7.3: Trigonometric Substitution
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- 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 √a2−x2, √a2+x2, and √x2−a2, 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 √a2−x2
Before developing a general strategy for integrals containing √a2−x2, consider the integral ∫√9−x2dx. 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
∫√9−x2dx=∫√9−(3sinθ)2⋅3cosθdθ.
After simplifying, we have
∫√9−x2dx=∫9√1−sin2θ⋅cosθdθ.
Letting 1−sin2θ=cos2θ, we now have
∫√9−x2dx=∫9√cos2θcosθdθ.
Assuming that cosθ≥0, we have
∫√9−x2dx=∫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 √a2−x2, we make the substitution x=asinθ and dx=acosθ. To see that this actually makes sense, consider the following argument: The domain of √a2−x2 is [−a,a]. Thus,
−a≤x≤a.
Consequently,
−1≤xa≤1.
Since the range of sinx over [−(π/2),π/2] is [−1,1], there is a unique angle θ satisfying −(π/2)≤θ≤π/2 so that sinθ=x/a, or equivalently, so that x=asinθ. If we substitute x=asinθ into √a2−x2, we get
√a2−x2=√a2−(asinθ)2Let x=asinθ where −π2≤θ≤π2.Simplify.=√a2−a2sin2θFactor out a2.=√a2(1−sin2θ)Substitute 1−sin2x=cos2x.=√a2cos2θTake the square root.=|acosθ|=acosθ
Since cosx≥0 on −π2≤θ≤π2 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 −π2≤θ≤π2. It is useful to observe that the expression √a2−x2 actually appears as the length of one side of the triangle. Last, should θ appear by itself, we use θ=sin−1(xa).

The essential part of this discussion is summarized in the following problem-solving strategy.
- 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 ∫1√a2−x2dx, ∫x√a2−x2dx, and ∫x√a2−x2dx, they can each be integrated directly either by formula or by a simple u-substitution.
- Make the substitution x=asinθ and dx=acosθdθ. Note: This substitution yields √a2−x2=acosθ.
- Simplify the expression.
- Evaluate the integral using techniques from the section on trigonometric integrals.
- 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=sin−1(xa).
The following example demonstrates the application of this problem-solving strategy.
Evaluate
∫√9−x2dx.
Solution
Begin by making the substitutions x=3sinθ and dx=3cosθdθ. Since sinθ=x3, we can construct the reference triangle shown in Figure 2.

Thus,
∫√9−x2dx=∫√9−(3sinθ)23cosθdθ Substitute x=3sinθ and dx=3cosθdθ.
=∫√9(1−sin2θ)⋅3cosθdθ Simplify.
=∫√9cos2θ⋅3cosθdθ Substitute cos2θ=1−sin2θ.
=∫3|cosθ|3cosθdθ Take the square root.
=∫9cos2θdθ Simplify. Since −π2≤θ≤π2,cosθ≥0 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.
=92sin−1(x3)+92⋅x3⋅√9−x23+C Substitute sin−1(x3)=θ and sinθ=x3.
Use the reference triangle to see that cosθ=√9−x23and make this substitution. Simplify.
=92sin−1(x3)+x√9−x22+C. Simplify.
Evaluate
∫√4−x2xdx.
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.

Thus,
∫√4−x2xdx=∫√4−(2sinθ)22sinθ2cosθdθ Substitute x=2sinθ and dx=2cosθdθ.
=∫2cos2θsinθdθ Substitute cos2θ=1−sin2θ and simplify.
=∫2(1−sin2θ)sinθdθ Substitute cos2θ=1−sin2θ.
=∫(2cscθ−2sinθ)dθ Separate the numerator, simplify, and use cscθ=1sinθ.
=2ln|cscθ−cotθ|+2cosθ+C Evaluate the integral.
=2ln|2x−√4−x2x|+√4−x2+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.
Evaluate ∫x3√1−x2dx two ways: first by using the substitution u=1−x2 and then by using a trigonometric substitution.
Method 1
Let u=1−x2 and hence x2=1−u. Thus, du=−2xdx. In this case, the integral becomes
∫x3√1−x2dx=−12∫x2√1−x2(−2xdx) Make the substitution.
=−12∫(1−u)√udu Expand the expression.
=−12∫(u1/2−u3/2)du Evaluate the integral.
=−12(23u3/2−25u5/2)+C Rewrite in terms of x.
=−13(1−x2)3/2+15(1−x2)5/2+C.
Method 2
Let x=sinθ. In this case, dx=cosθdθ. Using this substitution, we have
∫x3√1−x2dx=∫sin3θcos2θdθ
=∫(1−cos2θ)cos2θsinθdθ Let u=cosθ. Thus,du=−sinθdθ.
=∫(u4−u2)du
=15u5−13u3+C Substitute cosθ=u.
=15cos5θ−13cos3θ+C Use a reference triangle to see that cosθ=√1−x2.
=15(1−x2)5/2−13(1−x2)3/2+C.
Rewrite the integral ∫x3√25−x2dx 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=tan−1(x/a). The procedure for using this substitution is outlined in the following problem-solving strategy.
- 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.
- 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.)
- Simplify the expression.
- Evaluate the integral using techniques from the section on trigonometric integrals.
- 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=tan−1(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 x≤0.)

