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

    Zbulo përgjigjen

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

    Zbulo përgjigjen

    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.

    Zbulo përgjigjen

    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).

    Zbulo përgjigjen

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

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

    Zbulo përgjigjen

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

    Zbulo përgjigjen

    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 \([0,\frac{1}{2}].\)

    Zbulo përgjigjen

    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 \(\text{tan}\ \theta .\)

    Zbulo përgjigjen

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

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

    Zbulo përgjigjen

    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 \(x>2.\)

    Zbulo përgjigjen

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

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

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

    Zbulo përgjigjen

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

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

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

    Zbulo përgjigjen

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

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

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

    Zbulo përgjigjen

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

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

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

    Zbulo përgjigjen

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

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

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

    Zbulo përgjigjen

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

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

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

    Zbulo përgjigjen

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

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

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

    Zbulo përgjigjen

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

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

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

    Zbulo përgjigjen

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

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

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

    Zbulo përgjigjen

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

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

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

    Zbulo përgjigjen

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

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

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

    Zbulo përgjigjen

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

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

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

    Zbulo përgjigjen

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

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

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

    Zbulo përgjigjen

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

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

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

    Zbulo përgjigjen

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

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

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

    Zbulo përgjigjen

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

Mbaje punën tënde

Një llogari e lirë shtohet shënime në çdo mësim, një regjistrim të asaj që ju keni përfunduar, problemet tuaja të zgjidhura në një vend, dhe një mësues që ju mund të pyesni rreth kësaj faqeje. Matematika vetë është e hapur për të gjithë, të regjistruar apo jo.

Regjistrohu Hyr

Simbolet e përdorura këtu

Prek çdo simbol për përkufizimin e plotë, një fotografi dhe se çfarë do të thotë çdo shkronjë në të.

Pyetja që bëjnë njerëzit

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.

Pjesa e kësaj faqeje është adaptuar nga OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0), Wikipedia (CC BY-SA 4.0). E përmbledhur dhe ri-shkruar këtu; gabimet janë tona. Për shkak se burimi i tij është licencuar për përdorim jo komercial, ky mësim është i lirë, nuk shet asgjë dhe ndahet nën të njëjtën licencë.

Më shumë në 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