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Integrals Resulting in Inverse Trigonometric Functions

Integrate functions resulting in inverse trigonometric functions

Integrals that Result in Inverse Sine Functions

Let us begin this last section of the chapter with the three formulas. Along with these formulas, we use substitution to evaluate the integrals. We prove the formula for the inverse sine integral.

Condensed: the full section is in OpenStax Calculus Volume 2.

Integrals Resulting in Other Inverse Trigonometric Functions

There are six inverse trigonometric functions. However, only three integration formulas are noted in the rule on integration formulas resulting in inverse trigonometric functions because the remaining three are negative versions of the ones we use. The only difference is whether the integrand is positive or negative. Rather than memorizing three more formulas, if the integrand is negative, simply factor out −1 and evaluate the integral using one of the formulas already provided. To close this section, we examine one more formula: the integral resulting in the inverse tangent function.

Example

Try it.

Evaluate the integral \(\int \frac{1}{1+4{x}^{2}}dx.\)

Solution

Comparing this problem with the formulas stated in the rule on integration formulas resulting in inverse trigonometric functions, the integrand looks similar to the formula for \({\text{tan}}^{-1}u+C.\) So we use substitution, letting \(u=2x,\) then \(du=2dx\) and \(1\text{/}2du=dx.\) Then, we have

\[\frac{1}{2}\int \frac{1}{1+{u}^{2}}du=\frac{1}{2}\ {\text{tan}}^{-1}u+C=\frac{1}{2}\ {\text{tan}}^{-1}(2x)+C.\]
Example

Try it.

Evaluate the integral \(\int \frac{1}{9+{x}^{2}}dx.\)

Solution

Apply the formula with \(a=3.\) Then,

\[\int \frac{dx}{9+{x}^{2}}=\frac{1}{3}\ {\text{tan}}^{-1}(\frac{x}{3})+C.\]
Example

Try it.

Evaluate the definite integral \({\int }_{\sqrt{3}\text{/}3}^{\sqrt{3}}\frac{dx}{1+{x}^{2}}.\)

Solution

Use the formula for the inverse tangent. We have

\[\begin{array}{ll} \\ \\ {\int }_{\sqrt{3}\text{/}3}^{\sqrt{3}}\frac{dx}{1+{x}^{2}} & ={\text{tan}}^{-1}x{|}_{\sqrt{3}\text{/}3}^{\sqrt{3}} \\ & =[{\text{tan}}^{-1}(\sqrt{3})]-[{\text{tan}}^{-1}(\frac{\sqrt{3}}{3})] \\ & =\frac{\pi }{6}.\end{array}\]

Key Concepts

  • Formulas for derivatives of inverse trigonometric functions developed in Derivatives of Exponential and Logarithmic Functions lead directly to integration formulas involving inverse trigonometric functions.
  • Use the formulas listed in the rule on integration formulas resulting in inverse trigonometric functions to match up the correct format and make alterations as necessary to solve the problem.
  • Substitution is often required to put the integrand in the correct form.

Key Equations

Integrals That Produce Inverse Trigonometric Functions\(\int \frac{du}{\sqrt{{a}^{2}-{u}^{2}}}={\text{sin}}^{-1}(\frac{u}{a})+C\)
\(\int \frac{du}{{a}^{2}+{u}^{2}}=\frac{1}{a}\ {\text{tan}}^{-1}(\frac{u}{a})+C\)
\(\int \frac{du}{u\sqrt{{u}^{2}-{a}^{2}}}=\frac{1}{a}\ {\text{sec}}^{-1}(\frac{u}{a})+C\)

Integrals Resulting in Inverse Trigonometric Functions

In the following exercises, evaluate each integral in terms of an inverse trigonometric function.

In the following exercises, find each indefinite integral, using appropriate substitutions.

In the following exercises, solve for the antiderivative \(\int f\) of f with \(C=0,\) then use a calculator to graph f and the antiderivative over the given interval \([a,b].\) Identify a value of C such that adding C to the antiderivative recovers the definite integral \(F(x)={\int }_{a}^{x}f(t)dt.\)

In the following exercises, compute the antiderivative using appropriate substitutions.

In the following exercises, use a calculator to graph the antiderivative \(\int f\) with \(C=0\) over the given interval \([a,b].\) Approximate a value of C, if possible, such that adding C to the antiderivative gives the same value as the definite integral \(F(x)={\int }_{a}^{x}f(t)dt.\)

In the following exercises, compute each integral using appropriate substitutions.

