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Trigonometric Substitution

Solve integration problems involving the square root of a sum or difference of two squares.

Integrals Involving

Before developing a general strategy for integrals containing \(\sqrt{{a}^{2}-{x}^{2}},\) consider the integral \(\int \sqrt{9-{x}^{2}}dx.\) This integral cannot be evaluated using any of the techniques we have discussed so far. However, if we make the substitution \(x=3\ \text{sin}\ \theta ,\) we have \(dx=3\ \text{cos}\ \theta d\theta .\) After substituting into the integral, we have

\[\int \sqrt{9-{x}^{2}}\ dx={\int }^{\text{}}\sqrt{9-{(3\ \text{sin}\ \theta )}^{2}}3\ \text{cos}\ \theta d\theta .\]

After simplifying, we have

\[{\int }^{\text{}}\sqrt{9-{x}^{2}}\ dx={\int }^{\text{}}9\sqrt{1-{\text{sin}}^{2}\theta }\ \text{cos}\ \theta d\theta .\]

Letting \(1-{\text{sin}}^{2}\theta ={\text{cos}}^{2}\theta ,\) we now have

\[{\int }^{\text{}}\sqrt{9-{x}^{2}}\ dx={\int }^{\text{}}9\sqrt{{\text{cos}}^{2}\theta }\ \text{cos}\ \theta d\theta .\]

Assuming that \(\text{cos}\ \theta \ge 0,\) we have

\[{\int }^{\text{}}\sqrt{9-{x}^{2}}\ dx={\int }^{\text{}}9\ {\text{cos}}^{2}\theta d\theta .\]

At this point, we can evaluate the integral using the techniques developed for integrating powers and products of trigonometric functions. Before completing this example, let’s take a look at the general theory behind this idea.

To evaluate integrals involving \(\sqrt{{a}^{2}-{x}^{2}},\) we make the substitution \(x=a\ \text{sin}\ \theta\) and \(dx=a\ \text{cos}\ \theta .\) To see that this actually makes sense, consider the following argument: The domain of \(\sqrt{{a}^{2}-{x}^{2}}\) is \([\text{-}a,a].\) Thus, \(\text{-}a\le x\le a.\) Consequently, \(-1\le \frac{x}{a}\le 1.\) Since the range of \(\text{sin}\ x\) over \([\text{-}(\pi \text{/}2),\pi \text{/}2]\) is \([-1,1],\) there is a unique angle \(\theta\) satisfying \(\text{-}(\pi \text{/}2)\le \theta \le \pi \text{/}2\) so that \(\text{sin}\ \theta =x\text{/}a,\) or equivalently, so that \(x=a\ \text{sin}\ \theta .\) If we substitute \(x=a\ \text{sin}\ \theta\) into \(\sqrt{{a}^{2}-{x}^{2}},\) we get

\[\begin{array}{lllll}\sqrt{{a}^{2}-{x}^{2}} & =\sqrt{{a}^{2}-{(a\ \text{sin}\ \theta )}^{2}} & & & \text{Let}\ x=a\ \text{sin}\ \theta \ \text{where}\ -\frac{\pi }{2}\le \theta \le \frac{\pi }{2}.\ \text{Simplify.} \\ & =\sqrt{{a}^{2}-{a}^{2}{\text{sin}}^{2}\theta } & & & \text{Factor out}\ {a}^{2}. \\ & =\sqrt{{a}^{2}(1-{\text{sin}}^{2}\theta )} & & & \text{Substitute}\ 1-{\text{sin}}^{2}x={\text{cos}}^{2}x. \\ & =\sqrt{{a}^{2}{\text{cos}}^{2}\theta } & & & \text{Take the square root.} \\ & =|a\ \text{cos}\ \theta | & & & \\ & =a\ \text{cos}\ \theta . & & & \end{array}\]

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

Integrating Expressions Involving

For integrals containing \(\sqrt{{a}^{2}+{x}^{2},}\) let’s first consider the domain of this expression. Since \(\sqrt{{a}^{2}+{x}^{2}}\) is defined for all real values of \(x,\) we restrict our choice to those trigonometric functions that have a range of all real numbers. Thus, our choice is restricted to selecting either \(x=a\ \text{tan}\ \theta\) or \(x=a\ \text{cot}\ \theta .\) Either of these substitutions would actually work, but the standard substitution is \(x=a\ \text{tan}\ \theta\) or, equivalently, \(\text{tan}\ \theta =x\text{/}a.\) With this substitution, we make the assumption that \(\text{-}(\pi \text{/}2)<\theta <\pi \text{/}2,\) so that we also have \(\theta ={\text{tan}}^{-1}(x\text{/}a).\) The procedure for using this substitution is outlined in the following problem-solving strategy.

