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APEX Calculus

Section 6.4 Trigonometric Substitution

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 only evaluated special definite integrals which described nice, geometric shapes. For instance, we were able to evaluate
(6.4.1)339x2dx=9π2
as we recognized that f(x)=9x2 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).
Figure 6.4.1. Video introduction to Section 6.4
We start by demonstrating this method in evaluating the integral in Equation (6.4.1). After the example, we will generalize the method and give more examples.

Example 6.4.2. Using Trigonometric Substitution.

Evaluate 339x2dx.
Solution 1.
We begin by noting that 9(sin2(θ)+cos2(θ))=9, and hence 9cos2(θ)=99sin2(θ). If we let x=3sin(θ), then 9x2=99sin2(θ)=9cos2(θ).
Setting x=3sin(θ) gives dx=3cos(θ)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=3sin(θ)=3). Likewise, the bound of x=3 is replaced by the bound θ=π/2. Thus
339x2dx=π/2π/299sin2(θ)(3cos(θ))dθ=π/2π/239cos2(θ)cos(θ)dθ=π/2π/23|3cos(θ)|cos(θ)dθ.
On [π/2,π/2], cos(θ) is always positive, so we can drop the absolute value bars, then employ a power-reducing formula:
339x2dx=π/2π/29cos2(θ)dθ=π/2π/292(1+cos(2θ))dθ=92(θ+12sin(2θ))|π/2π/2=92π.
This matches our answer from before.
Solution 2. Video solution
We now describe in detail Trigonometric Substitution. This method excels when dealing with integrands that contain a2x2, x2a2 and x2+a2. The following Key Idea 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 6.4.3. Trigonometric Substitution.

  1. Integrands containing a2x2.
    Let x=asin(θ), dx=acos(θ)dθ. Thus θ=sin1(x/a), for π/2θπ/2. On this interval, cos(θ)0, so a2x2=acos(θ).
    Diagram showing trigonometric substitution with integrands containing square root of a^2-x^2.
    The diagram is of a right angled triangle. The perpendicular is marked x the base as a2x2. The hypotenuse is marked a. The angle opposite to the perpendicular is marked as θ.
    Figure 6.4.4.
  2. Integrands containing x2+a2.
    Let x=atan(θ), dx=asec2(θ)dθ. Thus θ=tan1(x/a), for π/2<θ<π/2. On this interval, sec(θ)>0, so x2+a2=asec(θ).
    Diagram showing trigonometric substitution with integrands containing square root of x^2+a^2.
    The diagram is of a right angled triangle. The perpendicular is marked x the base as a. The hypotenuse is marked x2+a2. The angle opposite to the perpendicular is marked as θ.
    Figure 6.4.5.
  3. Integrands containing x2a2.
    Let x=asec(θ), dx=asec(θ)tan(θ)dθ. Thus θ=sec1(x/a). If x/a1, then 0θ<π/2; if x/a1, then π/2<θπ. We restrict our work to where xa, so x/a1, and 0θ<π/2. On this interval, tan(θ)0, so x2a2=atan(θ).
    Diagram showing trigonometric substitution with integrands containing square root of x^2-a^2.
    The diagram is of a right angled triangle. The perpendicular is marked x2a2 the base as a. The hypotenuse is marked x. The angle opposite to the perpendicular is marked as θ.
    Figure 6.4.6.

Example 6.4.7. Using Trigonometric Substitution.

Evaluate 15+x2dx.
Solution 1.
Using Item 2 in Key Idea 6.4.3, we recognize a=5 and set x=5tan(θ). This makes dx=5sec2(θ)dθ. We will use the fact that 5+x2=5+5tan2(θ)=5sec2(θ)=5sec(θ). Substituting, we have:
15+x2dx=15+5tan2(θ)5sec2(θ)dθ=5sec2(θ)5sec(θ)dθ=sec(θ)dθ=ln|sec(θ)+tan(θ)|+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 Figure 6.4.5 helps. With x=5tan(θ), we have
tan(θ)=x5 and sec(θ)=x2+55.
This gives
15+x2dx=ln|sec(θ)+tan(θ)|+C=ln|x2+55+x5|+C.
We can leave this answer as is, or we can use a logarithmic identity to simplify it. Note:
ln|x2+55+x5|+C=ln|15(x2+5+x)|+C=ln|15|+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.)
Solution 2. Video solution

Example 6.4.8. Using Trigonometric Substitution.

Evaluate 4x21dx.
Solution 1.
We start by rewriting the integrand so that it looks like x2a2 for some value of a:
4x21=4(x214)=2x2(12)2.
So we have a=1/2, and following Part 3 of Key Idea 6.4.3, we set x=12sec(θ), and hence dx=12sec(θ)tan(θ)dθ. We now rewrite the integral with these substitutions:
4x21dx=2x2(12)2dx=214sec2(θ)14(12sec(θ)tan(θ))dθ=14(sec2(θ)1)(sec(θ)tan(θ))dθ=14tan2(θ)(sec(θ)tan(θ))dθ=12tan2(θ)sec(θ)dθ=12(sec2(θ)1)sec(θ)dθ=12(sec3(θ)sec(θ))dθ.
We integrated sec3(θ) in Example 6.3.11, finding its antiderivatives to be
sec3(θ)dθ=12(sec(θ)tan(θ)+ln|sec(θ)+tan(θ)|)+C.
Thus
4x21dx=12(sec3(θ)sec(θ))dθ=12(12(sec(θ)tan(θ)+ln|sec(θ)+tan(θ)|)ln|sec(θ)+tan(θ)|)+C=14(sec(θ)tan(θ)ln|sec(θ)+tan(θ)|)+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=12sec(θ), the reference triangle in Figure 6.4.6 shows that
tan(θ)=x21/4/(1/2)=2x21/4 and sec(θ)=2x.
Thus
14(sec(θ)tan(θ)ln|sec(θ)+tan(θ)|)+C=14(2x2x21/4ln|2x+2x21/4|)+C=14(4xx21/4ln|2x+2x21/4|)+C.
The final answer is given in the last line above, repeated here:
4x21dx=14(4xx21/4ln|2x+2x21/4|)+C.
Solution 2. Video solution

Example 6.4.9. Using Trigonometric Substitution.

