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Mathematics LibreTexts

7.3: Trigonometric Substitution

  • Gilbert Strang & Edwin “Jed” Herman
  • OpenStax

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

In this section, we explore integrals containing expressions of the form a2x2, a2+x2, and x2a2, where the values of a are positive. We have already encountered and evaluated integrals containing some expressions of this type, but many still remain inaccessible. The technique of trigonometric substitution comes in very handy when evaluating these integrals. This technique uses substitution to rewrite these integrals as trigonometric integrals.

Integrals Involving a2x2

Before developing a general strategy for integrals containing a2x2, consider the integral 9x2dx. This integral cannot be evaluated using any of the techniques we have discussed so far. However, if we make the substitution x=3sinθ, we have dx=3cosθdθ. After substituting into the integral, we have

9x2dx=9(3sinθ)23cosθdθ.

After simplifying, we have

9x2dx=91sin2θcosθdθ.

Letting 1sin2θ=cos2θ, we now have

9x2dx=9cos2θcosθdθ.

Assuming that cosθ0, we have

9x2dx=9cos2θdθ.

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 a2x2, we make the substitution x=asinθ and dx=acosθ. To see that this actually makes sense, consider the following argument: The domain of a2x2 is [a,a]. Thus,

axa.

Consequently,

1xa1.

Since the range of sinx over [(π/2),π/2] is [1,1], there is a unique angle θ satisfying (π/2)θπ/2 so that sinθ=x/a, or equivalently, so that x=asinθ. If we substitute x=asinθ into a2x2, we get

a2x2=a2(asinθ)2Let x=asinθ where π2θπ2.Simplify.=a2a2sin2θFactor out a2.=a2(1sin2θ)Substitute 1sin2x=cos2x.=a2cos2θTake the square root.=|acosθ|=acosθ

Since cosx0 on π2θπ2 and a>0,|acosθ|=acosθ. We can see, from this discussion, that by making the substitution x=asinθ, we are able to convert an integral involving a radical into an integral involving trigonometric functions. After we evaluate the integral, we can convert the solution back to an expression involving x. To see how to do this, let’s begin by assuming that 0<x<a. In this case, 0<θ<π2. Since sinθ=xa, we can draw the reference triangle in Figure 7.3.1 to assist in expressing the values of cosθ,tanθ, and the remaining trigonometric functions in terms of x. It can be shown that this triangle actually produces the correct values of the trigonometric functions evaluated at θ for all θ satisfying π2θπ2. It is useful to observe that the expression a2x2 actually appears as the length of one side of the triangle. Last, should θ appear by itself, we use θ=sin1(xa).

This figure is a right triangle. It has an angle labeled theta. This angle is opposite the vertical side. The hypotenuse is labeled a, the vertical leg is labeled x, and the horizontal leg is labeled as the square root of (a^2 – x^2). To the left of the triangle is the equation sin(theta) = x/a.
Figure 7.3.1: A reference triangle can help express the trigonometric functions evaluated at θ in terms of x.

The essential part of this discussion is summarized in the following problem-solving strategy.

Problem-Solving Strategy: Integrating Expressions Involving a2x2
  1. 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 1a2x2dx, xa2x2dx, and xa2x2dx, they can each be integrated directly either by formula or by a simple u-substitution.
  2. Make the substitution x=asinθ and dx=acosθdθ. Note: This substitution yields a2x2=acosθ.
  3. Simplify the expression.
  4. Evaluate the integral using techniques from the section on trigonometric integrals.
  5. Use the reference triangle from Figure 1 to rewrite the result in terms of x. You may also need to use some trigonometric identities and the relationship θ=sin1(xa).

The following example demonstrates the application of this problem-solving strategy.

Example 7.3.1: Integrating an Expression Involving a2x2

Evaluate

9x2dx.

Solution

Begin by making the substitutions x=3sinθ and dx=3cosθdθ. Since sinθ=x3, we can construct the reference triangle shown in Figure 2.

This figure is a right triangle. It has an angle labeled theta. This angle is opposite the vertical side. The hypotenuse is labeled 3, the vertical leg is labeled x, and the horizontal leg is labeled as the square root of (9 – x^2). To the left of the triangle is the equation sin(theta) = x/3.
Figure 7.3.2: A reference triangle can be constructed for Example 7.3.1.

Thus,

9x2dx=9(3sinθ)23cosθdθ  Substitute x=3sinθ and dx=3cosθdθ.

