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Other Strategies for Integration

Use a table of integrals to solve integration problems.

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 \(x=\text{tan}\ \theta ,\) we can arrive at

\[\int \frac{dx}{\sqrt{1+{x}^{2}}}=\text{ln}(x+\sqrt{{x}^{2}+1})+C.\]

However, using \(x=\text{sinh}\ \theta ,\) we obtained a different solution, namely,

\[\int \frac{dx}{\sqrt{1+{x}^{2}}}={\text{sinh}}^{-1}x+C.\]

We later showed algebraically that the two solutions are equivalent. That is, we showed that \({\text{sinh}}^{-1}x=\text{ln}(x+\sqrt{{x}^{2}+1}).\) 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

Try it.

Use the table formula

\[\int \frac{\sqrt{{a}^{2}-{u}^{2}}}{{u}^{2}}du=-\frac{\sqrt{{a}^{2}-{u}^{2}}}{u}-{\text{sin}}^{-1}\frac{u}{a}+C\]

to evaluate \(\int \frac{\sqrt{16-{e}^{2x}}}{{e}^{x}}dx.\)

Solution

If we look at integration tables, we see that several formulas contain expressions of the form \(\sqrt{{a}^{2}-{u}^{2}}.\) This expression is actually similar to \(\sqrt{16-{e}^{2x}},\) where \(a=4\) and \(u={e}^{x}.\) Keep in mind that we must also have \(du={e}^{x}\text{dx}.\) Multiplying the numerator and the denominator of the given integral by \({e}^{x}\) should help to put this integral in a useful form. Thus, we now have

\[\int \frac{\sqrt{16-{e}^{2x}}}{{e}^{x}}dx=\int \frac{\sqrt{16-{e}^{2x}}}{{e}^{2x}}{e}^{x}dx.\]

Substituting \(u={e}^{x}\) and \(du={e}^{x}\text{dx}\) produces \(\int \frac{\sqrt{{a}^{2}-{u}^{2}}}{{u}^{2}}du.\) From the integration table (#88 in Appendix A),

\[\int \frac{\sqrt{{a}^{2}-{u}^{2}}}{{u}^{2}}du=-\frac{\sqrt{{a}^{2}-{u}^{2}}}{u}-{\text{sin}}^{-1}\frac{u}{a}+C.\]

Thus,

\[\begin{array}{lllll}\int \frac{\sqrt{16-{e}^{2x}}}{{e}^{x}}dx & =\int \frac{\sqrt{16-{e}^{2x}}}{{e}^{2x}}{e}^{x}dx & & & \text{Substitute}\ u={e}^{x}\ \text{and}\ du={e}^{x}dx. \\ & =\int \frac{\sqrt{{4}^{2}-{u}^{2}}}{{u}^{2}}du & & & \text{Apply the formula using}\ a=4. \\ & =-\frac{\sqrt{{4}^{2}-{u}^{2}}}{u}-{\text{sin}}^{-1}\frac{u}{4}+C & & & \text{Substitute}\ u={e}^{x}. \\ & =-\frac{\sqrt{16-{e}^{2x}}}{{e}^{x}}-{\text{sin}}^{-1}(\frac{{e}^{x}}{4})+C. & & & \end{array}\]

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

Try it.

Use a computer algebra system to evaluate \(\int \frac{dx}{\sqrt{{x}^{2}-4}}.\) Compare this result with \(\text{ln}|\frac{\sqrt{{x}^{2}-4}}{2}+\frac{x}{2}|+C,\) a result we might have obtained if we had used trigonometric substitution.

Solution

Using Wolfram Alpha, we obtain

\[\int \frac{dx}{\sqrt{{x}^{2}-4}}=\text{ln}|\sqrt{{x}^{2}-4}+x|+C.\]

Notice that

\[\text{ln}|\frac{\sqrt{{x}^{2}-4}}{2}+\frac{x}{2}|+C=\text{ln}|\frac{\sqrt{{x}^{2}-4}+x}{2}|+C=\text{ln}|\sqrt{{x}^{2}-4}+x|-\text{ln}\ 2+C.\]

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.

Example

Try it.