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

Integrating Expressions Involving

The domain of the expression \(\sqrt{{x}^{2}-{a}^{2}}\) is \((\text{-}\infty ,\text{-}a]\cup [a,\text{+}\infty ).\) Thus, either \(x\le \text{-}a\) or \(x\ge a.\) Hence, \(\frac{x}{a}\le -1\) or \(\frac{x}{a}\ge 1.\) Since these intervals correspond to the range of \(\text{sec}\ \theta\) on the set \([0,\frac{\pi }{2})\cup (\frac{\pi }{2},\pi ],\) it makes sense to use the substitution \(\text{sec}\ \theta =\frac{x}{a}\) or, equivalently, \(x=a\ \text{sec}\ \theta ,\) where \(0\le \theta <\frac{\pi }{2}\) or \(\frac{\pi }{2}<\theta \le \pi .\) The corresponding substitution for \(dx\) is \(dx=a\ \text{sec}\ \theta \ \text{tan}\ \theta d\theta .\) The procedure for using this substitution is outlined in the following problem-solving strategy.

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

Key Concepts

  • For integrals involving \(\sqrt{{a}^{2}-{x}^{2}},\) use the substitution \(x=a\ \text{sin}\ \theta\) and \(dx=a\ \text{cos}\ \theta d\theta .\)
  • For integrals involving \(\sqrt{{a}^{2}+{x}^{2}},\) use the substitution \(x=a\ \text{tan}\ \theta\) and \(dx=a\ {\text{sec}}^{2}\theta d\theta .\)
  • For integrals involving \(\sqrt{{x}^{2}-{a}^{2}},\) substitute \(x=a\ \text{sec}\ \theta\) and \(dx=a\ \text{sec}\ \theta \ \text{tan}\ \theta d\theta .\)

Trigonometric Substitution

Simplify the following expressions by writing each one using a single trigonometric function.

Use the technique of completing the square to express each trinomial as the square of a binomial or the square of a binomial plus a constant.

Integrate using the method of trigonometric substitution. Express the final answer in terms of the variable.

In the following exercises, use the substitutions \(x=\text{sinh}\ \theta ,\text{cosh}\ \theta ,\) or \(\text{tanh}\ \theta .\) Express the final answers in terms of the variable x.

Use the technique of completing the square to evaluate the following integrals.

Solve the initial-value problem for y as a function of x.

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

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    Begin by making the substitutions \(x=3\ \text{sin}\ \theta\) and \(dx=3\ \text{cos}\ \theta d\theta .\) Since \(\text{sin}\ \theta =\frac{x}{3},\) we can construct the reference triangle shown in the following figure.