=9(1sin2θ)3cosθdθ Simplify.

=9cos2θ3cosθdθ Substitute cos2θ=1sin2θ.

=3|cosθ|3cosθdθ Take the square root.

=9cos2θdθ Simplify. Since π2θπ2,cosθ0 and |cosθ|=cosθ.

=9(12+12cos(2θ))dθ Use the strategy for integrating an even power of cosθ.

=92θ+94sin(2θ)+C Evaluate the integral.

=92θ+94(2sinθcosθ)+C  Substitute sin(2θ)=2sinθcosθ.

=92sin1(x3)+92x39x23+C Substitute sin1(x3)=θ and sinθ=x3.

Use the reference triangle to see that cosθ=9x23and make this substitution. Simplify.

=92sin1(x3)+x9x22+C. Simplify.

Example 7.3.2: Integrating an Expression Involving a2x2

Evaluate

4x2xdx.

Solution

First make the substitutions x=2sinθ and dx=2cosθdθ. Since sinθ=x2, we can construct the reference triangle shown in Figure 7.3.3.

This figure is a right triangle. It has an angle labeled theta. This angle is opposite the vertical side. The vertical leg is labeled x, and the horizontal leg is labeled as the square root of (4 – x^2). To the left of the triangle is the equation sin(theta) = x/2.
Figure 7.3.3: A reference triangle can be constructed for Example 7.3.2.

Thus,

4x2xdx=4(2sinθ)22sinθ2cosθdθ Substitute x=2sinθ and dx=2cosθdθ.

=2cos2θsinθdθ Substitute cos2θ=1sin2θ and simplify.

=2(1sin2θ)sinθdθ Substitute cos2θ=1sin2θ.

=(2cscθ2sinθ)dθ Separate the numerator, simplify, and use cscθ=1sinθ.

=2ln|cscθcotθ|+2cosθ+C Evaluate the integral.

=2ln|2x4x2x|+4x2+C. Use the reference triangle to rewrite the expression in terms of x and simplify.

In the next example, we see that we sometimes have a choice of methods.

Example 7.3.3: Integrating an Expression Involving a2x2 Two Ways

Evaluate x31x2dx two ways: first by using the substitution u=1x2 and then by using a trigonometric substitution.

Method 1

Let u=1x2 and hence x2=1u. Thus, du=2xdx. In this case, the integral becomes

x31x2dx=12x21x2(2xdx) Make the substitution.

=12(1u)udu Expand the expression.

=12(u1/2u3/2)du Evaluate the integral.

=12(23u3/225u5/2)+C Rewrite in terms of x.

=13(1x2)3/2+15(1x2)5/2+C.

Method 2

Let x=sinθ. In this case, dx=cosθdθ. Using this substitution, we have

x31x2dx=sin3θcos2θdθ

=(1cos2θ)cos2θsinθdθ Let u=cosθ. Thus,du=sinθdθ.

=(u4u2)du

=15u513u3+C Substitute cosθ=u.

=15cos5θ13cos3θ+C Use a reference triangle to see that cosθ=1x2.

=15(1x2)5/213(1x2)3/2+C.

Exercise 7.3.1

Rewrite the integral x325x2dx using the appropriate trigonometric substitution (do not evaluate the integral).

Hint

Substitute x=5sinθ and dx=5cosθdθ.

Answer

125sin3θdθ

Integrating Expressions Involving a2+x2

For integrals containing a2+x2,let’s first consider the domain of this expression. Since a2+x2 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=atanθ or x=acotθ. Either of these substitutions would actually work, but the standard substitution is x=atanθ or, equivalently, tanθ=x/a. With this substitution, we make the assumption that (π/2)<θ<π/2, so that we also have θ=tan1(x/a). The procedure for using this substitution is outlined in the following problem-solving strategy.

Problem-Solving Strategy: Integrating Expressions Involving a2+x2
  1. 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.
  2. Substitute x=atanθ and dx=asec2θdθ. This substitution yields a2+x2=a2+(atanθ)2=a2(1+tan2θ)=a2sec2θ=|asecθ|=asecθ. (Since π2<θ<π2 and secθ>0 over this interval, |asecθ|=asecθ.)
  3. Simplify the expression.
  4. Evaluate the integral using techniques from the section on trigonometric integrals.
  5. Use the reference triangle from Figure 7.3.4 to rewrite the result in terms of x. You may also need to use some trigonometric identities and the relationship θ=tan1(xa). (Note: The reference triangle is based on the assumption that x>0; however, the trigonometric ratios produced from the reference triangle are the same as the ratios for which x0.)
This figure is a right triangle. It has an angle labeled theta. This angle is opposite the vertical side. The hypotenuse is labeled the square root of (a^2+x^2), the vertical leg is labeled x, and the horizontal leg is labeled a. To the left of the triangle is the equation tan(theta) = x/a.
Figure 7.3.4: A reference triangle can be constructed to express the trigonometric functions evaluated at θ in terms of x.
Example 7.3.4: Integrating an Expression Involving a2+x2

Evaluate dx1+x2 and check the solution by differentiating.