In the following exercises, compute each definite integral.

Integrals that Result in Inverse Sine Functions

Let us begin this last section of the chapter with the three formulas. Along with these formulas, we use substitution to evaluate the integrals. We prove the formula for the inverse sine integral.

Condensed: the full section is in OpenStax Calculus Volume 1.

Integrals Resulting in Other Inverse Trigonometric Functions

There are six inverse trigonometric functions. However, only three integration formulas are noted in the rule on integration formulas resulting in inverse trigonometric functions because the remaining three are negative versions of the ones we use. The only difference is whether the integrand is positive or negative. Rather than memorizing three more formulas, if the integrand is negative, simply factor out −1 and evaluate the integral using one of the formulas already provided. To close this section, we examine one more formula: the integral resulting in the inverse tangent function.

Example

Try it.

Evaluate the integral \(\int \frac{1}{1+4{x}^{2}}dx.\)

Solution

Comparing this problem with the formulas stated in the rule on integration formulas resulting in inverse trigonometric functions, the integrand looks similar to the formula for \({\text{tan}}^{-1}u+C.\) So we use substitution, letting \(u=2x,\) then \(du=2dx\) and \(1\text{/}2du=dx.\) Then, we have

\[\frac{1}{2}\int \frac{1}{1+{u}^{2}}du=\frac{1}{2}\ {\text{tan}}^{-1}u+C=\frac{1}{2}\ {\text{tan}}^{-1}(2x)+C.\]
Example

Try it.

Evaluate the integral \(\int \frac{1}{9+{x}^{2}}dx.\)

Solution

Apply the formula with \(a=3.\) Then,

\[\int \frac{dx}{9+{x}^{2}}=\frac{1}{3}\ {\text{tan}}^{-1}(\frac{x}{3})+C.\]
Example

Try it.

Evaluate the definite integral \({\int }_{\sqrt{3}\text{/}3}^{\sqrt{3}}\frac{dx}{1+{x}^{2}}.\)

Solution

Use the formula for the inverse tangent. We have

\[\begin{array}{ll} \\ \\ {\int }_{\sqrt{3}\text{/}3}^{\sqrt{3}}\frac{dx}{1+{x}^{2}} & ={\text{tan}}^{-1}x{|}_{\sqrt{3}\text{/}3}^{\sqrt{3}} \\ & =[{\text{tan}}^{-1}(\sqrt{3})]-[{\text{tan}}^{-1}(\frac{\sqrt{3}}{3})] \\ & =\frac{\pi }{6}.\end{array}\]

Key Concepts

  • Formulas for derivatives of inverse trigonometric functions developed in Derivatives of Exponential and Logarithmic Functions lead directly to integration formulas involving inverse trigonometric functions.
  • Use the formulas listed in the rule on integration formulas resulting in inverse trigonometric functions to match up the correct format and make alterations as necessary to solve the problem.
  • Substitution is often required to put the integrand in the correct form.

Key Equations

Integrals That Produce Inverse Trigonometric Functions\(\int \frac{du}{\sqrt{{a}^{2}-{u}^{2}}}={\text{sin}}^{-1}(\frac{u}{a})+C\)
\(\int \frac{du}{{a}^{2}+{u}^{2}}=\frac{1}{a}\ {\text{tan}}^{-1}(\frac{u}{a})+C\)
\(\int \frac{du}{u\sqrt{{u}^{2}-{a}^{2}}}=\frac{1}{a}\ {\text{sec}}^{-1}(\frac{u}{a})+C\)

Integrals Resulting in Inverse Trigonometric Functions

In the following exercises, evaluate each integral in terms of an inverse trigonometric function.

In the following exercises, find each indefinite integral, using appropriate substitutions.

In the following exercises, solve for the antiderivative \(\int f\) of f with \(C=0,\) then use a calculator to graph f and the antiderivative over the given interval \([a,b].\) Identify a value of C such that adding C to the antiderivative recovers the definite integral \(F(x)={\int }_{a}^{x}f(t)dt.\)

In the following exercises, compute the antiderivative using appropriate substitutions.

