Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

4.0 Antiderivatives and Indefinite Integrals

  • Page ID
    10311
  • [ "stage:draft", "article:topic" ]

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    Integrals of Exponential Functions

    The exponential function is perhaps the most efficient function in terms of the operations of calculus. The exponential function, \(y=e^x\), is its own derivative and its own integral.

    Rule: Integrals of Exponential Functions

    Exponential functions can be integrated using the following formulas.

    \[∫e^xdx=e^x+C\]

    \[∫a^xdx=\dfrac{a^x}{\ln a}+C\]

    In this section we focus on integrals that result in inverse trigonometric functions. We have worked with these functions before. Recall, that trigonometric functions are not one-to-one unless the domains are restricted. When working with inverses of trigonometric functions, we always need to be careful to take these restrictions into account. Also, we previously developed formulas for derivatives of inverse trigonometric functions. The formulas developed there give rise directly to integration formulas involving inverse trigonometric functions.

    Integrals Involving Logarithmic Functions

    Integrating functions of the form \(f(x)=x^{−1}\) result in the absolute value of the natural log function, as shown in the following rule. Integral formulas for other logarithmic functions, such as \(f(x)=\ln x\) and \(f(x)=log_ax\), are also included in the rule.

    Rule: Integration Formulas Involving Logarithmic Functions

    The following formulas can be used to evaluate integrals involving logarithmic functions.

    \[\begin{align} ∫x^{−1}dx &=\ln |x|+C \\ ∫\ln x\,dx &= x\ln x−x+C =x (\ln x−1)+C \\ ∫log_a\,x\,dx &=\dfrac{x}{\ln a}(\ln x−1)+C \end{align}\]

    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.

    Rule: Integration Formulas Resulting in Inverse Trigonometric Functions

    The following integration formulas yield inverse trigonometric functions:

    \[ \begin{align} ∫\dfrac{du}{\sqrt{a^2−u^2}}&=\sin^{−1}\dfrac{u}{a}+C \\ ∫\dfrac{du}{a^2+u^2}&=\dfrac{1}{a}\tan^{−1}\dfrac{u}{a}+C \\ ∫\dfrac{du}{u\sqrt{u^2−a^2}}&=\dfrac{1}{a}sec^{−1}\dfrac{u}{a}+C \end{align}\]

    Proof:

    Let \( y=\sin^{−1}\dfrac{x}{a}.\) Then \( a \sin y=x\). Now let’s use implicit differentiation. We obtain

    \[ \dfrac{d}{dx}(a \sin y)=\dfrac{d}{dx}(x)\]

    \[ a\cos y\dfrac{dy}{dx}=1\]

    \[ \dfrac{dy}{dx}=\dfrac{1}{a\cos y}.\]

    For \( −\dfrac{π}{2}≤y≤\dfrac{π}{2},\cos y≥0.\) Thus, applying the Pythagorean identity \( \sin^2y+\cos^2y=1\), we have \( \cos y=\sqrt{1=\sin^2y}.\) This gives

    \[ \begin{align} \dfrac{1}{a \cos y} &=\dfrac{1}{a\sqrt{1−\sin^2y}} \\  &=\dfrac{1}{\sqrt{a^2−a^2 \sin^2y}} \\  &=\dfrac{1}{\sqrt{a^2−x^2}}. \end{align}\]

    Then for \( −a≤x≤a,\) we have

    \[ ∫\dfrac{1}{\sqrt{a^2−u^2}}du=sin^{−1}(\dfrac{u}{a})+C.\]

    Proof

    Let \( y=\sin^{−1}\dfrac{x}{a}.\) Then \( a \sin y=x\). Now let’s use implicit differentiation. We obtain

    \[ \dfrac{d}{dx}(a \sin y)=\dfrac{d}{dx}(x)\]

    \[ a\cos y\dfrac{dy}{dx}=1\]

    \[ \dfrac{dy}{dx}=\dfrac{1}{a\cos y}.\]

    For \( −\dfrac{π}{2}≤y≤\dfrac{π}{2},\cos y≥0.\) Thus, applying the Pythagorean identity \( \sin^2y+\cos^2y=1\), we have \( \cos y=\sqrt{1=\sin^2y}.\) This gives

    \[ \begin{align} \dfrac{1}{a \cos y} &=\dfrac{1}{a\sqrt{1−\sin^2y}} \\  &=\dfrac{1}{\sqrt{a^2−a^2 \sin^2y}} \\  &=\dfrac{1}{\sqrt{a^2−x^2}}. \end{align}\]