Evaluate \({\int }^{\text{}}{\text{sin}}^{3}x\ dx\) using a CAS. Compare the result to \(\frac{1}{3}{\text{cos}}^{3}x-\text{cos}\ x+C,\) the result we might have obtained using the technique for integrating odd powers of \(\text{sin}\ x\) discussed earlier in this chapter.

Solution

Using Wolfram Alpha, we obtain

\[\int {\text{sin}}^{3}x\ dx=\frac{1}{12}(\text{cos}(3x)-9\ \text{cos}\ x)+C.\]

This looks quite different from \(\frac{1}{3}{\text{cos}}^{3}x-\text{cos}\ x+C.\) To see that these antiderivatives are equivalent, we can make use of a few trigonometric identities:

\[\begin{array}{ll}\frac{1}{12}(\text{cos}(3x)-9\ \text{cos}\ x) & =\frac{1}{12}(\text{cos}(x+2x)-9\ \text{cos}\ x) \\ & =\frac{1}{12}(\text{cos}(x)\text{cos}(2x)-\text{sin}(x)\text{sin}(2x)-9\ \text{cos}\ x) \\ & =\frac{1}{12}(\text{cos}\ x(2\ {\text{cos}}^{2}x-1)-\text{sin}\ x(2\ \text{sin}\ x\ \text{cos}\ x)-9\ \text{cos}\ x) \\ & =\frac{1}{12}(2{\text{cos}}^{3}x-\text{cos}\ x-2\ \text{cos}\ x(1-{\text{cos}}^{2}x)-9\ \text{cos}\ x) \\ & =\frac{1}{12}(4{\text{cos}}^{3}x-12\ \text{cos}\ x) \\ & =\frac{1}{3}{\text{cos}}^{3}x-\text{cos}\ x.\end{array}\]

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.

Key Concepts

  • An integration table may be used to evaluate indefinite integrals.
  • A CAS (or computer algebra system) may be used to evaluate indefinite integrals.
  • It may require some effort to reconcile equivalent solutions obtained using different methods.

Other Strategies for Integration

Use a table of integrals to evaluate the following integrals.

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

Use a calculator or CAS to evaluate the following integrals.

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.

Use tables to perform the integration.

Ҳоло шумо Ҳеҷ ҳисобкунак инро ҳал намекунад, лекин қисмҳои он ҳисобшавандаанд. Яке аз инҳоро дар поён санҷед ё худи худро ворид кунед.

Машқ (40)

Ҳар як саволро аввал дар коғаз санҷед. Ҷавобро барои санҷиш нишон диҳед; ҷавобҳои санҷидашударо дар ҳалкунанда барои ҳар қадам кушода метавонед.

  1. Use the table formula

    \[\int \frac{\sqrt{{a}^{2}-{u}^{2}}}{{u}^{2}}du=-\frac{\sqrt{{a}^{2}-{u}^{2}}}{u}-{\text{sin}}^{-1}\frac{u}{a}+C\]

    to evaluate \(\int \frac{\sqrt{16-{e}^{2x}}}{{e}^{x}}dx.\)

    Ҷавоби ҷавобро нишон диҳед

    If we look at integration tables, we see that several formulas contain expressions of the form \(\sqrt{{a}^{2}-{u}^{2}}.\) This expression is actually similar to \(\sqrt{16-{e}^{2x}},\) where \(a=4\) and \(u={e}^{x}.\) Keep in mind that we must also have \(du={e}^{x}\text{dx}.\) Multiplying the numerator and the denominator of the given integral by \({e}^{x}\) should help to put this integral in a useful form. Thus, we now have

    \[\int \frac{\sqrt{16-{e}^{2x}}}{{e}^{x}}dx=\int \frac{\sqrt{16-{e}^{2x}}}{{e}^{2x}}{e}^{x}dx.\]

    Substituting \(u={e}^{x}\) and \(du={e}^{x}\text{dx}\) produces \(\int \frac{\sqrt{{a}^{2}-{u}^{2}}}{{u}^{2}}du.\) From the integration table (#88 in Appendix A),

    \[\int \frac{\sqrt{{a}^{2}-{u}^{2}}}{{u}^{2}}du=-\frac{\sqrt{{a}^{2}-{u}^{2}}}{u}-{\text{sin}}^{-1}\frac{u}{a}+C.\]