    Thus,

    \[\begin{array}{lllll}{\int }^{\text{}}\sqrt{9-{x}^{2}}\ dx & ={\int }^{\text{}}\sqrt{9-{(3\ \text{sin}\ \theta )}^{2}}3\ \text{cos}\ \theta d\theta & & & \text{Substitute}\ x=3\ \text{sin}\ \theta \ \text{and}\ dx=3\ \text{cos}\ \theta d\theta . \\ & ={\int }^{\text{}}\sqrt{9(1-{\text{sin}}^{2}\theta )}3\ \text{cos}\ \theta d\theta & & & \text{Simplify.} \\ & ={\int }^{\text{}}\sqrt{9\ {\text{cos}}^{2}\theta }3\ \text{cos}\ \theta d\theta & & & \text{Substitute}\ {\text{cos}}^{2}\theta =1-{\text{sin}}^{2}\theta . \\ & ={\int }^{\text{}}3|\text{cos}\ \theta |3\ \text{cos}\ \theta d\theta & & & \text{Take the square root.} \\ & ={\int }^{\text{}}9\ {\text{cos}}^{2}\theta d\theta & & & \begin{array}{l}\text{Simplify. Since}\ -\frac{\pi }{2}\le \theta \le \frac{\pi }{2},\ \text{cos}\ \theta \ge 0\ \text{and} \\ |\text{cos}\ \theta |=\text{cos}\ \theta .\end{array} \\ & ={\int }^{\text{}}9(\frac{1}{2}+\frac{1}{2}\text{cos}(2\theta ))d\theta & & & \begin{array}{l}\text{Use the strategy for integrating an even power} \\ \text{of}\ \text{cos}\ \theta .\end{array} \\ & =\frac{9}{2}\theta +\frac{9}{4}\text{sin}(2\theta )+C & & & \text{Evaluate the integral.} \\ & =\frac{9}{2}\theta +\frac{9}{4}(2\ \text{sin}\ \theta \ \text{cos}\ \theta )+C & & & \text{Substitute}\ \text{sin}(2\theta )=2\ \text{sin}\ \theta \ \text{cos}\ \theta . \\ & =\frac{9}{2}{\text{sin}}^{-1}(\frac{x}{3})+\frac{9}{2}\cdot \frac{x}{3}\cdot \frac{\sqrt{9-{x}^{2}}}{3}+C & & & \begin{array}{l}\text{Substitute}\ {\text{sin}}^{-1}(\frac{x}{3})=\theta \ \text{and}\ \text{sin}\ \theta =\frac{x}{3}.\ \text{Use} \\ \text{the reference triangle to see that} \\ \text{cos}\ \theta =\frac{\sqrt{9-{x}^{2}}}{3}\ \text{and make this substitution.}\end{array} \\ & =\frac{9}{2}{\text{sin}}^{-1}(\frac{x}{3})+\frac{x\sqrt{9-{x}^{2}}}{2}+C. & & & \text{Simplify.}\end{array}\]

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

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    First make the substitutions \(x=2\ \text{sin}\ \theta\) and \(dx=2\ \text{cos}\ \theta d\theta .\) Since \(\text{sin}\ \theta =\frac{x}{2},\) we can construct the reference triangle shown in the following figure.

    Thus,

    \[\begin{array}{lllll}\int \frac{\sqrt{4-{x}^{2}}}{x}dx & =\int \frac{\sqrt{4-{(2\ \text{sin}\ \theta )}^{2}}}{2\ \text{sin}\ \theta }2\ \text{cos}\ \theta d\theta & & & \text{Substitute}\ x=2\ \text{sin}\ \theta \ \text{and}\ \text{dx}=2\ \text{cos}\ \theta d\theta . \\ & =\int \frac{2\ {\text{cos}}^{2}\theta }{\text{sin}\ \theta }d\theta & & & \text{Substitute}\ {\text{cos}}^{2}\theta =1-{\text{sin}}^{2}\theta \ \text{and simplify.} \\ & =\int \frac{2(1-{\text{sin}}^{2}\theta )}{\text{sin}\ \theta }d\theta & & & \text{Substitute}\ {\text{sin}}^{2}\theta =1-{\text{cos}}^{2}\theta . \\ & ={\int }^{\text{}}(2\ \text{csc}\ \theta -2\ \text{sin}\ \theta )d\theta & & & \begin{array}{l}\text{Separate the numerator, simplify, and use} \\ \text{csc}\ \theta =\frac{1}{\text{sin}\ \theta }.\end{array} \\ & =2\ \text{ln}|\text{csc}\ \theta -\text{cot}\ \theta |+2\ \text{cos}\ \theta +C & & & \text{Evaluate the integral.} \\ & =2\ \text{ln}|\frac{2}{x}-\frac{\sqrt{4-{x}^{2}}}{x}|+\sqrt{4-{x}^{2}}+C. & & & \begin{array}{l}\text{Use the reference triangle to rewrite the} \\ \text{expression in terms of}\ x\ \text{and simplify.}\end{array}\end{array}\]

  3. Evaluate \({\int }^{\text{}}{x}^{3}\sqrt{1-{x}^{2}}\ dx\) two ways: first by using the substitution \(u=1-{x}^{2}\) and then by using a trigonometric substitution.

