6.4: Trigonometric Substitution
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In Section 5.2 we defined the definite integral as the signed area under the curve." In that section we had not yet learned the Fundamental Theorem of Calculus, so we evaluated special definite integrals which described nice, geometric shapes. For instance, we were able to evaluate
∫3−3√9−x2 dx=9π2
as we recognized that f(x)=√9−x2 described the upper half of a circle with radius 3.
We have since learned a number of integration techniques, including Substitution and Integration by Parts, yet we are still unable to evaluate the above integral without resorting to a geometric interpretation. This section introduces Trigonometric Substitution, a method of integration that fills this gap in our integration skill. This technique works on the same principle as Substitution as found in Section 6.1, though it can feel "backward." In Section 6.1, we set u=f(x), for some function f, and replaced f(x) with u. In this section, we will set x=f(θ), where f is a trigonometric function, then replace x with f(θ).
We start by demonstrating this method in evaluating the integral in 6.4.1. After the example, we will generalize the method and give more examples.
Example 6.4.1: Using Trigonometric Substitution
Evaluate ∫3−3√9−x2 dx.
Solution
We begin by noting that 9sin2θ+9cos2θ=9, and hence 9cos2θ=9−9sin2θ. If we let x=3sinθ, then 9−x2=9−9sin2θ=9cos2θ.
Setting x=3sinθ gives dx=3cosθ dθ. We are almost ready to substitute. We also wish to change our bounds of integration. The bound x=−3 corresponds to θ=−π/2 (for when θ=−π/2, x=3sinθ=−3). Likewise, the bound of x=3 is replaced by the bound θ=π/2. Thus
∫3−3√9−x2 dx=∫π/2−π/2√9−9sin2θ(3cosθ) dθ=∫π/2−π/23√9cos2θcosθ dθ=∫π/2−π/23|3cosθ|cosθ dθ.
On [−π/2,π/2], cosθ is always positive, so we can drop the absolute value bars, then employ a power--reducing formula:
=∫π/2−π/29cos2θ dθ=∫π/2−π/292(1+cos(2θ)) dθ=92(θ+12sin(2θ))|π/2−π/2=92π.
This matches our answer from before.
We now describe in detail Trigonometric Substitution. This method excels when dealing with integrands that contain √a2−x2, √x2−a2 and √x2+a2. The following Key Idea 13 outlines the procedure for each case, followed by more examples. Each right triangle acts as a reference to help us understand the relationships between x and θ.
Key Idea 13: Trigonometric Substitution
- For integrands containing √a2−x2:
Let x=asinθ, dx=acosθ dθ
Thus θ=sin−1(x/a), for −π/2≤θ≤π/2.
On this interval, cosθ≥0, so
√a2−x2=acosθ. - For integrands containing √x2+a2:
Let x=atanθ, dx=asec2θ dθ
Thus θ=tan−1(x/a), for −π/2<θ<π/2.
On this interval, secθ>0, so
√x2+a2=asecθ. - For integrands containing √x2−a2:
Let x=asecθ, dx=asecθtanθ dθ
Thus θ=sec−1(x/a). If x/a≥1, then 0≤θ<π/2; if x/a≤−1, then π/2<θ≤π.
We restrict our work to where x≥a, so x/a≥1, and 0≤θ<π/2.
On this interval, tanθ≥0, so
√x2−a2=atanθ.
Example 6.4.2: Using Trigonometric Substitution
Evaluate ∫1√5+x2 dx.
Solution
Using Key Idea 13(b), we recognize a=√5 and set x=√5tanθ. This makes dx=√5sec2θ dθ. We will use the fact that √5+x2=√5+5tan2θ=√5sec2θ=√5secθ. Substituting, we have:
∫1√5+x2 dx=∫1√5+5tan2θ√5sec2θ dθ=∫√5sec2θ√5secθ dθ=∫secθ dθ=ln|secθ+tanθ|+C.
While the integration steps are over, we are not yet done. The original problem was stated in terms of x, whereas our answer is given in terms of θ. We must convert back to x.
