Skip to Content
OpenStax Logo
Calculus Volume 2

3.5 Other Strategies for Integration

Calculus Volume 23.5 Other Strategies for Integration
Buy book
  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

  • 3.5.1. Use a table of integrals to solve integration problems.
  • 3.5.2. Use a computer algebra system (CAS) to solve integration problems.

In addition to the techniques of integration we have already seen, several other tools are widely available to assist with the process of integration. Among these tools are integration tables, which are readily available in many books, including the appendices to this one. Also widely available are computer algebra systems (CAS), which are found on calculators and in many campus computer labs, and are free online.

Tables of Integrals

Integration tables, if used in the right manner, can be a handy way either to evaluate or check an integral quickly. Keep in mind that when using a table to check an answer, it is possible for two completely correct solutions to look very different. For example, in Trigonometric Substitution, we found that, by using the substitution x=tanθ,x=tanθ, we can arrive at

dx1+x2=ln(x+x2+1)+C.dx1+x2=ln(x+x2+1)+C.

However, using x=sinhθ,x=sinhθ, we obtained a different solution—namely,

dx1+x2=sinh−1x+C.dx1+x2=sinh−1x+C.

We later showed algebraically that the two solutions are equivalent. That is, we showed that sinh−1x=ln(x+x2+1).sinh−1x=ln(x+x2+1). In this case, the two antiderivatives that we found were actually equal. This need not be the case. However, as long as the difference in the two antiderivatives is a constant, they are equivalent.

Example 3.36

Using a Formula from a Table to Evaluate an Integral

Use the table formula

a2u2u2du=a2u2usin−1ua+Ca2u2u2du=a2u2usin−1ua+C

to evaluate 16e2xexdx.16e2xexdx.

Solution

If we look at integration tables, we see that several formulas contain expressions of the form a2u2.a2u2. This expression is actually similar to 16e2x,16e2x, where a=4a=4 and u=ex.u=ex. Keep in mind that we must also have du=ex.du=ex. Multiplying the numerator and the denominator of the given integral by exex should help to put this integral in a useful form. Thus, we now have

16e2xexdx=16e2xe2xexdx.16e2xexdx=16e2xe2xexdx.

Substituting u=exu=ex and du=exdu=ex produces a2u2u2du.a2u2u2du. From the integration table (#88 in Appendix A),

a2u2u2du=a2u2usin−1ua+C.a2u2u2du=a2u2usin−1ua+C.

Thus,

16e2xexdx=16e2xe2xexdxSubstituteu=exanddu=exdx.=42u2u2duApply the formula usinga=4.=42u2usin−1u4+CSubstituteu=ex.=16e2xusin−1(ex4)+C.16e2xexdx=16e2xe2xexdxSubstituteu=exanddu=exdx.=42u2u2duApply the formula usinga=4.=42u2usin−1u4+CSubstituteu=ex.=16e2xusin−1(ex4)+C.

Computer Algebra Systems

If available, a CAS is a faster alternative to a table for solving an integration problem. Many such systems are widely available and are, in general, quite easy to use.

Example 3.37

Using a Computer Algebra System to Evaluate an Integral

Use a computer algebra system to evaluate dxx24.dxx24. Compare this result with ln|x242+x2|+C,ln|x242+x2|+C, a result we might have obtained if we had used trigonometric substitution.

Solution

Using Wolfram Alpha, we obtain

dxx24=ln|x24+x|+C.dxx24=ln|x24+x|+C.

Notice that

ln|x242+x2|+C=ln|x24+x2|+C=ln|x24+x|ln2+C.ln|x242+x2|+C=ln|x24+x2|+C=ln|x24+x|ln2+C.

Since these two antiderivatives differ by only a constant, the solutions are equivalent. We could have also demonstrated that each of these antiderivatives is correct by differentiating them.

Media

You can access an integral calculator for more examples.

Example 3.38

Using a CAS to Evaluate an Integral

Evaluate sin3xdxsin3xdx using a CAS. Compare the result to 13cos3xcosx+C,13cos3xcosx+C, the result we might have obtained using the technique for integrating odd powers of sinxsinx discussed earlier in this chapter.

Solution

Using Wolfram Alpha, we obtain

sin3xdx=112(cos(3x)9cosx)+C.sin3xdx=112(cos(3x)9cosx)+C.

This looks quite different from 13cos3xcosx+C.13cos3xcosx+C. To see that these antiderivatives are equivalent, we can make use of a few trigonometric identities:

112(cos(3x)9cosx)=112(cos(x+2x)9cosx)=112(cos(x)cos(2x)sin(x)sin(2x)9cosx)=112(cosx(2cos2x1)sinx(2sinxcosx)9cosx)=112(2cos3xcosx2cosx(1cos2x)9cosx)=112(4cos3x12cosx)=13cos3xcosx.112(cos(3x)9cosx)=112(cos(x+2x)9cosx)=112(cos(x)cos(2x)sin(x)sin(2x)9cosx)=112(cosx(2cos2x1)sinx(2sinxcosx)9cosx)=112(2cos3xcosx2cosx(1cos2x)9cosx)=112(4cos3x12cosx)=13cos3xcosx.

Thus, the two antiderivatives are identical.

We may also use a CAS to compare the graphs of the two functions, as shown in the following figure.