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    Method 1

    Let \(u=1-{x}^{2}\) and hence \({x}^{2}=1-u.\) Thus, \(du=-2x\ dx.\) In this case, the integral becomes

    \[\begin{array}{lllll}{\int }^{\text{}}{x}^{3}\sqrt{1-{x}^{2}}\ dx & =-\frac{1}{2}{\int }^{\text{}}{x}^{2}\sqrt{1-{x}^{2}}(-2x\ dx) & & & \text{Make the substitution.} \\ & =-\frac{1}{2}{\int }^{\text{}}(1-u)\sqrt{u}\ du & & & \text{Expand the expression.} \\ & =-\frac{1}{2}\int ({u}^{1\text{/}2}-{u}^{3\text{/}2})du & & & \text{Evaluate the integral.} \\ & =-\frac{1}{2}(\frac{2}{3}{u}^{3\text{/}2}-\frac{2}{5}{u}^{5\text{/}2})+C & & & \text{Rewrite in terms of}\ x. \\ & =-\frac{1}{3}{(1-{x}^{2})}^{3\text{/}2}+\frac{1}{5}{(1-{x}^{2})}^{5\text{/}2}+C. & & & \end{array}\]

    Method 2

    Let \(x=\text{sin}\ \theta .\) In this case, \(dx=\text{cos}\ \theta d\theta .\) Using this substitution, we have

    \[\begin{array}{lllll}{\int }^{\text{}}{x}^{3}\sqrt{1-{x}^{2}}\ dx & ={\int }^{\text{}}{\text{sin}}^{3}\theta \ {\text{cos}}^{2}\theta d\theta & & & \\ & ={\int }^{\text{}}(1-{\text{cos}}^{2}\theta ){\text{cos}}^{2}\theta \ \text{sin}\ \theta d\theta & & & \text{Let}\ u=\text{cos}\ \theta .\ \text{Thus,}\ du=\text{-}\text{sin}\ \theta d\theta . \\ & ={\int }^{\text{}}({u}^{4}-{u}^{2})du & & & \\ & =\frac{1}{5}{u}^{5}-\frac{1}{3}{u}^{3}+C & & & \text{Substitute}\ \text{cos}\ \theta =u. \\ & =\frac{1}{5}{\text{cos}}^{5}\theta -\frac{1}{3}{\text{cos}}^{3}\theta +C & & & \begin{array}{l}\text{Use a reference triangle to see that} \\ \text{cos}\ \theta =\sqrt{1-{x}^{2}}.\end{array} \\ & =\frac{1}{5}{(1-{x}^{2})}^{5\text{/}2}-\frac{1}{3}{(1-{x}^{2})}^{3\text{/}2}+C. & & & \end{array}\]

  4. Rewrite the integral \(\int \frac{{x}^{3}}{\sqrt{25-{x}^{2}}}dx\) using the appropriate trigonometric substitution (do not evaluate the integral).

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    \({\int }^{\text{}}125\ {\text{sin}}^{3}\theta d\theta\)

  5. Evaluate \(\int \frac{dx}{\sqrt{1+{x}^{2}}}\) and check the solution by differentiating.

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    Begin with the substitution \(x=\text{tan}\ \theta\) and \(dx={\text{sec}}^{2}\theta d\theta .\) Since \(\text{tan}\ \theta =x,\) draw the reference triangle in the following figure.

    Thus,

    \[\begin{array}{lllll}\int \frac{dx}{\sqrt{1+{x}^{2}}} & =\int \frac{{\text{sec}}^{2}\theta }{\text{sec}\ \theta }d\theta & & & \begin{array}{l}\text{Substitute}\ x=\text{tan}\ \theta \ \text{and}\ dx={\text{sec}}^{2}\theta d\theta .\ \text{This} \\ \text{substitution makes}\ \sqrt{1+{x}^{2}}=\text{sec}\ \theta .\ \text{Simplify.}\end{array} \\ & ={\int }^{\text{}}\text{sec}\ \theta d\theta & & & \text{Evaluate the integral.} \\ & =\text{ln}|\text{sec}\ \theta +\text{tan}\ \theta |+C & & & \begin{array}{l}\text{Use the reference triangle to express the result} \\ \text{in terms of}\ x.\end{array} \\ & =\text{ln}|\sqrt{1+{x}^{2}}+x|+C. & & & \end{array}\]

    To check the solution, differentiate:

    \[\begin{array}{ll}\frac{d}{dx}(\text{ln}|\sqrt{1+{x}^{2}}+x|) & =\frac{1}{\sqrt{1+{x}^{2}}+x}\cdot (\frac{x}{\sqrt{1+{x}^{2}}}+1) \\ & =\frac{1}{\sqrt{1+{x}^{2}}+x}\cdot \frac{x+\sqrt{1+{x}^{2}}}{\sqrt{1+{x}^{2}}} \\ & =\frac{1}{\sqrt{1+{x}^{2}}}.\end{array}\]

    Since \(\sqrt{1+{x}^{2}}+x>0\) for all values of \(x,\) we could rewrite \(\text{ln}|\sqrt{1+{x}^{2}}+x|+C=\text{ln}(\sqrt{1+{x}^{2}}+x)+C,\) if desired.