In the following exercises, use a calculator to graph the antiderivative \(\int f\) with \(C=0\) over the given interval \([a,b].\) Approximate a value of C, if possible, such that adding C to the antiderivative gives the same value as the definite integral \(F(x)={\int }_{a}^{x}f(t)dt.\)

In the following exercises, compute each integral using appropriate substitutions.

In the following exercises, compute each definite integral.

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Praktika (40)

Išbandykite kiekvieną ant popieriaus pirmą. Atskleisti atsakymą patikrinti; patikrintas tie gali būti atidarytas sprendėjas už kiekvieną žingsnį.

  1. Evaluate the definite integral \({\int }_{0}^{\frac{1}{2}}\frac{dx}{\sqrt{1-{x}^{2}}}.\)

    Atskleisti atsakymą

    We can go directly to the formula for the antiderivative in the rule on integration formulas resulting in inverse trigonometric functions, and then evaluate the definite integral. We have

    \[\begin{array}{ll} \\ \\ {\int }_{0}^{\frac{1}{2}}\frac{dx}{\sqrt{1-{x}^{2}}} & ={\text{sin}}^{-1}x{|}_{0}^{\frac{1}{2}} \\ & ={\text{sin}}^{-1}\frac{1}{2}-{\text{sin}}^{-1}0 \\ & =\frac{\pi }{6}-0 \\ & =\frac{\pi }{6}.\end{array}\]

  2. Evaluate the integral \(\int \frac{dx}{\sqrt{1-16{x}^{2}}}.\)

    Atskleisti atsakymą

    \(\frac{1}{4}\ {\text{sin}}^{-1}(4x)+C\)

  3. Evaluate the integral \(\int \frac{dx}{\sqrt{4-9{x}^{2}}}.\)

    Atskleisti atsakymą

    Substitute \(u=3x.\) Then \(du=3dx\) and we have

    \[\int \frac{dx}{\sqrt{4-9{x}^{2}}}=\frac{1}{3}\int \frac{du}{\sqrt{4-{u}^{2}}}.\]

    Applying the formula with \(a=2,\) we obtain

    \[\begin{array}{ll}\int \frac{dx}{\sqrt{4-9{x}^{2}}} & =\frac{1}{3}\int \frac{du}{\sqrt{4-{u}^{2}}} \\ \\ & =\frac{1}{3}{\text{sin}}^{-1}(\frac{u}{2})+C \\ & =\frac{1}{3}{\text{sin}}^{-1}(\frac{3x}{2})+C.\end{array}\]

  4. Find the indefinite integral using an inverse trigonometric function and substitution for \(\int \frac{dx}{\sqrt{9-{x}^{2}}}.\)

    Atskleisti atsakymą

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

  5. Evaluate the definite integral \({\int }_{0}^{\sqrt{3}\text{/}2}\frac{du}{\sqrt{1-{u}^{2}}}.\)

    Atskleisti atsakymą

    The format of the problem matches the inverse sine formula. Thus,

    \[\begin{array}{ll} \\ \\ {\int }_{0}^{\sqrt{3}\text{/}2}\frac{du}{\sqrt{1-{u}^{2}}} & ={\text{sin}}^{-1}u{|}_{0}^{\sqrt{3}\text{/}2} \\ & =[{\text{sin}}^{-1}(\frac{\sqrt{3}}{2})]-[{\text{sin}}^{-1}(0)] \\ & =\frac{\pi }{3}.\end{array}\]

  6. Evaluate the integral \(\int \frac{1}{1+4{x}^{2}}dx.\)

    Atskleisti atsakymą

    Comparing this problem with the formulas stated in the rule on integration formulas resulting in inverse trigonometric functions, the integrand looks similar to the formula for \({\text{tan}}^{-1}u+C.\) So we use substitution, letting \(u=2x,\) then \(du=2dx\) and \(1\text{/}2du=dx.\) Then, we have

    \[\frac{1}{2}\int \frac{1}{1+{u}^{2}}du=\frac{1}{2}\ {\text{tan}}^{-1}u+C=\frac{1}{2}\ {\text{tan}}^{-1}(2x)+C.\]

  7. Use substitution to find the antiderivative \(\int \frac{dx}{25+4{x}^{2}}.\)

    Atskleisti atsakymą

    \(\frac{1}{10}\ {\text{tan}}^{-1}(\frac{2x}{5})+C\)

