cosx.cosx. For integrals of this type, the identities

sin2x=1212cos(2x)=1cos(2x)2sin2x=1212cos(2x)=1cos(2x)2

and

cos2x=12+12cos(2x)=1+cos(2x)2cos2x=12+12cos(2x)=1+cos(2x)2

are invaluable. These identities are sometimes known as power-reducing identities and they may be derived from the double-angle identity cos(2x)=cos2xsin2xcos(2x)=cos2xsin2x and the Pythagorean identity cos2x+sin2x=1.cos2x+sin2x=1.

Example 3.10

Integrating an Even Power of sinxsinx

Evaluate sin2xdx.sin2xdx.

Solution

To evaluate this integral, let’s use the trigonometric identity sin2x=1212cos(2x).sin2x=1212cos(2x). Thus,

sin2xdx=(1212cos(2x))dx=12x14sin(2x)+C.sin2xdx=(1212cos(2x))dx=12x14sin(2x)+C.
Checkpoint 3.7

Evaluate cos2xdx.cos2xdx.

The general process for integrating products of powers of sinxsinx and cosxcosx is summarized in the following set of guidelines.

Problem-Solving Strategy: Integrating Products and Powers of sin x and cos x

To integrate cosjxsinkxdxcosjxsinkxdx use the following strategies:

  1. If kk is odd, rewrite sinkx=sink1xsinxsinkx=sink1xsinx and use the identity sin2x=1cos2xsin2x=1cos2x to rewrite sink1xsink1x in terms of cosx.cosx. Integrate using the substitution u=cosx.u=cosx. This substitution makes du=sinxdx.du=sinxdx.
  2. If jj is odd, rewrite cosjx=cosj1xcosxcosjx=cosj1xcosx and use the identity cos2x=1sin2xcos2x=1sin2x to rewrite cosj1xcosj1x in terms of sinx.sinx. Integrate using the substitution u=sinx.u=sinx. This substitution makes du=cosxdx.du=cosxdx. (Note: If both jj and kk are odd, either strategy 1 or strategy 2 may be used.)
  3. If both jj and kk are even, use sin2x=(1/2)(1/2)cos(2x)sin2x=(1/2)(1/2)cos(2x) and cos2x=(1/2)+(1/2)cos(2x).cos2x=(1/2)+(1/2)cos(2x). After applying these formulas, simplify and reapply strategies 1 through 3 as appropriate.

Example 3.11

Integrating cosjxsinkxdxcosjxsinkxdx where k is Odd

Evaluate cos8xsin5xdx.cos8xsin5xdx.

Solution

Since the power on sinxsinx is odd, use strategy 1. Thus,

cos8xsin5xdx=cos8xsin4xsinxdxBreak offsinx.=cos8x(sin2x)2sinxdxRewritesin4x=(sin2x)2.=cos8x(1cos2x)2sinxdxSubstitutesin2x=1cos2x.=u8(1u2)2(du)Letu=cosxanddu=sinxdx.=(u8+2u10u12)duExpand.=19u9+211u11113u13+CEvaluate the integral.=19cos9x+211cos11x113cos13x+C.Substituteu=cosx.cos8xsin5xdx=cos8xsin4xsinxdxBreak offsinx.=cos8x(sin2x)2sinxdxRewritesin4x=(sin2x)2.=cos8x(1cos2x)2sinxdxSubstitutesin2x=1cos2x.=u8(1u2)2(du)Letu=cosxanddu=sinxdx.=(u8+2u10u12)duExpand.=19u9+211u11113u13+CEvaluate the integral.=19cos9x+211cos11x113cos13x+C.Substituteu=cosx.

Example 3.12

Integrating cosjxsinkxdxcosjxsinkxdx where k and j are Even

Evaluate sin4xdx.sin4xdx.

Solution

Since the power on sinxsinx is even (k=4)(k=4) and the power on cosxcosx is even (j=0),(j=0), we must use strategy 3. Thus,

sin4xdx=(sin2x)2dxRewritesin4x=(sin2x)2.=(1212cos(2x))2dxSubstitutesin2x=1212cos(2x).=(1412cos(2x)+14cos2(2x))dxExpand(1212cos(2x))2.=(1412cos(2x)+14(12+12cos(4x))dx.sin4xdx=(sin2x)2dxRewritesin4x=(sin2x)2.=(1212cos(2x))2dxSubstitutesin2x=1212cos(2x).=(1412cos(2x)+14cos2(2x))dxExpand(1212cos(2x))2.=(1412cos(2x)+14(12+12cos(4x))dx.

