This is the last week of new integration techniques. Together with substitution (Week 9), integration by parts, and trigonometric integrals (Week 12), trigonometric substitution and partial fractions round out the standard toolkit for integrating elementary functions by hand. The week closes by asking what happens when the definite integral's own requirements — a finite interval, a bounded integrand — are relaxed.
13.1 Why Trigonometric Substitution
Expressions like √(a2−x2), √(a2+x2), or √(x2−a2) resist ordinary substitution (Week 9 §9.6) because no algebraic u simplifies the square root. Replacing x with a trigonometric function of a new variable θ makes the Pythagorean identity (Week 1, Eq. 1.8) collapse the square root into a single trig function.
13.2 The Three Standard Substitutions
| Expression | Substitution | Identity used |
|---|---|---|
| √(a2−x2) | x=a sin θ | 1−sin2θ=cos2θ |
| √(a2+x2) | x=a tan θ | 1+tan2θ=sec2θ |
| √(x2−a2) | x=a sec θ | sec2θ−1=tan2θ |
These are exactly Week 1's Eq. 1.8 and Eq. 1.9, chosen so that the substitution makes the expression under the root a perfect square. After integrating in θ, a reference triangle converts the answer back to x.
13.3 Worked Trigonometric Substitution
Example 13.1 — The a²−x² case
Evaluate ∫ dx√(4−x2).
Step 1 — Substitute x=2 sin θ (a=2)
dx=2 cos θ dθ; √(4−x2) = √(4−4sin2θ) = 2 cos θ.
Step 2 — Substitute and simplify
∫ 2 cos θ dθ2 cos θ = ∫ dθ = θ + C
Step 3 — Convert back to x using x=2 sin θ, i.e. θ=arcsin(x/2)
arcsin(x/2) + C
13.4 Partial Fractions: Setting Up the Decomposition
A proper rational function P(x)/Q(x) (numerator degree less than denominator degree) can be split into a sum of simpler fractions once Q(x) is factored. If the numerator's degree is not less, perform polynomial long division first.
13.5 Partial Fractions: Distinct Linear Factors
For each distinct linear factor (x−r) of Q(x), include a term A/(x−r) with an unknown constant A, found by clearing denominators and matching coefficients (or substituting convenient values of x).
Example 13.2 — Distinct linear factors
Evaluate ∫ 3x+5x2−x−2 dx.
Step 1 — Factor the denominator and set up the decomposition
x2−x−2 = (x−2)(x+1). 3x+5(x−2)(x+1) = Ax−2 + Bx+1
Step 2 — Clear denominators and solve for A, B
3x+5 = A(x+1) + B(x−2). At x=2: 11=3A ⇒ A=11/3. At x=−1: 2=−3B ⇒ B=−2/3.
Step 3 — Integrate each simple term (Eq. 11.7)
(11/3)ln|x−2| − (2/3)ln|x+1| + C
13.6 Partial Fractions: Repeated and Irreducible Factors
A repeated linear factor (x−r)k requires k terms, A1/(x−r) + A2/(x−r)2 + … + Ak/(x−r)k. An irreducible quadratic factor x2+bx+c (no real roots) requires a term (Bx+C)/(x2+bx+c), whose integral typically needs completing the square and then an arctangent form or trigonometric substitution (§13.2).
13.7 Worked Partial Fractions Example
Example 13.3 — A repeated factor
Evaluate ∫ x+1x(x−1)2 dx.
Step 1 — Set up the decomposition
x+1x(x−1)2 = Ax + Bx−1 + C(x−1)2
Step 2 — Clear denominators and solve
x+1 = A(x−1)2 + Bx(x−1) + Cx. At x=0: 1=A. At x=1: 2=C. Matching x2 coefficients: 0=A+B ⇒ B=−1.
Step 3 — Integrate term by term
ln|x| − ln|x−1| − 2/(x−1) + C
13.8 Improper Integrals: Infinite Limits of Integration
The definite integral (Week 7) was defined only for a finite interval [a,b]. Extending to an infinite interval defines the result as a limit:
The improper integral converges if this limit exists (as a finite number) and diverges otherwise.