  8. Evaluate the integral \(\int \frac{1}{9+{x}^{2}}dx.\)

    Atskleisti atsakymą

    Apply the formula with \(a=3.\) Then,

    \[\int \frac{dx}{9+{x}^{2}}=\frac{1}{3}\ {\text{tan}}^{-1}(\frac{x}{3})+C.\]

  9. Evaluate the integral \(\int \frac{dx}{16+{x}^{2}}.\)

    Atskleisti atsakymą

    \(\frac{1}{4}\ {\text{tan}}^{-1}(\frac{x}{4})+C\)

  10. Evaluate the definite integral \({\int }_{\sqrt{3}\text{/}3}^{\sqrt{3}}\frac{dx}{1+{x}^{2}}.\)

    Atskleisti atsakymą

    Use the formula for the inverse tangent. We have

    \[\begin{array}{ll} \\ \\ {\int }_{\sqrt{3}\text{/}3}^{\sqrt{3}}\frac{dx}{1+{x}^{2}} & ={\text{tan}}^{-1}x{|}_{\sqrt{3}\text{/}3}^{\sqrt{3}} \\ & =[{\text{tan}}^{-1}(\sqrt{3})]-[{\text{tan}}^{-1}(\frac{\sqrt{3}}{3})] \\ & =\frac{\pi }{6}.\end{array}\]

  11. Evaluate the definite integral \({\int }_{0}^{2}\frac{dx}{4+{x}^{2}}.\)

    Atskleisti atsakymą

    \(\frac{\pi }{8}\)

  12. \({\int }_{0}^{\sqrt{3}\text{/}2}\frac{dx}{\sqrt{1-{x}^{2}}}\)

    Atskleisti atsakymą

    \({\text{sin}}^{-1}x{|}_{0}^{\sqrt{3}\text{/}2}=\frac{\pi }{3}\)

  13. \({\int }_{-1\text{/}2}^{1\text{/}2}\frac{dx}{\sqrt{1-{x}^{2}}}\)

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

    Atskleisti atsakymą

    \({\text{tan}}^{-1}x{|}_{\sqrt{3}}^{1}=-\frac{\pi }{12}\)

  15. \({\int }_{1\text{/}\sqrt{3}}^{\sqrt{3}}\frac{dx}{1+{x}^{2}}\)

  16. \({\int }_{\frac{2}{\sqrt{3}}}^{\sqrt{2}}\frac{dx}{|x|\sqrt{{x}^{2}-1}}\)

    Atskleisti atsakymą

    \({\text{sec}}^{-1}{|}_{\frac{2}{\sqrt{3}}}^{\sqrt{2}}=\frac{\pi }{4}-\frac{\pi }{6}=\frac{\pi }{12}\)

  17. \({\int }_{\sqrt{2}}^{2}\frac{dx}{|x|\sqrt{{x}^{2}-1}}\)

  18. \(\int \frac{dx}{\sqrt{9-{x}^{2}}}\)

    Atskleisti atsakymą

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

  19. \(\int \frac{dx}{\sqrt{1-16{x}^{2}}}\)

  20. \(\int \frac{dx}{9+{x}^{2}}\)

    Atskleisti atsakymą

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

  21. \(\int \frac{dx}{25+16{x}^{2}}\)

  22. \(\int \frac{dx}{|x|\sqrt{{x}^{2}-9}}\)

    Atskleisti atsakymą

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

  23. \(\int \frac{dx}{|x|\sqrt{4{x}^{2}-16}}\)

  24. Explain the relationship \(\text{-}{\text{cos}}^{-1}t+C=\int \frac{dt}{\sqrt{1-{t}^{2}}}={\text{sin}}^{-1}t+C.\) Is it true, in general, that \({\text{cos}}^{-1}t=\text{-}{\text{sin}}^{-1}t?\)

    Atskleisti atsakymą

    \(\text{cos}(\frac{\pi }{2}-\theta )=\text{sin}\ \theta .\) So, \({\text{sin}}^{-1}t=\frac{\pi }{2}-{\text{cos}}^{-1}t.\) They differ by a constant.