    Thus,

    \[\begin{array}{lllll}\int \frac{\sqrt{16-{e}^{2x}}}{{e}^{x}}dx & =\int \frac{\sqrt{16-{e}^{2x}}}{{e}^{2x}}{e}^{x}dx & & & \text{Substitute}\ u={e}^{x}\ \text{and}\ du={e}^{x}dx. \\ & =\int \frac{\sqrt{{4}^{2}-{u}^{2}}}{{u}^{2}}du & & & \text{Apply the formula using}\ a=4. \\ & =-\frac{\sqrt{{4}^{2}-{u}^{2}}}{u}-{\text{sin}}^{-1}\frac{u}{4}+C & & & \text{Substitute}\ u={e}^{x}. \\ & =-\frac{\sqrt{16-{e}^{2x}}}{{e}^{x}}-{\text{sin}}^{-1}(\frac{{e}^{x}}{4})+C. & & & \end{array}\]

  2. Use a computer algebra system to evaluate \(\int \frac{dx}{\sqrt{{x}^{2}-4}}.\) Compare this result with \(\text{ln}|\frac{\sqrt{{x}^{2}-4}}{2}+\frac{x}{2}|+C,\) a result we might have obtained if we had used trigonometric substitution.

    Ҷавоби ҷавобро нишон диҳед

    Using Wolfram Alpha, we obtain

    \[\int \frac{dx}{\sqrt{{x}^{2}-4}}=\text{ln}|\sqrt{{x}^{2}-4}+x|+C.\]

    Notice that

    \[\text{ln}|\frac{\sqrt{{x}^{2}-4}}{2}+\frac{x}{2}|+C=\text{ln}|\frac{\sqrt{{x}^{2}-4}+x}{2}|+C=\text{ln}|\sqrt{{x}^{2}-4}+x|-\text{ln}\ 2+C.\]

    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.

  3. Evaluate \({\int }^{\text{}}{\text{sin}}^{3}x\ dx\) using a CAS. Compare the result to \(\frac{1}{3}{\text{cos}}^{3}x-\text{cos}\ x+C,\) the result we might have obtained using the technique for integrating odd powers of \(\text{sin}\ x\) discussed earlier in this chapter.

    Ҷавоби ҷавобро нишон диҳед

    Using Wolfram Alpha, we obtain

    \[\int {\text{sin}}^{3}x\ dx=\frac{1}{12}(\text{cos}(3x)-9\ \text{cos}\ x)+C.\]

    This looks quite different from \(\frac{1}{3}{\text{cos}}^{3}x-\text{cos}\ x+C.\) To see that these antiderivatives are equivalent, we can make use of a few trigonometric identities:

    \[\begin{array}{ll}\frac{1}{12}(\text{cos}(3x)-9\ \text{cos}\ x) & =\frac{1}{12}(\text{cos}(x+2x)-9\ \text{cos}\ x) \\ & =\frac{1}{12}(\text{cos}(x)\text{cos}(2x)-\text{sin}(x)\text{sin}(2x)-9\ \text{cos}\ x) \\ & =\frac{1}{12}(\text{cos}\ x(2\ {\text{cos}}^{2}x-1)-\text{sin}\ x(2\ \text{sin}\ x\ \text{cos}\ x)-9\ \text{cos}\ x) \\ & =\frac{1}{12}(2{\text{cos}}^{3}x-\text{cos}\ x-2\ \text{cos}\ x(1-{\text{cos}}^{2}x)-9\ \text{cos}\ x) \\ & =\frac{1}{12}(4{\text{cos}}^{3}x-12\ \text{cos}\ x) \\ & =\frac{1}{3}{\text{cos}}^{3}x-\text{cos}\ x.\end{array}\]

    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.