The reference triangle given in Key Idea 13(b) helps. With x=√5tanθ, we have
tanθ=x√5andsecθ=√x2+5√5.
This gives
∫1√5+x2 dx=ln|secθ+tanθ|+C=ln|√x2+5√5+x√5|+C.
We can leave this answer as is, or we can use a logarithmic identity to simplify it. Note:
ln|√x2+5√5+x√5|+C=ln|1√5(√x2+5+x)|+C=ln|1√5|+ln|√x2+5+x|+C=ln|√x2+5+x|+C,
where the ln(1/√5) term is absorbed into the constant C. (In Section 6.6 we will learn another way of approaching this problem.)
Example 6.4.3: Using Trigonometric Substitution
Evaluate ∫√4x2−1 dx.
Solution
We start by rewriting the integrand so that it looks like √x2−a2 for some value of a:
√4x2−1=√4(x2−14)=2√x2−(12)2
So we have a=1/2, and following Key Idea 13(c), we set x=12secθ, and hence dx=12secθtanθ dθ.
We now rewrite the integral with these substitutions:
∫√4x2−1 dx=∫2√x2−(12)2 dx=∫2√14sec2θ−14(12secθtanθ) dθ=∫√14(sec2θ−1)(secθtanθ) dθ=∫√14tan2θ(secθtanθ) dθ=∫12tan2θsecθ dθ=12∫(sec2θ−1)secθ dθ=12∫(sec3θ−secθ) dθ.
We integrated sec3θ in Example 6.3.6, finding its antiderivatives to be
∫sec3θ dθ=12(secθtanθ+ln|secθ+tanθ|)+C.
Thus
∫√4x2−1 dx=12∫(sec3θ−secθ) dθ=12(12(secθtanθ+ln|secθ+tanθ|)−ln|secθ+tanθ|)+C=14(secθtanθ−ln|secθ+tanθ|)+C.
We are not yet done. Our original integral is given in terms of x, whereas our final answer, as given, is in terms of θ. We need to rewrite our answer in terms of x. With a=1/2, and x=12secθ, the reference triangle in Key Idea 13(c) shows that
tanθ=√x2−1/4/(1/2)=2√x2−1/4andsecθ=2x.
Thus
14(secθtanθ−ln|secθ+tanθ|)+C=14(2x⋅2√x2−1/4−ln|2x+2√x2−1/4|)+C=14(4x√x2−1/4−ln|2x+2√x2−1/4|)+C.
The final answer is given in the last line above, repeated here:
∫√4x2−1 dx=14(4x√x2−1/4−ln|2x+2√x2−1/4|)+C.
Example 6.4.4: Using Trigonometric Substitution
Evaluate ∫√4−x2x2 dx.
Solution
We use Key Idea 13(a) with a=2, x=2sinθ, dx=2cosθ and hence √4−x2=2cosθ. This gives
∫√4−x2x2 dx=∫2cosθ4sin2θ(2cosθ) dθ=∫cot2θ dθ=∫(csc2θ−1) dθ=−cotθ−θ+C.
We need to rewrite our answer in terms of x. Using the reference triangle found in Key Idea 13(a), we have cotθ=√4−x2/x and θ=sin−1(x/2). Thus
∫√4−x2x2 dx=−√4−x2x−sin−1(x2)+C.
Trigonometric Substitution can be applied in many situations, even those not of the form √a2−x2, √x2−a2 or √x2+a2. In the following example, we apply it to an integral we already know how to handle.
Example 6.4.5: Using Trigonometric Substitution
Evaluate ∫1x2+1 dx.
Solution
We know the answer already as tan−1x+C. We apply Trigonometric Substitution here to show that we get the same answer without inherently relying on knowledge of the derivative of the arctangent function.
Using Key Idea 13(b), let x=tanθ, dx=sec2θ dθ and note that x2+1=tan2θ+1=sec2θ. Thus
∫1x2+1 dx=∫1sec2θsec2θ dθ=∫1 dθ=θ+C.