  6. Use the substitution \(x=\text{sinh}\ \theta\) to evaluate \(\int \frac{dx}{\sqrt{1+{x}^{2}}}.\)

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    Because \(\text{sinh}\ \theta\) has a range of all real numbers, and \(1+{\text{sinh}}^{2}\theta ={\text{cosh}}^{2}\theta ,\) we may also use the substitution \(x=\text{sinh}\ \theta\) to evaluate this integral. In this case, \(dx=\text{cosh}\ \theta d\theta .\) Consequently,

    \[\begin{array}{lllll}\int \frac{dx}{\sqrt{1+{x}^{2}}} & =\int \frac{\text{cosh}\ \theta }{\sqrt{1+{\text{sinh}}^{2}\theta }}d\theta & & & \begin{array}{l}\text{Substitute}\ x=\text{sinh}\ \theta \ \text{and}\ dx=\text{cosh}\ \theta d\theta . \\ \text{Substitute}\ 1+{\text{sinh}}^{2}\theta ={\text{cosh}}^{2}\theta .\end{array} \\ & =\int \frac{\text{cosh}\ \theta }{\sqrt{{\text{cosh}}^{2}\theta }}d\theta & & & \sqrt{{\text{cosh}}^{2}\theta }=|\text{cosh}\ \theta | \\ & =\int \frac{\text{cosh}\ \theta }{|\text{cosh}\ \theta |}d\theta & & & |\text{cosh}\ \theta |=\text{cosh}\ \theta \ \text{since}\ \text{cosh}\ \theta >0\ \text{for all}\ \theta . \\ & =\int \frac{\text{cosh}\ \theta }{\text{cosh}\ \theta }d\theta & & & \text{Simplify.} \\ & ={\int }^{\text{}}1d\theta & & & \text{Evaluate the integral.} \\ & =\theta +C & & & \text{Since}\ x=\text{sinh}\ \theta ,\ \text{we know}\ \theta ={\text{sinh}}^{-1}x. \\ & ={\text{sinh}}^{-1}x+C. & & & \end{array}\]

  7. Find the length of the curve \(y={x}^{2}\) over the interval \([0,\frac{1}{2}].\)

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    Because \(\frac{dy}{dx}=2x,\) the arc length is given by

    \[{\int }_{0}^{1\text{/}2}\sqrt{1+{(2x)}^{2}}\ dx={\int }_{0}^{1\text{/}2}\sqrt{1+4{x}^{2}}\ dx.\]

    To evaluate this integral, use the substitution \(x=\frac{1}{2}\text{tan}\ \theta\) and \(dx=\frac{1}{2}{\text{sec}}^{2}\theta d\theta .\) We also need to change the limits of integration. If \(x=0,\) then \(\theta =0\) and if \(x=\frac{1}{2},\) then \(\theta =\frac{\pi }{4}.\) Thus,

    \[\begin{array}{lllll}{\int }_{0}^{1\text{/}2}\sqrt{1+4{x}^{2}}\ dx & ={\int }_{0}^{\pi \text{/}4}\sqrt{1+{\text{tan}}^{2}\theta }\frac{1}{2}{\text{sec}}^{2}\theta d\theta & & & \begin{array}{l}\text{After substitution,} \\ \sqrt{1+4{x}^{2}}=\text{tan}\ \theta .\ \text{Substitute} \\ 1+{\text{tan}}^{2}\theta ={\text{sec}}^{2}\theta \ \text{and simplify.}\end{array} \\ & =\frac{1}{2}{\int }_{0}^{\pi \text{/}4}{\text{sec}}^{3}\theta d\theta & & & \begin{array}{l}\text{We derived this integral in the} \\ \text{previous section.}\end{array} \\ & =\frac{1}{2}(\frac{1}{2}\text{sec}\ \theta \ \text{tan}\ \theta +\frac{1}{2}\text{ln}|\text{sec}\ \theta +\text{tan}\ \theta |)|{}_{\begin{array}{l} \\ 0\end{array}}^{\begin{array}{l}\pi \text{/}4 \\ \end{array}} & & & \text{Evaluate and simplify.} \\ & =\frac{1}{4}(\sqrt{2}+\text{ln}(\sqrt{2}+1)). & & & \end{array}\]