  4. Use a CAS to evaluate \(\int \frac{dx}{\sqrt{{x}^{2}+4}}.\)

    Ҷавоби ҷавобро нишон диҳед

    Possible solutions include \({\text{sinh}}^{-1}(\frac{x}{2})+C\) and \(\text{ln}|\sqrt{{x}^{2}+4}+x|+C.\)

  5. \(\int _{0}^{4}\frac{x}{\sqrt{1+2x}}dx\)

  6. \(\int \frac{x+3}{{x}^{2}+2x+2}dx\)

    Ҷавоби ҷавобро нишон диҳед

    \(\frac{1}{2}\text{ln}|{x}^{2}+2x+2|+2\ \text{arctan}(x+1)+C\)

  7. \(\int {x}^{3}\sqrt{1+2{x}^{2}}\ dx\)

  8. \(\int \frac{1}{\sqrt{{x}^{2}+6x}}dx\)

    Ҷавоби ҷавобро нишон диҳед

    \({\text{cosh}}^{-1}(\frac{x+3}{3})+C\)

  9. \(\int \frac{x}{x+1}dx\)

  10. \(\int x\cdot {2}^{{x}^{2}}dx\)

    Ҷавоби ҷавобро нишон диҳед

    \(\frac{{2}^{{x}^{2}-1}}{\text{ln}\ 2}+C\)

  11. \(\int \frac{1}{4{x}^{2}+25}dx\)

  12. \(\int \frac{dy}{\sqrt{4-{y}^{2}}}\)

    Ҷавоби ҷавобро нишон диҳед

    \(\text{arcsin}(\frac{y}{2})+C\)

  13. \(\int {\text{sin}}^{3}(2x)\text{cos}(2x)dx\)

  14. \(\int \text{csc}(2w)\text{cot}(2w)dw\)

    Ҷавоби ҷавобро нишон диҳед

    \(-\frac{1}{2}\text{csc}(2w)+C\)

  15. \(\int {2}^{y}dy\)

  16. \({\int }_{0}^{1}\frac{3x\ dx}{\sqrt{{x}^{2}+8}}\)

    Ҷавоби ҷавобро нишон диҳед

    \(9-6\sqrt{2}\)

  17. \({\int }_{-1\text{/}4}^{1\text{/}4}{\text{sec}}^{2}(\pi x)\text{tan}(\pi x)dx\)

  18. \({\int }_{0}^{\pi \text{/}2}{\text{tan}}^{2}(\frac{x}{2})dx\)

    Ҷавоби ҷавобро нишон диҳед

    \(2-\frac{\pi }{2}\)

  19. \(\int {\text{cos}}^{3}x\ dx\)

  20. \(\int {\text{tan}}^{5}(3x)dx\)

    Ҷавоби ҷавобро нишон диҳед

    \(\frac{1}{12}{\text{tan}}^{4}(3x)-\frac{1}{6}{\text{tan}}^{2}(3x)+\frac{1}{3}\text{ln}|\text{sec}(3x)|+C\)

  21. \(\int {\text{sin}}^{2}y\ {\text{cos}}^{3}ydy\)

  22. [T] \(\int \frac{dw}{1+\text{sec}(\frac{w}{2})}\)

    Ҷавоби ҷавобро нишон диҳед

    \(2\ \text{cot}(\frac{w}{2})-2\ \text{csc}(\frac{w}{2})+w+C\)

  23. [T] \(\int \frac{dw}{1-\text{cos}(7w)}\)

  24. [T] \({\int }_{0}^{t}\frac{dt}{4\ \text{cos}\ t+3\ \text{sin}\ t}\)

    Ҷавоби ҷавобро нишон диҳед

    \(\frac{1}{5}\text{ln}|\frac{2(5+4\ \text{sin}\ t-3\ \text{cos}\ t)}{4\ \text{cos}\ t+3\ \text{sin}\ t}|\)

  25. [T] \(\int \frac{\sqrt{{x}^{2}-9}}{3x}dx\)

  26. [T] \(\int \frac{dx}{{x}^{1\text{/}2}+{x}^{1\text{/}3}}\)

    Ҷавоби ҷавобро нишон диҳед

    \(6{x}^{1\text{/}6}-3{x}^{1\text{/}3}+2\sqrt{x}-6\ \text{ln}[1+{x}^{1\text{/}6}]+C\)

  27. [T] \(\int \frac{dx}{x\sqrt{x-1}}\)