Since cos2(2x)cos2(2x) has an even power, substitute cos2(2x)=12+12cos(4x):cos2(2x)=12+12cos(4x):

=(3812cos(2x)+18cos(4x))dxSimplify.=38x14sin(2x)+132sin(4x)+CEvaluate the integral.=(3812cos(2x)+18cos(4x))dxSimplify.=38x14sin(2x)+132sin(4x)+CEvaluate the integral.
Checkpoint 3.8

Evaluate cos3xdx.cos3xdx.

Checkpoint 3.9

Evaluate cos2(3x)dx.cos2(3x)dx.

In some areas of physics, such as quantum mechanics, signal processing, and the computation of Fourier series, it is often necessary to integrate products that include sin(ax),sin(ax), sin(bx),sin(bx), cos(ax),cos(ax), and cos(bx).cos(bx). These integrals are evaluated by applying trigonometric identities, as outlined in the following rule.

Rule: Integrating Products of Sines and Cosines of Different Angles

To integrate products involving sin(ax),sin(ax), sin(bx),sin(bx), cos(ax),cos(ax), and cos(bx),cos(bx), use the substitutions

sin(ax)sin(bx)=12cos((ab)x)12cos((a+b)x)sin(ax)sin(bx)=12cos((ab)x)12cos((a+b)x)
3.3
sin(ax)cos(bx)=12sin((ab)x)+12sin((a+b)x)sin(ax)cos(bx)=12sin((ab)x)+12sin((a+b)x)
3.4
cos(ax)cos(bx)=12cos((ab)x)+12cos((a+b)x)cos(ax)cos(bx)=12cos((ab)x)+12cos((a+b)x)
3.5

These formulas may be derived from the sum-of-angle formulas for sine and cosine.

Example 3.13

Evaluating sin(ax)cos(bx)dxsin(ax)cos(bx)dx

Evaluate sin(5x)cos(3x)dx.sin(5x)cos(3x)dx.

Solution

Apply the identity sin(5x)cos(3x)=12sin(2x)12cos(8x).sin(5x)cos(3x)=12sin(2x)12cos(8x). Thus,

sin(5x)cos(3x)dx=12sin(2x)12cos(8x)dx=14cos(2x)116sin(8x)+C.sin(5x)cos(3x)dx=12sin(2x)12cos(8x)dx=14cos(2x)116sin(8x)+C.
Checkpoint 3.10

Evaluate cos(6x)cos(5x)dx.cos(6x)cos(5x)dx.

Integrating Products and Powers of tanx and secx

Before discussing the integration of products and powers of tanxtanx and secx,secx, it is useful to recall the integrals involving tanxtanx and secxsecx we have already learned:

  1. sec2xdx=tanx+Csec2xdx=tanx+C
  2. secxtanxdx=secx+Csecxtanxdx=secx+C
  3. tanxdx=ln|secx|+Ctanxdx=ln|secx|+C
  4. secxdx=ln|secx+tanx|+C.secxdx=ln|secx+tanx|+C.

For most integrals of products and powers of tanxtanx and secx,secx, we rewrite the expression we wish to integrate as the sum or difference of integrals of the form tanjxsec2xdxtanjxsec2xdx or secjxtanxdx.secjxtanxdx. As we see in the following example, we can evaluate these new integrals by using u-substitution.

Example 3.14

Evaluating secjxtanxdxsecjxtanxdx

Evaluate sec5xtanxdx.sec5xtanxdx.

Solution

Start by rewriting sec5xtanxsec5xtanx as sec4xsecxtanx.sec4xsecxtanx.

sec5xtanxdx=sec4xsecxtanxdxLetu=secx;then,du=secxtanxdx.=u4duEvaluate the integral.=15u5+CSubstitutesecx=u.=15sec5x+Csec5xtanxdx=sec4xsecxtanxdxLetu=secx;then,du=secxtanxdx.=u4duEvaluate the integral.=15u5+CSubstitutesecx=u.=15sec5x+C

Media

You can read some interesting information at this website to learn about a common integral involving the secant.

Checkpoint 3.11

Evaluate tan5xsec2xdx.tan5xsec2xdx.

We now take a look at the various strategies for integrating products and powers of secxsecx and tanx.tanx.