  25. Explain the relationship \({\text{sec}}^{-1}t+C=\int \frac{dt}{|t|\sqrt{{t}^{2}-1}}=\text{-}{\text{csc}}^{-1}t+C.\) Is it true, in general, that \({\text{sec}}^{-1}t=\text{-}{\text{csc}}^{-1}t?\)

  26. Explain what is wrong with the following integral: \({\int }_{1}^{2}\frac{dt}{\sqrt{1-{t}^{2}}}.\)

    Atskleisti atsakymą

    \(\sqrt{1-{t}^{2}}\) is not defined as a real number when \(t>1.\)

  27. Explain what is wrong with the following integral: \({\int }_{-1}^{1}\frac{dt}{|t|\sqrt{{t}^{2}-1}}.\)

  28. [T] \(\int \frac{1}{\sqrt{9-{x}^{2}}}dx\) over \([-3,3]\)

    Atskleisti atsakymą

    The antiderivative is \({\text{sin}}^{-1}(\frac{x}{3})+C.\) Taking \(C=\frac{\pi }{2}\) recovers the definite integral.

  29. [T] \(\int \frac{9}{9+{x}^{2}}dx\) over \([-6,6]\)

  30. [T] \(\int \frac{\text{cos}\ x}{4+{\text{sin}}^{2}x}dx\) over \([-6,6]\)

    Atskleisti atsakymą

    The antiderivative is \(\frac{1}{2}\ {\text{tan}}^{-1}(\frac{\text{sin}\ x}{2})+C.\) Taking \(C=\frac{1}{2}\ {\text{tan}}^{-1}(\frac{\text{sin}(6)}{2})\) recovers the definite integral.

  31. [T] \(\int \frac{{e}^{x}}{1+{e}^{2x}}dx\) over \([-6,6]\)

  32. \(\int \frac{{\text{sin}}^{-1}tdt}{\sqrt{1-{t}^{2}}}\)

    Atskleisti atsakymą

    \(\frac{1}{2}{({\text{sin}}^{-1}t)}^{2}+C\)

  33. \(\int \frac{dt}{{\text{sin}}^{-1}t\sqrt{1-{t}^{2}}}\)

  34. \(\int \frac{{\text{tan}}^{-1}(2t)}{1+4{t}^{2}}dt\)

    Atskleisti atsakymą

    \(\frac{1}{4}{({\text{tan}}^{-1}(2t))}^{2}+C\)

  35. \(\int \frac{t{\text{tan}}^{-1}({t}^{2})}{1+{t}^{4}}dt\)

  36. \(\int \frac{{\text{sec}}^{-1}(\frac{t}{2})}{|t|\sqrt{{t}^{2}-4}}dt\)

    Atskleisti atsakymą

    \(\frac{1}{4}({\text{sec}}^{-1}{(\frac{t}{2})}^{2})+C\)

  37. \(\int \frac{t{\text{sec}}^{-1}({t}^{2})}{{t}^{2}\sqrt{{t}^{4}-1}}dt\)

  38. [T] \(\int \frac{1}{x\sqrt{{x}^{2}-4}}dx\) over \([2,6]\)

    Atskleisti atsakymą

    The antiderivative is \(\frac{1}{2}\ {\text{sec}}^{-1}(\frac{x}{2})+C.\) Taking \(C=0\) recovers the definite integral over \([2,6].\)

  39. [T] \(\int \frac{1}{(2x+2)\sqrt{x}}dx\) over \([0,6]\)

  40. [T] \(\int \frac{(\text{sin}\ x+x\ \text{cos}\ x)}{1+{x}^{2}{\text{sin}}^{2}x}dx\) over \([-6,6]\)

    Atskleisti atsakymą

    The general antiderivative is \({\text{tan}}^{-1}(x\ \text{sin}\ x)+C.\) Taking \(C=\text{-}{\text{tan}}^{-1}(6\ \text{sin}(6))\) recovers the definite integral.

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Kaip vartoti: Integrals Resulting in Inverse Trigonometric Functions

  1. Integrate functions resulting in inverse trigonometric functions
  2. Formulas for derivatives of inverse trigonometric functions developed in
  3. Use the formulas listed in the rule on integration formulas resulting in inverse trigonometric functions to match up the correct format and make alterations as necessary to solve the problem.
  4. Substitution is often required to put the integrand in the correct form.

Klausimai, kuriuos klausia žmonės

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.

Šio puslapio dalys pritaikytos nuo OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0), OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). Čia yra įtikinamų ir iš naujo paaiškintų klaidų. Kadangi jo šaltinis yra licencijuotas nekomerciniam naudojimui, ši pamoka yra nemokama, nieko neparduoda ir dalijasi pagal tą pačią licenciją.

Daugiau informacijos 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