  28. [T] \(\int {x}^{3}\text{sin}\ x\ dx\)

    Ҷавоби ҷавобро нишон диҳед

    \(\text{-}{x}^{3}\text{cos}\ x+3{x}^{2}\text{sin}\ x+6x\ \text{cos}\ x-6\ \text{sin}\ x+C\)

  29. [T] \(\int x\sqrt{{x}^{4}-9}\ dx\)

  30. [T] \(\int \frac{x}{1+{e}^{\text{-}{x}^{2}}}dx\)

    Ҷавоби ҷавобро нишон диҳед

    \(\frac{1}{2}({x}^{2}+\text{ln}|1+{e}^{\text{-}{x}^{2}}|)+C\)

  31. [T] \(\int \frac{\sqrt{3-5x}}{2x}dx\)

  32. [T] \(\int \frac{dx}{x\sqrt{x-1}}\)

    Ҷавоби ҷавобро нишон диҳед

    \(2\ \text{arctan}(\sqrt{x-1})+C\)

  33. [T] \(\int {e}^{x}{\text{cos}}^{-1}({e}^{x})dx\)

  34. [T] \({\int }_{0}^{\pi \text{/}4}\text{cos}(2x)dx\)

    Ҷавоби ҷавобро нишон диҳед

    \(0.5=\frac{1}{2}\)

  35. [T] \({\int }_{0}^{1}x\cdot {e}^{\text{-}{x}^{2}}dx\)

  36. [T] \({\int }_{0}^{8}\frac{2x}{\sqrt{{x}^{2}+36}}dx\)

    Ҷавоби ҷавобро нишон диҳед

    8.0

  37. [T] \({\int }_{0}^{2\text{/}\sqrt{3}}\frac{1}{4+9{x}^{2}}dx\)

  38. [T] \(\int \frac{dx}{{x}^{2}+4x+13}\)

    Ҷавоби ҷавобро нишон диҳед

    \(\frac{1}{3}\text{arctan}(\frac{1}{3}(x+2))+C\)

  39. [T] \(\int \frac{dx}{1+\text{sin}\ x}\)

  40. \(\int \frac{dx}{{x}^{2}+2x+10}\)

    Ҷавоби ҷавобро нишон диҳед

    \(\frac{1}{3}\text{arctan}(\frac{x+1}{3})+C\)

Кор кардан

Дар ҳисоби ройгон ба ҳар як дарс қайдҳо илова карда мешаванд, қайди он, ки шумо чӣ кор кардаед, масъалаҳои ҳалшуда дар як ҷо ва муаллиме, ки шумо метавонед дар бораи ин саҳифа пурсед. Математика барои ҳама кушода аст, хоҳ ворид шуда бошад ё на.

Бақайдгирӣ Ворид шудан

Символы, используемые здесь

Барои муайянкунии пурраи маъно, тасвир ва маънои ҳар як ҳарф дар он, ба ҳар як аломат пахш кунед.

Чӣ тавр: Other Strategies for Integration

  1. Use a table of integrals to solve integration problems.
  2. Use a computer algebra system (CAS) to solve integration problems.
  3. An integration table may be used to evaluate indefinite integrals.
  4. A CAS (or computer algebra system) may be used to evaluate indefinite integrals.
  5. It may require some effort to reconcile equivalent solutions obtained using different methods.

Саволҳои маъмул

What is a derivative in one sentence?

The slope of the graph at a point, the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate: the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function, a polynomial times an exponential, log or trig function.

Қисми ин саҳифа аз OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). Дар ин ҷо барои мо чизе нест, ки мо ба он бирасем. Азбаски манбаъ барои истифодаи ғайрикоммерсиалӣ иҷозатнома дода шудааст, ин дарс ройгон аст, чизе намефурӯшад ва дар асоси иҷозатномаи якхела тақсим карда мешавад.

& Иловаи забон Calculus

LimitsDerivativesIntegralsDefinite integralsTaylor seriesSeries and sumsMaxima and minimaThe chain ruleImplicit differentiationRelated rates and optimisationIntegration techniques: substitution, parts, partial fractionsApplications of integration: area, volume, arc lengthInfinite series and convergence tests