Problem-Solving Strategy: Integrating tankxsecjxdxtankxsecjxdx

To integrate tankxsecjxdx,tankxsecjxdx, use the following strategies:

  1. If jj is even and j2,j2, rewrite secjx=secj2xsec2xsecjx=secj2xsec2x and use sec2x=tan2x+1sec2x=tan2x+1 to rewrite secj2xsecj2x in terms of tanx.tanx. Let u=tanxu=tanx and du=sec2x.du=sec2x.
  2. If kk is odd and j1,j1, rewrite tankxsecjx=tank1xsecj1xsecxtanxtankxsecjx=tank1xsecj1xsecxtanx and use tan2x=sec2x1tan2x=sec2x1 to rewrite tank1xtank1x in terms of secx.secx. Let u=secxu=secx and du=secxtanxdx.du=secxtanxdx. (Note: If jj is even and kk is odd, then either strategy 1 or strategy 2 may be used.)
  3. If kk is odd where k3k3 and j=0,j=0, rewrite tankx=tank2xtan2x=tank2x(sec2x1)=tank2xsec2xtank2x.tankx=tank2xtan2x=tank2x(sec2x1)=tank2xsec2xtank2x. It may be necessary to repeat this process on the tank2xtank2x term.
  4. If kk is even and jj is odd, then use tan2x=sec2x1tan2x=sec2x1 to express tankxtankx in terms of secx.secx. Use integration by parts to integrate odd powers of secx.secx.

Example 3.15

Integrating tankxsecjxdxtankxsecjxdx when jj is Even

Evaluate tan6xsec4xdx.tan6xsec4xdx.

Solution

Since the power on secxsecx is even, rewrite sec4x=sec2xsec2xsec4x=sec2xsec2x and use sec2x=tan2x+1sec2x=tan2x+1 to rewrite the first sec2xsec2x in terms of tanx.tanx. Thus,

tan6xsec4xdx=tan6x(tan2x+1)sec2xdxLetu=tanxanddu=sec2x.=u6(u2+1)duExpand.=(u8+u6)duEvaluate the integral.=19u9+17u7+CSubstitutetanx=u.=19tan9x+17tan7x+C.tan6xsec4xdx=tan6x(tan2x+1)sec2xdxLetu=tanxanddu=sec2x.=u6(u2+1)duExpand.=(u8+u6)duEvaluate the integral.=19u9+17u7+CSubstitutetanx=u.=19tan9x+17tan7x+C.

Example 3.16

Integrating tankxsecjxdxtankxsecjxdx when kk is Odd

Evaluate tan5xsec3xdx.tan5xsec3xdx.

Solution

Since the power on tanxtanx is odd, begin by rewriting tan5xsec3x=tan4xsec2xsecxtanx.tan5xsec3x=tan4xsec2xsecxtanx. Thus,

tan5xsec3x=tan4xsec2xsecxtanx.Writetan4x=(tan2x)2.tan5xsec3xdx=(tan2x)2sec2xsecxtanxdxUsetan2x=sec2x1.=(sec2x1)2sec2xsecxtanxdxLetu=secxanddu=secxtanxdx.=(u21)2u2duExpand.=(u62u4+u2)duIntegrate.=17u725u5+13u3+CSubstitutesecx=u.=17sec7x25sec5x+13sec3x+C.tan5xsec3x=tan4xsec2xsecxtanx.Writetan4x=(tan2x)2.tan5xsec3xdx=(tan2x)2sec2xsecxtanxdxUsetan2x=sec2x1.=(sec2x1)2sec2xsecxtanxdxLetu=secxanddu=secxtanxdx.=(u21)2u2duExpand.=(u62u4+u2)duIntegrate.=17u725u5+13u3+CSubstitutesecx=u.=17sec7x25sec5x+13sec3x+C.

Example 3.17

Integrating tankxdxtankxdx where kk is Odd and k3k3

Evaluate tan3xdx.tan3xdx.

Solution

Begin by rewriting tan3x=tanxtan2x=tanx(sec2x1)=tanxsec2xtanx.tan3x=tanxtan2x=tanx(sec2x1)=tanxsec2xtanx. Thus,

tan3xdx=(tanxsec2xtanx)dx=tanxsec2xdxtanxdx=12tan2xln|secx|+C.tan3xdx=(tanxsec2xtanx)dx=tanxsec2xdxtanxdx=12tan2xln|secx|+C.

For the first integral, use the substitution u=tanx.u=tanx. For the second integral, use the formula.

Example 3.18

Integrating sec3xdxsec3xdx

Integrate sec3xdx.sec3xdx.