This is the graph of a periodic function. The waves have an amplitude of approximately 0.7 and a period of approximately 10. The graph represents the functions y = cos^3(x)/3 – cos(x) and y = 1/12(cos(3x)-9cos(x). The graph is the same for both functions.
Figure 3.12 The graphs of y=13cos3xcosxy=13cos3xcosx and y=112(cos(3x)9cosx)y=112(cos(3x)9cosx) are identical.
Checkpoint 3.21

Use a CAS to evaluate dxx2+4.dxx2+4.

Section 3.5 Exercises

Use a table of integrals to evaluate the following integrals.

244.

04x1+2xdx04x1+2xdx

245.

x+3x2+2x+2dxx+3x2+2x+2dx

246.

x31+2x2dxx31+2x2dx

247.

1x2+6xdx1x2+6xdx

248.

xx+1dxxx+1dx

249.

x·2x2dxx·2x2dx

250.

14x2+25dx14x2+25dx

251.

dy4y2dy4y2

252.

sin3(2x)cos(2x)dxsin3(2x)cos(2x)dx

253.

csc(2w)cot(2w)dwcsc(2w)cot(2w)dw

254.

2ydy2ydy

255.

013xdxx2+8013xdxx2+8

256.

−1/41/4sec2(πx)tan(πx)dx−1/41/4sec2(πx)tan(πx)dx

257.

0π/2tan2(x2)dx0π/2tan2(x2)dx

258.

cos3xdxcos3xdx

259.

tan5(3x)dxtan5(3x)dx

260.

sin2ycos3ydysin2ycos3ydy

Use a CAS to evaluate the following integrals. Tables can also be used to verify the answers.

261.

[T] dw1+sec(w2)dw1+sec(w2)

262.

[T] dw1cos(7w)dw1cos(7w)

263.

[T] 0tdt4cost+3sint0tdt4cost+3sint

264.

[T] x293xdxx293xdx

265.

[T] dxx1/2+x1/3dxx1/2+x1/3

266.

[T] dxxx1dxxx1

267.

[T] x3sinxdxx3sinxdx

268.

[T] xx49dxxx49dx

269.

[T] x1+ex2dxx1+ex2dx

270.

[T] 35x2xdx35x2xdx

271.

[T] dxxx1dxxx1

272.

[T] excos−1(ex)dxexcos−1(ex)dx

Use a calculator or CAS to evaluate the following integrals.

273.

[T] 0π/4cos(2x)dx0π/4cos(2x)dx

274.

[T] 01x·ex2dx01x·ex2dx

275.

[T] 082xx2+36dx082xx2+36dx

276.

[T] 02/314+9x2dx02/314+9x2dx

277.

[T] dxx2+4x+13dxx2+4x+13

278.

[T] dx1+sinxdx1+sinx

Use tables to evaluate the integrals. You may need to complete the square or change variables to put the integral into a form given in the table.

279.

dxx2+2x+10dxx2+2x+10

280.

dxx26xdxx26x

281.

exe2x4dxexe2x4dx

282.

cosxsin2x+2sinxdxcosxsin2x+2sinxdx

283.

arctan(x3)x4dxarctan(x3)x4dx

284.

ln|x|arcsin(ln|x|)xdxln|x|arcsin(ln|x|)xdx

Use tables to perform the integration.

285.

dxx2+16dxx2+16

286.

3x2x+7dx3x2x+7dx

287.

dx1cos(4x)dx1cos(4x)

288.

dx4x+1dx4x+1

289.

Find the area bounded by y(4+25x2)=5,x=0,y=0,andx=4.y(4+25x2)=5,x=0,y=0,andx=4. Use a table of integrals or a CAS.

290.

The region bounded between the curve y=11+cosx,0.3x1.1,y=11+cosx,0.3x1.1, and the x-axis is revolved about the x-axis to generate a solid. Use a table of integrals to find the volume of the solid generated. (Round the answer to two decimal places.)

291.

Use substitution and a table of integrals to find the area of the surface generated by revolving the curve y=ex,0x3,y=ex,0x3, about the x-axis. (Round the answer to two decimal places.)

292.

[T] Use an integral table and a calculator to find the area of the surface generated by revolving the curve y=x22,0x1,y=x22,0x1, about the x-axis. (Round the answer to two decimal places.)

293.

[T] Use a CAS or tables to find the area of the surface generated by revolving the curve y=cosx,0xπ2,y=cosx,0xπ2, about the x-axis. (Round the answer to two decimal places.)

294.

Find the length of the curve y=x24y=x24 over [0,8].[0,8].

295.

Find the length of the curve y=exy=ex over [0,ln(2)].[0,ln(2)].

296.

Find the area of the surface formed by revolving the graph of y=2xy=2x over the interval [0,9][0,9] about the x-axis.

297.

Find the average value of the function f(x)=1x2+1f(x)=1x2+1 over the interval [−3,3].[−3,3].

298.

Approximate the arc length of the curve y=tan(πx)y=tan(πx) over the interval [0,14].[0,14]. (Round the answer to three decimal places.)

Citation/Attribution

Want to cite, share, or modify this book? This book is Creative Commons Attribution-NonCommercial-ShareAlike License 4.0 and you must attribute OpenStax.

Attribution information
  • If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution:
    Access for free at https://openstax.org/books/calculus-volume-2/pages/1-introduction
  • If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution:
    Access for free at https://openstax.org/books/calculus-volume-2/pages/1-introduction
Citation information

© Mar 30, 2016 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike License 4.0 license. The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.