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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.

이제 너 계산기는 이것을 해결하지 않지만, 그 조각은 계산 가능합니다. 아래의 하나를 시도하거나 자신의 것을 입력하십시오.

연습 (40)

먼저 종이에 각각을 시도해 보세요. 확인하기 위해 답을 드러내세요. 확인된 답은 각 단계의 솔버에서 열 수 있습니다.

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

    답을 드러내세요

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

    답을 드러내세요

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

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

    답을 드러내세요

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

    답을 드러내세요

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

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

    답을 드러내세요

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

    답을 드러내세요

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

    답을 드러내세요

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

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

    답을 드러내세요

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

    답을 드러내세요

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

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

    답을 드러내세요

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

    답을 드러내세요

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

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

    답을 드러내세요

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

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

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

    답을 드러내세요

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

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

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

    답을 드러내세요

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

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

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

    답을 드러내세요

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

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

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

    답을 드러내세요

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

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

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

    답을 드러내세요

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

  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 \({\text{cos}}^{-1}t=\text{-}{\text{sin}}^{-1}t?\)

    답을 드러내세요

    \(\text{cos}(\frac{\pi }{2}-\theta )=\text{sin}\ \theta .\) 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: \({\int }_{1}^{2}\frac{dt}{\sqrt{1-{t}^{2}}}.\)

    답을 드러내세요

    \(\sqrt{1-{t}^{2}}\) 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 \([-3,3]\)

    답을 드러내세요

    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 \([-6,6]\)

    답을 드러내세요

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

    답을 드러내세요

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

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

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

    답을 드러내세요

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

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

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

    답을 드러내세요

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

  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 \([2,6]\)

    답을 드러내세요

    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 \([-6,6]\)

    답을 드러내세요

    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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어떻게: 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.

사람들이 묻는 질문

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.

이 페이지의 일부는 다음에서 변경되었습니다. OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0), OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). 여기서 압축하고 다시 설명; 오류는 우리의. 소스가 비상업적 사용을 위해 라이선스되어 있기 때문에, 이 레슨은 무료이며, 아무것도 판매하지 않으며, 동일한 라이선스로 공유됩니다.

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