Solution

This integral requires integration by parts. To begin, let u=secxu=secx and dv=sec2x.dv=sec2x. These choices make du=secxtanxdu=secxtanx and v=tanx.v=tanx. Thus,

sec3xdx=secxtanxtanxsecxtanxdx=secxtanxtan2xsecxdxSimplify.=secxtanx(sec2x1)secxdxSubstitutetan2x=sec2x1.=secxtanx+secxdxsec3xdxRewrite.=secxtanx+ln|secx+tanx|sec3xdx.Evaluatesecxdx.sec3xdx=secxtanxtanxsecxtanxdx=secxtanxtan2xsecxdxSimplify.=secxtanx(sec2x1)secxdxSubstitutetan2x=sec2x1.=secxtanx+secxdxsec3xdxRewrite.=secxtanx+ln|secx+tanx|sec3xdx.Evaluatesecxdx.

We now have

sec3xdx=secxtanx+ln|secx+tanx|sec3xdx.sec3xdx=secxtanx+ln|secx+tanx|sec3xdx.

Since the integral sec3xdxsec3xdx has reappeared on the right-hand side, we can solve for sec3xdxsec3xdx by adding it to both sides. In doing so, we obtain

2sec3xdx=secxtanx+ln|secx+tanx|.2sec3xdx=secxtanx+ln|secx+tanx|.

Dividing by 2, we arrive at

sec3xdx=12secxtanx+12ln|secx+tanx|+C.sec3xdx=12secxtanx+12ln|secx+tanx|+C.
Checkpoint 3.12

Evaluate tan3xsec7xdx.tan3xsec7xdx.

Reduction Formulas

Evaluating secnxdxsecnxdx for values of nn where nn is odd requires integration by parts. In addition, we must also know the value of secn2xdxsecn2xdx to evaluate secnxdx.secnxdx. The evaluation of tannxdxtannxdx also requires being able to integrate tann2xdx.tann2xdx. To make the process easier, we can derive and apply the following power reduction formulas. These rules allow us to replace the integral of a power of secxsecx or tanxtanx with the integral of a lower power of secxsecx or tanx.tanx.

Rule: Reduction Formulas for secnxdxsecnxdx and tannxdxtannxdx

secnxdx=1n1secn2xtanx+n2n1secn2xdxsecnxdx=1n1secn2xtanx+n2n1secn2xdx
3.6
tannxdx=1n1tann1xtann2xdxtannxdx=1n1tann1xtann2xdx
3.7

The first power reduction rule may be verified by applying integration by parts. The second may be verified by following the strategy outlined for integrating odd powers of tanx.tanx.

Example 3.19

Revisiting sec3xdxsec3xdx

Apply a reduction formula to evaluate sec3xdx.sec3xdx.

Solution

By applying the first reduction formula, we obtain

sec3xdx=12secxtanx+12secxdx=12secxtanx+12ln|secx+tanx|+C.sec3xdx=12secxtanx+12secxdx=12secxtanx+12ln|secx+tanx|+C.

Example 3.20

Using a Reduction Formula

Evaluate tan4xdx.tan4xdx.

Solution

Applying the reduction formula for tan4xdxtan4xdx we have

tan4xdx=13tan3xtan2xdx=13tan3x(tanxtan0xdx)Apply the reduction formula totan2xdx.=13tan3xtanx+1dxSimplify.=13tan3xtanx+x+C.Evaluate1dx.tan4xdx=13tan3xtan2xdx=13tan3x(tanxtan0xdx)Apply the reduction formula totan2xdx.=13tan3xtanx+1dxSimplify.=13tan3xtanx+x+C.Evaluate1dx.
Checkpoint 3.13

Apply the reduction formula to sec5xdx.sec5xdx.

Section 3.2 Exercises

Fill in the blank to make a true statement.

69.

sin2x+_______=1sin2x+_______=1

70.

sec2x1=_______sec2x1=_______

Use an identity to reduce the power of the trigonometric function to a trigonometric function raised to the first power.

71.

sin2x=_______sin2x=_______

72.

cos2x=_______cos2x=_______

Evaluate each of the following integrals by u-substitution.

73.

sin3xcosxdxsin3xcosxdx

74.

cosxsinxdxcosxsinxdx

75.

tan5(2x)sec2(2x)dxtan5(2x)sec2(2x)dx

76.

sin7(2x)cos(2x)dxsin7(2x)cos(2x)dx

77.

tan(x2)sec2(x2)dxtan(x2)sec2(x2)dx

78.

tan2xsec2xdxtan2xsec2xdx

Compute the following integrals using the guidelines for integrating powers of trigonometric functions. Use a CAS to check the solutions. (Note: Some of the problems may be done using techniques of integration learned previously.)