  8. Rewrite \({\int }^{\text{}}{x}^{3}\sqrt{{x}^{2}+4}\ dx\) by using a substitution involving \(\text{tan}\ \theta .\)

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    \({\int }^{\text{}}32\ {\text{tan}}^{3}\theta \ {\text{sec}}^{3}\theta d\theta\)

  9. Find the area of the region between the graph of \(f(x)=\sqrt{{x}^{2}-9}\) and the x-axis over the interval \([3,5].\)

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    First, sketch a rough graph of the region described in the problem, as shown in the following figure.

    We can see that the area is \(A={\int }_{3}^{5}\sqrt{{x}^{2}-9}\ dx.\) To evaluate this definite integral, substitute \(x=3\ \text{sec}\ \theta\) and \(dx=3\ \text{sec}\ \theta \ \text{tan}\ \theta d\theta .\) We must also change the limits of integration. If \(x=3,\) then \(3=3\ \text{sec}\ \theta\) and hence \(\theta =0.\) If \(x=5,\) then \(\theta ={\text{sec}}^{-1}(\frac{5}{3}).\) After making these substitutions and simplifying, we have

    \[\begin{array}{lllll}\text{Area} & ={\int }_{3}^{5}\sqrt{{x}^{2}-9}\ dx & & & \\ & ={\int }_{0}^{{\text{sec}}^{-1}(5\text{/}3)}9\ {\text{tan}}^{2}\theta \ \text{sec}\ \theta d\theta & & & \text{Use}\ {\text{tan}}^{2}\theta =1-{\text{sec}}^{2}\theta . \\ & ={\int }_{0}^{{\text{sec}}^{-1}(5\text{/}3)}9({\text{sec}}^{2}\theta -1)\text{sec}\ \theta d\theta & & & \text{Expand.} \\ & ={\int }_{0}^{{\text{sec}}^{-1}(5\text{/}3)}9({\text{sec}}^{3}\theta -\text{sec}\ \theta )d\theta & & & \text{Evaluate the integral.} \\ & =(\frac{9}{2}\text{ln}|\text{sec}\ \theta +\text{tan}\ \theta |+\frac{9}{2}\text{sec}\ \theta \ \text{tan}\ \theta )-9\ \text{ln}|\text{sec}\ \theta +\text{tan}\ \theta ||{}_{\begin{array}{l} \\ 0\end{array}}^{\begin{array}{l}{\text{sec}}^{-1}(5\text{/}3) \\ \end{array}} & & & \text{Simplify.} \\ & =\frac{9}{2}\text{sec}\ \theta \ \text{tan}\ \theta -\frac{9}{2}\text{ln}|\text{sec}\ \theta +\text{tan}\ \theta ||{}_{\begin{array}{l} \\ 0\end{array}}^{\begin{array}{l}{\text{sec}}^{-1}(5\text{/}3) \\ \end{array}} & & & \begin{array}{l}\text{Evaluate. Use}\ \text{sec}({\text{sec}}^{-1}\frac{5}{3})=\frac{5}{3} \\ \text{and}\ \text{tan}({\text{sec}}^{-1}\frac{5}{3})=\frac{4}{3}.\end{array} \\ & =\frac{9}{2}\cdot \frac{5}{3}\cdot \frac{4}{3}-\frac{9}{2}\text{ln}|\frac{5}{3}+\frac{4}{3}|-(\frac{9}{2}\cdot 1\cdot 0-\frac{9}{2}\text{ln}|1+0|) & & & \\ & =10-\frac{9}{2}\text{ln}\ 3. & & & \end{array}\]

  10. Evaluate \(\int \frac{dx}{\sqrt{{x}^{2}-4}}.\) Assume that \(x>2.\)

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    \(\text{ln}|\frac{x}{2}+\frac{\sqrt{{x}^{2}-4}}{2}|+C\)

  11. \(4-4\ {\text{sin}}^{2}\theta\)

  12. \(9\ {\text{sec}}^{2}\theta -9\)

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    \(9\ {\text{tan}}^{2}\theta\)

  13. \({a}^{2}+{a}^{2}{\text{tan}}^{2}\theta\)

  14. \({a}^{2}+{a}^{2}{\text{sinh}}^{2}\theta\)