Evaluate 4x2x2dx.
Solution 1.
We use Part 1 of Key Idea 6.4.3 with a=2, x=2sin(θ), dx=2cos(θ) and hence 4x2=2cos(θ). This gives
4x2x2dx=2cos(θ)4sin2(θ)(2cos(θ))dθ=cot2(θ)dθ=(csc2(θ)1)dθ=cot(θ)θ+C.
We need to rewrite our answer in terms of x. Using the reference triangle found in Figure 6.4.4, we have cot(θ)=4x2/x and θ=sin1(x/2). Thus
4x2x2dx=4x2xsin1(x2)+C.
Solution 2. Video solution
Trigonometric Substitution can be applied in many situations, even those not of the form a2x2, x2a2 or x2+a2. In the following example, we apply it to an integral we already know how to handle.

Example 6.4.10. Using Trigonometric Substitution.

Evaluate 1x2+1dx.
Solution.
We know the answer already as tan1(x)+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 Part 2 of Key Idea 6.4.3, let x=tan(θ), dx=sec2(θ)dθ and note that x2+1=tan2(θ)+1=sec2(θ). Thus
1x2+1dx=1sec2(θ)sec2(θ)dθ=1dθ=θ+C.
Since x=tan(θ), θ=tan1(x), and we conclude that 1x2+1dx=tan1(x)+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 6.4.11. Using Trigonometric Substitution.

Evaluate 1(x2+6x+10)2dx.
Solution 1.
We start by completing the square, then make the substitution u=x+3, followed by the trigonometric substitution of u=tan(θ):
1(x2+6x+10)2dx=1((x+3)2+1)2dx=1(u2+1)2du.
Now make the substitution u=tan(θ), du=sec2(θ)dθ:
=1(tan2(θ)+1)2sec2(θ)dθ=1(sec2(θ))2sec2(θ)dθ=cos2(θ)dθ.
Applying a power reducing formula, we have
=(12+12cos(2θ))dθ(6.4.2)=12θ+14sin(2θ)+C.
We need to return to the variable x. As u=tan(θ), θ=tan1(u). Using the identity sin(2θ)=2sin(θ)cos(θ) and using the reference triangle found in Figure 6.4.5, we have
14sin(2θ)=12uu2+11u2+1=12uu2+1.
Finally, we return to x with the substitution u=x+3. We start with the expression in Equation (6.4.2):
12θ+14sin(2θ)+C=12tan1(u)+12uu2+1+C=12tan1(x+3)+x+32(x2+6x+10)+C.
Stating our final result in one line,
1(x2+6x+10)2dx=12tan1(x+3)+x+32(x2+6x+10)+C.
Solution 2. Video solution
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 6.4.12. Definite integration and Trigonometric Substitution.

Evaluate 05x2x2+25dx.
Solution 1.
Using Part 2 of Key Idea 6.4.3, we set x=5tan(θ), dx=5sec2(θ)dθ, and note that x2+25=5sec(θ). As we substitute, we can also change the bounds of integration.
The lower bound of the original integral is x=0. As x=5tan(θ), we solve for θ and find θ=tan1(x/5). Thus the new lower bound is θ=tan1(0)=0. The original upper bound is x=5, thus the new upper bound is θ=tan1(5/5)=π/4.
Thus we have
05x2x2+25dx=0π/425tan2(θ)5sec(θ)5sec2(θ)dθ=250π/4tan2(θ)sec(θ)dθ.
We encountered this indefinite integral in Example 6.4.8 where we found
tan2(θ)sec(θ)dθ=12(sec(θ)tan(θ)ln|sec(θ)+tan(θ)|).
So
250π/4tan2(θ)sec(θ)dθ=252(sec(θ)tan(θ)ln|sec(θ)+tan(θ)|)|0π/4=252(2ln(2+1))6.661.
Solution 2. Video solution
The following equalities are very useful when evaluating integrals using Trigonometric Substitution.

Key Idea 6.4.13. Useful Equalities with Trigonometric Substitution.

  1. sin(2θ)=2sin(θ)cos(θ)
  2. cos(2θ)=cos2(θ)sin2(θ)=2cos2(θ)1=12sin2(θ)
  3. sec3(θ)dθ=12(sec(θ)tan(θ)+ln|sec(θ)+tan(θ)|)+C
  4. cos2(θ)dθ=12(1+cos(2θ))dθ=12(θ+sin(θ)cos(θ))+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.

Exercises Exercises

Terms and Concepts

1.
2.
If one uses Trigonometric Substitution on an integrand containing 9x2, then one should set x=.
3.
Consider the Pythagorean Identity sin2(θ)+cos2(θ)=1.
  1. What identity is obtained when both sides are divided by cos2(θ)?
  2. Use the new identity to simplify 9tan2(θ)+9.

Problems

Exercise Group.
Apply Trigonometric Substitution to evaluate the indefinite integral.
Exercise Group.
Evaluate the indefinite integral. Trigonometric Substitution may not be required.
24.
x2(1x2)3/2dx
Exercise Group.
Evaluate the definite integral by making the proper trigonometric substitution and changing the bounds of integration. (Note: the corresponding indefinite integrals appeared previously in the Section 6.4 exercises.)
30.
331(x2+1)2dx
32.
11x21x2dx
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