Solution

Begin with the substitution x=tanθ and dx=sec2θdθ. Since tanθ=x, draw the reference triangle in Figure 7.3.5.

This figure is a right triangle. It has an angle labeled theta. This angle is opposite the vertical side. The hypotenuse is labeled the square root of (1+x^2), the vertical leg is labeled x, and the horizontal leg is labeled 1. To the left of the triangle is the equation tan(theta) = x/1.
Figure 7.3.5: The reference triangle for Example 7.3.4.

Thus,

dx1+x2=sec2θsecθdθSubstitute x=tanθ and dx=sec2θdθ.This substitution makes 1+x2=secθ. Simplify.=secθdθEvaluate the integral.=ln|secθ+tanθ|+CUse the reference triangle to express the result in terms of x.=ln|1+x2+x|+C

To check the solution, differentiate:

ddx(ln|1+x2+x|)=11+x2+x(x1+x2+1)=11+x2+xx+1+x21+x2=11+x2.

Since 1+x2+x>0 for all values of x, we could rewrite ln|1+x2+x|+C=ln(1+x2+x)+C, if desired.

Example 7.3.5: Evaluating dx1+x2 Using a Different Substitution

Use the substitution x=sinhθ to evaluate dx1+x2.

Solution

Because sinhθ has a range of all real numbers, and 1+sinh2θ=cosh2θ, we may also use the substitution x=sinhθ to evaluate this integral. In this case, dx=coshθdθ. Consequently,

dx1+x2=coshθ1+sinh2θdθSubstitute x=sinhθ and dx=coshθdθ.Substitute 1+sinh2θ=cosh2θ.=coshθcosh2θdθSince cosh2θ=|coshθ|=coshθ|coshθ|dθ|coshθ|=coshθ since coshθ>0 for all θ.=coshθcoshθdθSimplify.=1dθEvaluate the integral.=θ+CSince x=sinhθ, we know θ=sinh1x.=sinh1x+C.

Analysis

This answer looks quite different from the answer obtained using the substitution x=tanθ. To see that the solutions are the same, set y=sinh1x. Thus, sinhy=x. From this equation we obtain:

eyey2=x.

After multiplying both sides by 2ey and rewriting, this equation becomes:

e2y2xey1=0.

Use the quadratic equation to solve for ey:

ey=2x±4x2+42.

Simplifying, we have:

ey=x±x2+1.

Since xx2+1<0, it must be the case that ey=x+x2+1. Thus,

y=ln(x+x2+1).

Last, we obtain

sinh1x=ln(x+x2+1).

After we make the final observation that, since x+x2+1>0,

ln(x+x2+1)=ln1+x2+x,

we see that the two different methods produced equivalent solutions.

Example 7.3.6: Finding an Arc Length

Find the length of the curve y=x2 over the interval [0,12].

Solution

Because dydx=2x, the arc length is given by

1/201+(2x)2dx=1/201+4x2dx.

To evaluate this integral, use the substitution x=12tanθ and dx=12sec2θdθ. We also need to change the limits of integration. If x=0, then θ=0 and if x=12, then θ=π4. Thus,

1/201+4x2dx=π/401+tan2θ12sec2θdθ After substitution,1+4x2=secθ. (Substitute 1+tan2θ=sec2θ and simplify.)

=12π/40sec3θdθ We derived this integral in the previous section.

=12(12secθtanθ+12ln|secθ+tanθ|)π/40 Evaluate and simplify.

=14(2+ln(2+1)).

Exercise 7.3.2

Rewrite x3x2+4dx by using a substitution involving tanθ.

Hint

Use x=2tanθ and dx=2sec2θdθ.