Since x=tanθ, θ=tan−1x, and we conclude that ∫1x2+1 dx=tan−1x+C.
The next example is similar to the previous one in that it does not involve a square--root. It shows how several techniques and identities can be combined to obtain a solution.
Example 6.4.6: Using Trigonometric Substitution
Evaluate ∫1(x2+6x+10)2 dx.
Solution
We start by completing the square, then make the substitution u=x+3, followed by the trigonometric substitution of u=tanθ:
∫1(x2+6x+10)2 dx=∫1((x+3)2+1)2 dx=∫1(u2+1)2 du.
Now make the substitution u=tanθ, du=sec2θ dθ:
=∫1(tan2θ+1)2sec2θ dθ=∫1(sec2θ)2sec2θ dθ=∫cos2θ dθ.
Applying a power reducing formula, we have
=∫(12+12cos(2θ)) dθ=12θ+14sin(2θ)+C.
We need to return to the variable x. As u=tanθ, θ=tan−1u. Using the identity sin(2θ)=2sinθcosθ and using the reference triangle found in Key Idea 13(b), we have
14sin(2θ)=12u√u2+1⋅1√u2+1=12uu2+1.
Finally, we return to x with the substitution u=x+3. We start with the expression in Equation (???):
12θ+14sin(2θ)+C=12tan−1u+12uu2+1+C=12tan−1(x+3)+x+32(x2+6x+10)+C.
Stating our final result in one line,
∫1(x2+6x+10)2 dx=12tan−1(x+3)+x+32(x2+6x+10)+C.
Our last example returns us to definite integrals, as seen in our first example. Given a definite integral that can be evaluated using Trigonometric Substitution, we could first evaluate the corresponding indefinite integral (by changing from an integral in terms of x to one in terms of θ, then converting back to x) and then evaluate using the original bounds. It is much more straightforward, though, to change the bounds as we substitute.
Example 6.4.7: Definite integration and Trigonometric Substitution
Evaluate ∫50x2√x2+25 dx.
Solution
Using Key Idea 13(b), we set x=5tanθ, dx=5sec2θ dθ, and note that √x2+25=5secθ. As we substitute, we can also change the bounds of integration.
The lower bound of the original integral is x=0. As x=5tanθ, we solve for θ and find θ=tan−1(x/5). Thus the new lower bound is θ=tan−1(0)=0. The original upper bound is x=5, thus the new upper bound is θ=tan−1(5/5)=π/4.
Thus we have
∫50x2√x2+25 dx=∫π/4025tan2θ5secθ5sec2θ dθ=25∫π/40tan2θsecθ dθ.
We encountered this indefinite integral in Example 6.4.3 where we found
∫tan2θsecθ dθ=12(secθtanθ−ln|secθ+tanθ|).
So
25∫π/40tan2θsecθ dθ=252(secθtanθ−ln|secθ+tanθ|)|π/40=252(√2−ln(√2+1))≈6.661.
The following equalities are very useful when evaluating integrals using Trigonometric Substitution.
Ket Idea 14: Useful Equalities with Trigonometric Substitution
- sin(2θ)=2sinθcosθ
- cos(2θ)=cos2θ−sin2θ=2cos2θ−1=1−2sin2θ
- ∫sec3θ dθ=12(secθtanθ+ln|secθ+tanθ|)+C
- ∫cos2θ dθ=∫12(1+cos(2θ)) dθ=12(θ+sinθcosθ)+C.
The next section introduces Partial Fraction Decomposition, which is an algebraic technique that turns "complicated" fractions into sums of "simpler" fractions, making integration easier.
Contributors and Attributions
Gregory Hartman (Virginia Military Institute). Contributions were made by Troy Siemers and Dimplekumar Chalishajar of VMI and Brian Heinold of Mount Saint Mary's University. This content is copyrighted by a Creative Commons Attribution - Noncommercial (BY-NC) License. http://www.apexcalculus.com/
Integrated by Justin Marshall.