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    \({a}^{2}{\text{cosh}}^{2}\theta\)

  15. \(16\ {\text{cosh}}^{2}\theta -16\)

  16. \(4{x}^{2}-4x+1\)

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    \(4{(x-\frac{1}{2})}^{2}\)

  17. \(2{x}^{2}-8x+3\)

  18. \(\text{-}{x}^{2}-2x+4\)

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    \(\text{-}{(x+1)}^{2}+5\)

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

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

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    \(\text{ln}|\frac{\sqrt{{x}^{2}-{a}^{2}}+x}{a}|+C\)

  21. \(\int \sqrt{4-{x}^{2}}\ dx\)

  22. \(\int \frac{dx}{\sqrt{1+9{x}^{2}}}\)

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    \(\frac{1}{3}\text{ln}|\sqrt{9{x}^{2}+1}+3x|+C\)

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

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

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

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

  26. \(\int \sqrt{{x}^{2}+9}dx\)

    답을 드러내세요

    \(9[\frac{x\sqrt{{x}^{2}+9}}{18}+\frac{1}{2}ln|\frac{\sqrt{{x}^{2}+9}}{3}+\frac{x}{3}|]+C\)

  27. \(\int \frac{\sqrt{{x}^{2}-25}}{x}dx\)

  28. \(\int \frac{{\theta }^{3}d\theta }{\sqrt{9-{\theta }^{2}}}\)

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    \(-\frac{1}{3}\sqrt{9-{\theta }^{2}}(18+{\theta }^{2})+C\)

  29. \(\int \frac{dx}{\sqrt{{x}^{6}-{x}^{2}}}\)

  30. \(\int \sqrt{{x}^{6}-{x}^{8}}dx\)

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    \(\frac{(-1+{x}^{2})(2+3{x}^{2})\sqrt{{x}^{6}-{x}^{8}}}{15{x}^{3}}+C\)

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

  32. \(\int \frac{dx}{{({x}^{2}-9)}^{3\text{/}2}}\)

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    \(-\frac{x}{9\sqrt{-9+{x}^{2}}}+C\)

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

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

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

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

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

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

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

  38. \(\int {(1-{x}^{2})}^{3\text{/}2}dx\)

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    \(\frac{1}{8}(x(5-2{x}^{2})\sqrt{1-{x}^{2}}+3\ \text{arcsin}\ x)+C\)

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

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

    답을 드러내세요

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

Symbols used here

\int f(x)\,dx,\ \int_a^b
integral
Antiderivative (indefinite) or signed area from a to b (definite).
\sqrt{x},\ \sqrt[n]{x}
square root, n-th root
The non-negative number whose square (n-th power) is x.
\infty
infinity
Not a number: "grows without bound" in limits and intervals.
\pi
pi
Ratio of a circle's circumference to its diameter, 3.14159…
\theta
theta
The usual name for an angle.
A \cup B,\ A \cap B,\ A \setminus B
union, intersection, difference
In either; in both; in A but not B.
f'(x),\ \frac{dy}{dx}
derivative
Instantaneous rate of change; slope of the graph.
\leq,\ \geq
less/greater than or equal
Inequalities that allow equality; < and > exclude it.
\sup,\ \inf
supremum, infimum
Least upper bound, greatest lower bound.
e
Euler's number
2.71828…, the base whose exponential is its own derivative.
\sin,\ \cos,\ \tan
sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.
\log_b x,\ \ln x
logarithm, natural log
The exponent b must be raised to for x; ln uses base e.
\sum_{k=1}^{n} a_k
summation
Add a_k for k = 1 up to n.
\lim_{x \to a} f(x)
limit
The value f(x) approaches as x approaches a.
y',\ y''
first and second derivative of y
Prime notation for derivatives with respect to x (or t).
C,\ C_1,\ C_2
arbitrary constants
Constants of integration fixed by initial conditions.

How to: Trigonometric Substitution

  1. Solve integration problems involving the square root of a sum or difference of two squares.
  2. It is a good idea to make sure the integral cannot be evaluated easily in another way. For example, although this method can be applied to integrals of the form
  3. Make the substitution
  4. Simplify the expression.
  5. Evaluate the integral using techniques from the section on trigonometric integrals.
  6. Use the reference triangle from
  7. Check to see whether the integral can be evaluated easily by using another method. In some cases, it is more convenient to use an alternative method.
  8. Substitute

Questions people ask

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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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.

에 더 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