Answer

32tan3θsec3θdθ

Integrating Expressions Involving x2a2

The domain of the expression x2a2 is (,a][a,+). Thus, either xa or xa. Hence, xa1 or xa1. Since these intervals correspond to the range of secθ on the set [0,π2)(π2,π], it makes sense to use the substitution secθ=xa or, equivalently, x=asecθ, where 0θ<π2 or π2<θπ. The corresponding substitution for dx is dx=asecθtanθdθ. The procedure for using this substitution is outlined in the following problem-solving strategy.

Problem-Solving Strategy: Integrals Involving x2a2
  1. Check to see whether the integral cannot be evaluated using another method. If so, we may wish to consider applying an alternative technique.
  2. Substitute x=asecθ and dx=asecθtanθdθ. This substitution yields x2a2=(asecθ)2a2=a2(sec2θ1)=a2tan2θ=|atanθ|. For xa,|atanθ|=atanθ and for xa,|atanθ|=atanθ.
  3. Simplify the expression.
  4. Evaluate the integral using techniques from the section on trigonometric integrals.
  5. Use the reference triangles from Figure 7.3.6 to rewrite the result in terms of x.
  6. You may also need to use some trigonometric identities and the relationship θ=sec1(xa). (Note: We need both reference triangles, since the values of some of the trigonometric ratios are different depending on whether x>a or x<a.)
This figure has two right triangles. The first triangle is in the first quadrant of the xy coordinate system and has an angle labeled theta. This angle is opposite the vertical side. The hypotenuse is labeled x, the vertical leg is labeled the square root of (x^2-a^2), and the horizontal leg is labeled a. The horizontal leg is on the x-axis. To the left of the triangle is the equation sec(theta) = x/a, x>a. There are also the equations sin(theta)= the square root of (x^2-a^2)/x, cos(theta) = a/x, and tan(theta) = the square root of (x^2-a^2)/a. The second triangle is in the second quadrant, with the hypotenuse labeled –x. The horizontal leg is labeled –a and is on the negative x-axis. The vertical leg is labeled the square root of (x^2-a^2). To the right of the triangle is the equation sec(theta) = x/a, x<-a. There are also the equations sin(theta)= the negative square root of (x^2-a^2)/x, cos(theta) = a/x, and tan(theta) = the negative square root of (x^2-a^2)/a.
Figure 7.3.6: Use the appropriate reference triangle to express the trigonometric functions evaluated at θ in terms of x.
Example 7.3.7: Finding the Area of a Region

Find the area of the region between the graph of f(x)=x29 and the x-axis over the interval [3,5].

Solution

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

This figure is the graph of the function f(x) = the square root of (x^2-9). It is an increasing curve that starts on the x-axis at 3 and is in the first quadrant. Under the curve above the x-axis is a shaded region bounded to the right at x = 5.
Figure 7.3.7: Calculating the area of the shaded region requires evaluating an integral with a trigonometric substitution.

We can see that the area is A=53x29dx. To evaluate this definite integral, substitute x=3secθ and dx=3secθtanθdθ. We must also change the limits of integration. If x=3, then 3=3secθ and hence θ=0. If x=5, then θ=sec1(53). After making these substitutions and simplifying, we have

Area=53x29dx

=sec1(5/3)09tan2θsecθdθ Use tan2θ=sec2θ1.

=sec1(5/3)09(sec2θ1)secθdθ Expand.

=sec1(5/3)09(sec3θsecθ)dθ Evaluate the integral.

=(92ln|secθ+tanθ|+92secθtanθ)9ln|secθ+tanθ|sec1(5/3)0 Simplify.

=92secθtanθ92ln|secθ+tanθ|sec1(5/3)0 Evaluate. Use sec(sec153)=53 and tan(sec153)=43.

=92534392ln53+43(921092ln|1+0|)

=1092ln3

Exercise 7.3.3

Evaluate dxx24.Assume that x>2.

Hint

Substitute x=2secθ and dx=2secθtanθdθ.

Answer

ln|x2+x242|+C

Key Concepts

  • For integrals involving a2x2, use the substitution x=asinθ and dx=acosθdθ.
  • For integrals involving a2+x2, use the substitution x=atanθ and dx=asec2θdθ.
  • For integrals involving x2a2, substitute x=asecθ and dx=asecθtanθdθ.

Glossary

trigonometric substitution
an integration technique that converts an algebraic integral containing expressions of the form a2x2, a2+x2, or x2a2 into a trigonometric integral

This page titled 7.3: Trigonometric Substitution is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Gilbert Strang & Edwin “Jed” Herman (OpenStax) via source content that was edited to the style and standards of the LibreTexts platform.

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