79.

sin3xdxsin3xdx

80.

cos3xdxcos3xdx

81.

sinxcosxdxsinxcosxdx

82.

cos5xdxcos5xdx

83.

sin5xcos2xdxsin5xcos2xdx

84.

sin3xcos3xdxsin3xcos3xdx

85.

sinxcosxdxsinxcosxdx

86.

sinxcos3xdxsinxcos3xdx

87.

secxtanxdxsecxtanxdx

88.

tan(5x)dxtan(5x)dx

89.

tan2xsecxdxtan2xsecxdx

90.

tanxsec3xdxtanxsec3xdx

91.

sec4xdxsec4xdx

92.

cotxdxcotxdx

93.

cscxdxcscxdx

94.

tan3xsecxdxtan3xsecxdx

For the following exercises, find a general formula for the integrals.

95.

sin2axcosaxdxsin2axcosaxdx

96.

sinaxcosaxdx.sinaxcosaxdx.

Use the double-angle formulas to evaluate the following integrals.

97.

0πsin2xdx0πsin2xdx

98.

0πsin4xdx0πsin4xdx

99.

cos23xdxcos23xdx

100.

sin2xcos2xdxsin2xcos2xdx

101.

sin2xdx+cos2xdxsin2xdx+cos2xdx

102.

sin2xcos2(2x)dxsin2xcos2(2x)dx

For the following exercises, evaluate the definite integrals. Express answers in exact form whenever possible.

103.

02πcosxsin2xdx02πcosxsin2xdx

104.

0πsin3xsin5xdx0πsin3xsin5xdx

105.

0πcos(99x)sin(101x)dx0πcos(99x)sin(101x)dx

106.

ππcos2(3x)dxππcos2(3x)dx

107.

02πsinxsin(2x)sin(3x)dx02πsinxsin(2x)sin(3x)dx

108.

04πcos(x/2)sin(x/2)dx04πcos(x/2)sin(x/2)dx

109.

π/6π/3cos3xsinxdxπ/6π/3cos3xsinxdx (Round this answer to three decimal places.)

110.

π/3π/3sec2x1dxπ/3π/3sec2x1dx

111.

0π/21cos(2x)dx0π/21cos(2x)dx

112.

Find the area of the region bounded by the graphs of the equations y=sinx,y=sin3x,x=0,andx=π2.y=sinx,y=sin3x,x=0,andx=π2.

113.

Find the area of the region bounded by the graphs of the equations y=cos2x,y=sin2x,x=π4,andx=π4.y=cos2x,y=sin2x,x=π4,andx=π4.

114.

A particle moves in a straight line with the velocity function v(t)=sin(ωt)cos2(ωt).v(t)=sin(ωt)cos2(ωt). Find its position function x=f(t)x=f(t) if f(0)=0.f(0)=0.

115.

Find the average value of the function f(x)=sin2xcos3xf(x)=sin2xcos3x over the interval [π,π].[π,π].

For the following exercises, solve the differential equations.

116.

dydx=sin2x.dydx=sin2x. The curve passes through point (0,0).(0,0).

117.

dydθ=sin4(πθ)dydθ=sin4(πθ)

118.

Find the length of the curve y=ln(cscx),π4xπ2.y=ln(cscx),π4xπ2.

119.

Find the length of the curve y=ln(sinx),π3xπ2.y=ln(sinx),π3xπ2.

120.

Find the volume generated by revolving the curve y=cos(3x)y=cos(3x) about the x-axis, 0xπ36.0xπ36.

For the following exercises, use this information: The inner product of two functions f and g over [a,b][a,b] is defined by f(x)·g(x)=f,g=abf·gdx.f(x)·g(x)=f,g=abf·gdx. Two distinct functions f and g are said to be orthogonal if f,g=0.f,g=0.

121.

Show that {sin(2x),cos(3x)}{sin(2x),cos(3x)} are orthogonal over the interval [π,π].[π,π].

122.

Evaluate ππsin(mx)cos(nx)dx.ππsin(mx)cos(nx)dx.

123.

Integrate y=tanxsec4x.y=tanxsec4x.

For each pair of integrals, determine which one is more difficult to evaluate. Explain your reasoning.

124.

sin456xcosxdxsin456xcosxdx or sin2xcos2xdxsin2xcos2xdx

125.

tan350xsec2xdxtan350xsec2xdx or tan350xsecxdxtan350xsecxdx

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