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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.
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Nhazi (40)
Nwapụta n'ime akwụkwọ n'oge mbụ. Gosi nzaghachi ka ịlelee ya; ndị ahụ e gosipụtara nwere ike ịbụ ndị a mepere n'ime onye nrụpụta maka nzọụkwụ ọbụla.
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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 \(\int \frac{\sqrt{16-{e}^{2x}}}{{e}^{x}}dx.\)
Gosi nzaghachi
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,
\[\begin{array}{lllll}\int \frac{\sqrt{16-{e}^{2x}}}{{e}^{x}}dx & =\int \frac{\sqrt{16-{e}^{2x}}}{{e}^{2x}}{e}^{x}dx & & & \text{Substitute}\ u={e}^{x}\ \text{and}\ du={e}^{x}dx. \\ & =\int \frac{\sqrt{{4}^{2}-{u}^{2}}}{{u}^{2}}du & & & \text{Apply the formula using}\ a=4. \\ & =-\frac{\sqrt{{4}^{2}-{u}^{2}}}{u}-{\text{sin}}^{-1}\frac{u}{4}+C & & & \text{Substitute}\ u={e}^{x}. \\ & =-\frac{\sqrt{16-{e}^{2x}}}{{e}^{x}}-{\text{sin}}^{-1}(\frac{{e}^{x}}{4})+C. & & & \end{array}\] -
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.
Gosi nzaghachi
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.
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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.
Gosi nzaghachi
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.
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Use a CAS to evaluate \(\int \frac{dx}{\sqrt{{x}^{2}+4}}.\)
Gosi nzaghachi
Possible solutions include \({\text{sinh}}^{-1}(\frac{x}{2})+C\) and \(\text{ln}|\sqrt{{x}^{2}+4}+x|+C.\)
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\(\int _{0}^{4}\frac{x}{\sqrt{1+2x}}dx\)
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\(\int \frac{x+3}{{x}^{2}+2x+2}dx\)
Gosi nzaghachi
\(\frac{1}{2}\text{ln}|{x}^{2}+2x+2|+2\ \text{arctan}(x+1)+C\)
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\(\int {x}^{3}\sqrt{1+2{x}^{2}}\ dx\)
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\(\int \frac{1}{\sqrt{{x}^{2}+6x}}dx\)
Gosi nzaghachi
\({\text{cosh}}^{-1}(\frac{x+3}{3})+C\)
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\(\int \frac{x}{x+1}dx\)
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\(\int x\cdot {2}^{{x}^{2}}dx\)
Gosi nzaghachi
\(\frac{{2}^{{x}^{2}-1}}{\text{ln}\ 2}+C\)
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\(\int \frac{1}{4{x}^{2}+25}dx\)
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\(\int \frac{dy}{\sqrt{4-{y}^{2}}}\)
Gosi nzaghachi
\(\text{arcsin}(\frac{y}{2})+C\)
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\(\int {\text{sin}}^{3}(2x)\text{cos}(2x)dx\)
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\(\int \text{csc}(2w)\text{cot}(2w)dw\)
Gosi nzaghachi
\(-\frac{1}{2}\text{csc}(2w)+C\)
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\(\int {2}^{y}dy\)
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\({\int }_{0}^{1}\frac{3x\ dx}{\sqrt{{x}^{2}+8}}\)
Gosi nzaghachi
\(9-6\sqrt{2}\)
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\({\int }_{-1\text{/}4}^{1\text{/}4}{\text{sec}}^{2}(\pi x)\text{tan}(\pi x)dx\)
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\({\int }_{0}^{\pi \text{/}2}{\text{tan}}^{2}(\frac{x}{2})dx\)
Gosi nzaghachi
\(2-\frac{\pi }{2}\)
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\(\int {\text{cos}}^{3}x\ dx\)
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\(\int {\text{tan}}^{5}(3x)dx\)
Gosi nzaghachi
\(\frac{1}{12}{\text{tan}}^{4}(3x)-\frac{1}{6}{\text{tan}}^{2}(3x)+\frac{1}{3}\text{ln}|\text{sec}(3x)|+C\)
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\(\int {\text{sin}}^{2}y\ {\text{cos}}^{3}ydy\)
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[T] \(\int \frac{dw}{1+\text{sec}(\frac{w}{2})}\)
Gosi nzaghachi
\(2\ \text{cot}(\frac{w}{2})-2\ \text{csc}(\frac{w}{2})+w+C\)
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[T] \(\int \frac{dw}{1-\text{cos}(7w)}\)
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[T] \({\int }_{0}^{t}\frac{dt}{4\ \text{cos}\ t+3\ \text{sin}\ t}\)
Gosi nzaghachi
\(\frac{1}{5}\text{ln}|\frac{2(5+4\ \text{sin}\ t-3\ \text{cos}\ t)}{4\ \text{cos}\ t+3\ \text{sin}\ t}|\)
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[T] \(\int \frac{\sqrt{{x}^{2}-9}}{3x}dx\)
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[T] \(\int \frac{dx}{{x}^{1\text{/}2}+{x}^{1\text{/}3}}\)
Gosi nzaghachi
\(6{x}^{1\text{/}6}-3{x}^{1\text{/}3}+2\sqrt{x}-6\ \text{ln}[1+{x}^{1\text{/}6}]+C\)
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[T] \(\int \frac{dx}{x\sqrt{x-1}}\)
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[T] \(\int {x}^{3}\text{sin}\ x\ dx\)
Gosi nzaghachi
\(\text{-}{x}^{3}\text{cos}\ x+3{x}^{2}\text{sin}\ x+6x\ \text{cos}\ x-6\ \text{sin}\ x+C\)
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[T] \(\int x\sqrt{{x}^{4}-9}\ dx\)
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[T] \(\int \frac{x}{1+{e}^{\text{-}{x}^{2}}}dx\)
Gosi nzaghachi
\(\frac{1}{2}({x}^{2}+\text{ln}|1+{e}^{\text{-}{x}^{2}}|)+C\)
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[T] \(\int \frac{\sqrt{3-5x}}{2x}dx\)
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[T] \(\int \frac{dx}{x\sqrt{x-1}}\)
Gosi nzaghachi
\(2\ \text{arctan}(\sqrt{x-1})+C\)
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[T] \(\int {e}^{x}{\text{cos}}^{-1}({e}^{x})dx\)
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[T] \({\int }_{0}^{\pi \text{/}4}\text{cos}(2x)dx\)
Gosi nzaghachi
\(0.5=\frac{1}{2}\)
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[T] \({\int }_{0}^{1}x\cdot {e}^{\text{-}{x}^{2}}dx\)
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[T] \({\int }_{0}^{8}\frac{2x}{\sqrt{{x}^{2}+36}}dx\)
Gosi nzaghachi
8.0
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[T] \({\int }_{0}^{2\text{/}\sqrt{3}}\frac{1}{4+9{x}^{2}}dx\)
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[T] \(\int \frac{dx}{{x}^{2}+4x+13}\)
Gosi nzaghachi
\(\frac{1}{3}\text{arctan}(\frac{1}{3}(x+2))+C\)
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[T] \(\int \frac{dx}{1+\text{sin}\ x}\)
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\(\int \frac{dx}{{x}^{2}+2x+10}\)
Gosi nzaghachi
\(\frac{1}{3}\text{arctan}(\frac{x+1}{3})+C\)
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Olee otú: Other Strategies for Integration
- Use a table of integrals to solve integration problems.
- Use a computer algebra system (CAS) to solve integration problems.
- 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.
Ajụjụ ndị mmadụ jụrụ
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.
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