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Trigonometric substitution

In mathematics, a trigonometric substitution replaces a trigonometric function with another expression. In calculus, trigonometric substitutions are a technique for evaluating integrals.

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 \((\text{-}\infty ,\text{-}a]\cup [a,+\infty ).\) 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.

Trigonometric substitution

In mathematics, a trigonometric substitution replaces a trigonometric function with another expression. In calculus, trigonometric substitutions are a technique for evaluating integrals. In this case, a radical expression is replaced with a trigonometric one. Trigonometric identities may help simplify the answer.

In the case of a definite integral, this substitution changes the interval of integration. Alternatively, the antiderivative of the integrand may be applied to the original interval.

Examples of Case III

Integrals such as

\[\int\frac{dx}{x^2 - a^2}\]

can also be evaluated by partial fractions rather than trigonometric substitutions. However, the integral

\[\int\sqrt{x^2 - a^2}\, dx\]

cannot. In this case, an appropriate substitution is: \[x = a \sec\theta,\, dx = a \sec\theta\tan\theta\, d\theta, \, \theta = \arcsec\frac{x}{a},\]

where \(a > 0\) so that \(\sqrt{a^2}=a,\) and \(0 \le \theta < \frac{\pi}{2}\) by assuming \(x > 0,\) so that \(\tan \theta \ge 0\) and \(\sqrt{\tan^2 \theta} = \tan \theta.\)

Then, \[\begin{aligned} \int\sqrt{x^2 - a^2}\, dx &= \int\sqrt{a^2 \sec^2\theta - a^2} \cdot a \sec\theta\tan\theta\, d\theta \\ &= \int\sqrt{a^2 (\sec^2\theta - 1)} \cdot a \sec\theta\tan\theta\, d\theta \\ &= \int\sqrt{a^2 \tan^2\theta} \cdot a \sec\theta\tan\theta\, d\theta \\ &= \int a^2 \sec\theta\tan^2\theta\, d\theta \\ &= a^2 \int (\sec\theta)(\sec^2\theta - 1)\, d\theta \\ &= a^2 \int (\sec^3\theta - \sec\theta)\, d\theta. \end{aligned}\]

Condensed: the full section is in Wikipedia.

Substitutions that eliminate trigonometric functions

Substitution can be used to remove trigonometric functions.

For instance,

\[\begin{aligned} \int f(\sin(x), \cos(x))\, dx &=\int\frac1{\pm\sqrt{1-u^2}} f\left(u,\pm\sqrt{1-u^2}\right)\, du && u=\sin (x) \\[6pt] \int f(\sin(x), \cos(x))\, dx &=\int\frac{1}{\mp\sqrt{1-u^2}} f\left(\pm\sqrt{1-u^2},u\right)\, du && u=\cos (x) \\[6pt] \int f(\sin(x), \cos(x))\, dx &=\int\frac2{1+u^2} f \left(\frac{2u}{1+u^2},\frac{1-u^2}{1+u^2}\right)\, du && u=\tan\left (\frac{x}{2} \right ) \\[6pt] \end{aligned}\]

The last substitution is known as the Weierstrass substitution, which makes use of tangent half-angle formulas.

For example,

\[\begin{aligned} \int\frac{4 \cos x}{(1+\cos x)^3}\, dx &= \int\frac2{1+u^2}\frac{4\left(\frac{1-u^2}{1+u^2}\right)}{\left(1+\frac{1-u^2}{1+u^2}\right)^3}\, du = \int (1-u^2)(1+u^2)\, du \\&= \int (1-u^4)\,du = u - \frac{u^5}{5} + C = \tan \frac{x}{2} - \frac{1}{5} \tan^5 \frac{x}{2} + C. \end{aligned}\]

Hyperbolic substitution

Substitutions of hyperbolic functions can also be used to simplify integrals.

For example, to integrate \(1/\sqrt{a^2+x^2}\), introduce the substitution \(x=a\sinh{u}\) (and hence \(dx=a\cosh u \,du\)), then use the identity \(\cosh^2 (x) - \sinh^2 (x) = 1\) to find:

\[\begin{aligned} \int \frac{dx}{\sqrt{a^2+x^2}} &= \int \frac{a\cosh u \,du}{\sqrt{a^2+a^2\sinh^2 u}} \\[6pt] &=\int \frac{\cosh{u} \,du}{\sqrt{1+\sinh^2{u}}} \\[6pt] &=\int \frac{\cosh{u}}{\cosh u} \,du \\[6pt] &=u+C \\[6pt] &=\sinh^{-1}{\frac{x}{a}} + C. \end{aligned}\]

If desired, this result may be further transformed using other identities, such as using the relation \(\sinh^{-1}{z} = \operatorname{arsinh}{z} = \ln(z + \sqrt{z^2 + 1})\): \[\begin{aligned} \sinh^{-1}{\frac{x}{a}} + C &=\ln\left(\frac{x}{a} + \sqrt{\frac{x^2}{a^2} + 1}\,\right) + C \\[6pt] &=\ln\left(\frac{x + \sqrt{x^2+a^2}}{a}\,\right) + C. \end{aligned}\]

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

    Vis svaret

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

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

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

    Vis svaret

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

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

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

    Vis svaret

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

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

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

    Vis svaret

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

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

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

    Vis svaret

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

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

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

    Vis svaret

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

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

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

    Vis svaret

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

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

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

    Vis svaret

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

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

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

    Vis svaret

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

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

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

    Vis svaret

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

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

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

    Vis svaret

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

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

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

    Vis svaret

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

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

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

    Vis svaret

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

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

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

    Vis svaret

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

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

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

    Vis svaret

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

Fortsett ditt eget arbeid

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Spørsmål folk stiller

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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Mer i 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