Other Strategies for Integration

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

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

   បង្ហាញ​ចម្លើយ

   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.

   បង្ហាញ​ចម្លើយ

   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.

   បង្ហាញ​ចម្លើយ

   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 \(\int \frac{dx}{\sqrt{{x}^{2}+4}}.\)

   បង្ហាញ​ចម្លើយ

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

   បង្ហាញ​ចម្លើយ

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

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

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

   បង្ហាញ​ចម្លើយ

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

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

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

    បង្ហាញ​ចម្លើយ

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

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

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

    បង្ហាញ​ចម្លើយ

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

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

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

    បង្ហាញ​ចម្លើយ

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

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

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

    បង្ហាញ​ចម្លើយ

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

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

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

    បង្ហាញ​ចម្លើយ

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

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

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

    បង្ហាញ​ចម្លើយ

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

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

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

    បង្ហាញ​ចម្លើយ

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

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

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

    បង្ហាញ​ចម្លើយ

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

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

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

    បង្ហាញ​ចម្លើយ

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

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

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

    បង្ហាញ​ចម្លើយ

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

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

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

    បង្ហាញ​ចម្លើយ

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

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

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

    បង្ហាញ​ចម្លើយ

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

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

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

    បង្ហាញ​ចម្លើយ

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

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

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

    បង្ហាញ​ចម្លើយ

    8.0

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

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

    បង្ហាញ​ចម្លើយ

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

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

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

    បង្ហាញ​ចម្លើយ

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

ព្យាយាម​របស់​អ្នក​ផ្ទាល់

Parts of this page are adapted from OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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