Example 13.4 — An infinite limit of integration
Evaluate ∫1∞ (1/x2) dx.
Step 1 — Rewrite as a limit (Eq. 13.1)
limb→∞ ∫1b x−2 dx
Step 2 — Evaluate the definite integral for fixed b
[−x−1]1b = −1/b + 1
Step 3 — Take the limit as b→∞ (Week 3, §3.8)
limb→∞ (1−1/b) = 1. The integral converges to 1.
13.9 Improper Integrals: Unbounded Integrands
If f is unbounded near an endpoint (say f(x)→±∞ as x→b−), the integral over [a,b] is defined as
Example 13.5 — An unbounded integrand
Evaluate ∫01 (1/√x) dx.
Step 1 — Note the integrand is unbounded as x→0⁺, rewrite as a limit
limc→0+ ∫c1 x−1/2 dx
Step 2 — Evaluate the definite integral for fixed c
[2x1/2]c1 = 2 − 2√c
Step 3 — Take the limit as c→0⁺
limc→0+ (2−2√c) = 2. The integral converges to 2, despite the integrand growing without bound.
13.10 Convergence and Divergence
Not every improper integral converges. A standard comparison family:
The dividing line at p=1 is ∫1∞ (1/x)dx, which diverges since it equals limb→∞ ln b = ∞ (Week 11, Eq. 11.7) — the natural logarithm grows too slowly to converge, but any positive power above 1 grows just fast enough in the denominator.
13.11 Worked Capstone Example
Example 13.6 — Trigonometric substitution inside an improper integral
Evaluate ∫01 dx√(1−x2), noting the integrand is unbounded at x=1.
Step 1 — Rewrite as a limit (§13.9)
limc→1− ∫0c dx√(1−x2)
Step 2 — Antidifferentiate using trigonometric substitution (§13.2-13.3, a=1)
By the method of Example 13.1, ∫ dx√(1−x2) = arcsin x + C.
Step 3 — Evaluate and take the limit
[arcsin x]0c = arcsin c − 0. As c→1−, arcsin c → π/2.
The integral converges to π/2.
13.12 Nomenclature
| Symbol | Meaning |
|---|---|
| improper integral | A definite integral with an infinite limit of integration or an unbounded integrand |
| converges / diverges | The defining limit exists (finite) / does not exist or is infinite |
| partial fractions | Decomposing a rational function into a sum of simpler fractions |
| reference triangle | A right triangle used to convert a trigonometric substitution's answer back to the original variable |
13.13 Problems
- Trig substitution. Evaluate ∫ dx/√(x2+9).
- Trig substitution, sec. Evaluate ∫ dx/(x√(x2−4)) for x>2.
- Partial fractions, distinct linear. Evaluate ∫ (2x+1)/[(x−1)(x+3)] dx.
- Partial fractions, repeated. Evaluate ∫ 1/[x2(x+1)] dx.
- Partial fractions, requiring long division first. Evaluate ∫ (x2)/(x−1) dx.
- Improper, infinite limit. Evaluate ∫0∞ e−x dx.
- Improper, divergence. Show that ∫1∞ (1/√x) dx diverges, using Eq. 13.3.
- Improper, unbounded integrand. Evaluate ∫01 (1/x2) dx, and explain why it diverges even though [0,1] is bounded.
- Both endpoints improper. Evaluate ∫−∞∞ e−|x| dx, by splitting at 0.
- Combined. Evaluate ∫2∞ 1/[x(x−1)] dx, using partial fractions before taking the limit.
13.14 References
- Thomas, G. B., Weir, M. D. and Hass, J. Thomas' Calculus, SI Units, 15th ed. Pearson, 2024. — §8.4 (trigonometric substitutions), §8.5 (integration of rational functions by partial fractions), §8.8 (improper integrals).
- Pearson MyLab Math — the Week 13 homework set is assigned directly from the sections above; see the Calendar for the due date.