Evaluate ∫dx√1+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.

Thus,
∫dx√1+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|)=1√1+x2+x⋅(x√1+x2+1)=1√1+x2+x⋅x+√1+x2√1+x2=1√1+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.
Use the substitution x=sinhθ to evaluate ∫dx√1+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,
∫dx√1+x2=∫coshθ√1+sinh2θdθSubstitute x=sinhθ and dx=coshθdθ.Substitute 1+sinh2θ=cosh2θ.=∫coshθ√cosh2θdθSince √cosh2θ=|coshθ|=∫coshθ|coshθ|dθ|coshθ|=coshθ since coshθ>0 for all θ.=∫coshθcoshθdθSimplify.=∫1dθEvaluate the integral.=θ+CSince x=sinhθ, we know θ=sinh−1x.=sinh−1x+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=sinh−1x. Thus, sinhy=x. From this equation we obtain:
ey−e−y2=x.
After multiplying both sides by 2ey and rewriting, this equation becomes:
e2y−2xey−1=0.
Use the quadratic equation to solve for ey:
ey=2x±√4x2+42.
Simplifying, we have:
ey=x±√x2+1.
Since x−√x2+1<0, it must be the case that ey=x+√x2+1. Thus,
y=ln(x+√x2+1).
Last, we obtain
sinh−1x=ln(x+√x2+1).
After we make the final observation that, since x+√x2+1>0,
ln(x+√x2+1)=ln∣√1+x2+x∣,
we see that the two different methods produced equivalent solutions.
Find the length of the curve y=x2 over the interval [0,12].
Solution
Because dydx=2x, the arc length is given by
∫1/20√1+(2x)2dx=∫1/20√1+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,
∫1/20√1+4x2dx=∫π/40√1+tan2θ⋅12sec2θdθ After substitution,√1+4x2=secθ. (Substitute 1+tan2θ=sec2θ and simplify.)
=12∫π/40sec3θdθ We derived this integral in the previous section.
=12(12secθtanθ+12ln|secθ+tanθ|)∣π/40 Evaluate and simplify.
=14(√2+ln(√2+1)).
Rewrite ∫x3√x2+4dx by using a substitution involving tanθ.
- Hint
-
Use x=2tanθ and dx=2sec2θdθ.
- Answer
-
∫32tan3θsec3θdθ
Integrating Expressions Involving √x2−a2
The domain of the expression √x2−a2 is (−∞,−a]∪[a,+∞). Thus, either x≤−a or x≥a. Hence, xa≤−1 or xa≥1. 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 0≤θ<π2 or π2<θ≤π. The corresponding substitution for dx is dx=asecθtanθdθ. The procedure for using this substitution is outlined in the following problem-solving strategy.
- Check to see whether the integral cannot be evaluated using another method. If so, we may wish to consider applying an alternative technique.
- Substitute x=asecθ and dx=asecθtanθdθ. This substitution yields √x2−a2=√(asecθ)2−a2=√a2(sec2θ−1)=√a2tan2θ=|atanθ|. For x≥a,|atanθ|=atanθ and for x≤−a,|atanθ|=−atanθ.
- Simplify the expression.
- Evaluate the integral using techniques from the section on trigonometric integrals.
- Use the reference triangles from Figure 7.3.6 to rewrite the result in terms of x.
- You may also need to use some trigonometric identities and the relationship θ=sec−1(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.)

Find the area of the region between the graph of f(x)=√x2−9 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.

We can see that the area is A=∫53√x2−9dx. 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=sec−1(53). After making these substitutions and simplifying, we have
Area=∫53√x2−9dx
=∫sec−1(5/3)09tan2θsecθdθ Use tan2θ=sec2θ−1.
=∫sec−1(5/3)09(sec2θ−1)secθdθ Expand.
=∫sec−1(5/3)09(sec3θ−secθ)dθ Evaluate the integral.
=(92ln|secθ+tanθ|+92secθtanθ)−9ln|secθ+tanθ|∣sec−1(5/3)0 Simplify.
=92secθtanθ−92ln|secθ+tanθ|∣sec−1(5/3)0 Evaluate. Use sec(sec−153)=53 and tan(sec−153)=43.
=92⋅53⋅43−92ln∣53+43∣−(92⋅1⋅0−92ln|1+0|)
=10−92ln3
Evaluate ∫dx√x2−4.Assume that x>2.
- Hint
-
Substitute x=2secθ and dx=2secθtanθdθ.
- Answer
-
ln|x2+√x2−42|+C
Key Concepts
- For integrals involving √a2−x2, use the substitution x=asinθ and dx=acosθdθ.
- For integrals involving √a2+x2, use the substitution x=atanθ and dx=asec2θdθ.
- For integrals involving √x2−a2, substitute x=asecθ and dx=asecθtanθdθ.
Glossary
- trigonometric substitution
- an integration technique that converts an algebraic integral containing expressions of the form √a2−x2, √a2+x2, or √x2−a2 into a trigonometric integral