    Then for \( −a≤x≤a,\) we have

    \[ ∫\dfrac{1}{\sqrt{a^2−u^2}}du=sin^{−1}(\dfrac{u}{a})+C.\]

    Example \( \PageIndex{1}\): Evaluating an  INDefinite Integral Using Inverse Trigonometric Functions

    Evaluate the indefinite integral

    \[ \int \dfrac{dx}{\sqrt{1− 9x^2}}. \nonumber\]

    Solution

    We can go directly to the formula for the antiderivative in the rule on integration formulas resulting in inverse trigonometric functions. We have

    \[ ∫\dfrac{dx}{\sqrt{1−9x^2}}=\frac{1}{3}\sin^{−1}(3x) +C\nonumber\]

    Exercise \(\PageIndex{1}\)

    Find the antiderivative of

    \[ ∫\dfrac{dx}{\sqrt{1−16x^2}}.\nonumber\]

    Hint

    Substitute \( u=4x\)

    Answer

    \[ \dfrac{1}{4}sin^{−1}(4x)+C\nonumber\]

    Example \( \PageIndex{2}\): Finding an Antiderivative Involving an Inverse Trigonometric Function

    Evaluate the integral

    \[ ∫\dfrac{dx}{\sqrt{4−9x^2}}.\nonumber\]

    Solution

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

    \[ ∫\dfrac{dx}{\sqrt{4−9x^2}}=\dfrac{1}{3}∫\dfrac{du}{\sqrt{4−u^2}}.\nonumber\]

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

    \[ ∫\dfrac{dx}{\sqrt{4−9x^2}}=\dfrac{1}{3}∫\dfrac{du}{\sqrt{4−u^2}}=\dfrac{1}{3}sin^{−1}(\dfrac{u}{2})+C=\dfrac{1}{3}sin^{−1}(\dfrac{3x}{2})+C.\nonumber\]

    Exercise \(\PageIndex{2}\)

    Find the indefinite integral using an inverse trigonometric function and substitution for \( ∫\dfrac{dx}{\sqrt{9−x^2}}\).

    Hint

    Use the formula in the rule on integration formulas resulting in inverse trigonometric functions.

    Answer

    \[ \sin^{−1}(\dfrac{x}{3})+C\]

    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 \( \PageIndex{4}\): Finding an Antiderivative Involving the Inverse Tangent Function

    Find an antiderivative of \( ∫\dfrac{1}{1+4x^2}dx.\)

    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 \( tan^{−1}u+C\). So we use substitution, letting \( u=2x\), then \( du=2dx\) and \( 1/2du=dx.\)Then, we have

    \( \dfrac{1}{2}∫\dfrac{1}{1+u^2}du=\dfrac{1}{2}\tan^{−1}u+C=\dfrac{1}{2}\tan^{−1}(2x)+C.\)

    Exercise \(\PageIndex{3}\)

    Use substitution to find the antiderivative of \( ∫\dfrac{dx}{25+4x^2}.\)

    Hint

    Use the solving strategy from Example \( \PageIndex{4}\) and the rule on integration formulas resulting in inverse trigonometric functions.

    Answer

    \[ \dfrac{1}{10}tan^{−1}(\dfrac{2x}{5})+C\]

    Example \( \PageIndex{5}\): Applying the Integration Formulas

    Find the antiderivative of \( ∫\dfrac{1}{9+x^2}dx.\)

    Solution

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

    \[ ∫\dfrac{dx}{9+x^2}=\dfrac{1}{3}\tan^{−1}(\dfrac{x}{3})+C.\]

    Exercise \(\PageIndex{4}\)

    Find the antiderivative of \( ∫\dfrac{dx}{16+x^2}\).

    Hint

    Follow the steps in Example.

    Answer

    \[ \dfrac{1}{4}\tan^{−1}(\dfrac{x}{4})+C\]

     

    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

    \( ∫\dfrac{du}{\sqrt{a^2−u^2}}=sin^{−1}(\dfrac{u}{a})+C\)

    \( ∫\dfrac{du}{a^2+u^2}=\dfrac{1}{a}tan^{−1}(\dfrac{u}{a})+C\)

    \( ∫\dfrac{du}{u\sqrt{u^2−a^2}}=\dfrac{1}{a}sec^{−1}(\dfrac{u}{a})+C\)

    Contributors

    • Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY 3/0 license. Download for free at http://cnx.org.