- Apply substitution and integration by parts
- Integrate rational functions with partial fractions
- Decide whether an improper integral converges
- Compute volumes of revolution and work with integrals
The table of integrals handles simple functions, but what about ∫ x eˣ dx, ∫ ln x dx or ∫ dx/(x² − 1)? No formula in the table fits. Integration has no universal recipe like the rules of differentiation, yet a small toolbox covers most cases: substitution reverses the chain rule, integration by parts reverses the product rule, and partial fractions split a complicated fraction into simple pieces.
Substitution
- uthe new variable (the inner function)
- duthe differential of the new variable
In a definite integral change the limits as well: x = a becomes u = g(a), and x = b becomes u = g(b).
a) ∫ 2x(x² + 1)⁵ dx b) ∫ tan x dx c) ∫₀¹ x · e^(x²) dx
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b) tan x = sin x / cos x. u = cos x, du = −sin x dx: ∫ (−du/u) = −ln|u| + C = −ln|cos x| + C.
c) u = x², du = 2x dx, so x dx = du/2. Limits: x = 0 ⇒ u = 0, x = 1 ⇒ u = 1.
(1/2) ∫₀¹ eᵘ du = (1/2)(e − 1) ≈ 0.859.
Integration by parts
Start from the product rule: (uv)′ = u′v + uv′. Integrating both sides gives uv = ∫ v du + ∫ u dv. Rearranged, it becomes a formula that trades one integral for another:
- uthe part you differentiate: du = u′ dx
- dvthe part you integrate: v = ∫ dv
Integration by parts. For a definite integral: ∫ₐᵇ u dv = [uv]ₐᵇ − ∫ₐᵇ v du.
a) ∫ x eˣ dx b) ∫ ln x dx c) ∫₀^π x sin x dx
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∫ x eˣ dx = x eˣ − ∫ eˣ dx = eˣ(x − 1) + C.
b) u = ln x (L), dv = dx ⇒ du = dx/x, v = x.
∫ ln x dx = x ln x − ∫ x · (1/x) dx = x ln x − x + C.
c) u = x, dv = sin x dx ⇒ du = dx, v = −cos x.
[−x cos x]₀^π + ∫₀^π cos x dx = (π − 0) + [sin x]₀^π = π + 0 = π.
Partial fractions and trigonometric integrals
A rational function P(x)/Q(x) whose numerator has lower degree than its denominator can be split into simple fractions according to the factors of Q: a linear factor (x − a) gives A/(x − a); a repeated factor (x − a)² gives A/(x − a) + B/(x − a)²; an irreducible quadratic x² + px + q gives (Ax + B)/(x² + px + q). Each piece integrates with the table: ∫ A/(x − a) dx = A ln|x − a| + C.
- a, bdistinct roots of the denominator (a ≠ b)
- A, Bconstants to be found
- P(x)a polynomial of degree less than 2
Partial fractions for two distinct linear factors; then ∫ = A ln|x − a| + B ln|x − b| + C.
Compute ∫ (5x − 1)/(x² − x − 2) dx.
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(5x − 1)/((x − 2)(x + 1)) = A/(x − 2) + B/(x + 1).
x = 2: A = (5 · 2 − 1)/(2 + 1) = 9/3 = 3. x = −1: B = (5 · (−1) − 1)/(−1 − 2) = −6/(−3) = 2.
Check: 3(x + 1) + 2(x − 2) = 5x − 1. ✓
∫ = 3 ln|x − 2| + 2 ln|x + 1| + C.
- Even powers: reduce the power with sin² x = (1 − cos 2x)/2 and cos² x = (1 + cos 2x)/2. Hence ∫ sin² x dx = x/2 − sin 2x/4 + C and ∫₀^π sin² x dx = π/2.
- An odd power: split off one factor and substitute: ∫ sin³ x dx = ∫ (1 − cos² x) sin x dx; u = cos x ⇒ −cos x + cos³ x/3 + C.
- Products: turn them into sums with sin α · cos β = (1/2)[sin(α − β) + sin(α + β)] and similar formulas.
Improper integrals
- ∞an infinite limit: compute with a finite b first, then let b → ∞
- limif the limit is finite the integral converges, otherwise it diverges
The same idea works for a function that is unbounded at an endpoint: ∫₀¹ f(x) dx = lim (t→0⁺) ∫ₜ¹ f(x) dx.
- pthe exponent; the borderline case p = 1 diverges in both integrals
The p-integrals: the benchmark for comparing other integrals.
a) ∫₁^∞ dx/x² b) ∫₁^∞ dx/x c) ∫₀¹ dx/√x d) ∫₀^∞ e⁻ˣ dx
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b) lim (b→∞) [ln x]₁ᵇ = lim ln b = ∞: diverges.
c) lim (t→0⁺) [2√x]ₜ¹ = 2 − 0 = 2: converges, although the integrand is unbounded near 0.
d) lim (b→∞) [−e⁻ˣ]₀ᵇ = lim (1 − e⁻ᵇ) = 1: converges.
Applications: volumes of revolution and work
- f(x)the radius of the disc at position x
- π (f(x))² dxthe volume of a thin disc of thickness dx
The disc method: the region under y = f(x), a ≤ x ≤ b, rotates about the x-axis.
a) Find the volume of the solid obtained by rotating the region under y = √x, 0 ≤ x ≤ 4, about the x-axis.
b) Derive the formula for the volume of a ball.
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b) A ball of radius R comes from rotating the half-circle y = √(R² − x²), −R ≤ x ≤ R:
V = π ∫ (R² − x²) dx (from −R to R) = π [R²x − x³/3] = π (2R³ − 2R³/3) = (4/3)πR³.
We recover the familiar school formula, which is a good check.
- F(x)a variable force, N
- xthe position along the motion, m
- Wwork, J (joules)
Work of a variable force: the integral version of “force × distance”.
A spring with stiffness k = 200 N/m obeys Hooke's law F = kx. How much work is needed to stretch it from 0 to 10 cm, and then from 10 cm to 20 cm?
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0 → 0.1 m: W = 200 · 0.1²/2 = 1 J.
0.1 → 0.2 m: W = 100 · (0.2² − 0.1²) = 100 · 0.03 = 3 J.
The second 10 cm cost three times as much as the first, because the force grows with the extension.
Key points
- Substitution: ∫ f(g(x)) g′(x) dx = ∫ f(u) du; look for a function together with its derivative.
- By parts: ∫ u dv = uv − ∫ v du; choose u in LIATE order.
- Partial fractions: factor the denominator, split the fraction, each piece gives a logarithm (divide first if the numerator's degree is too high).
- An improper integral is defined as a limit; ∫₁^∞ dx/xᵖ converges only for p > 1.
- Volume of revolution V = π ∫ f² dx; work of a variable force = ∫ F dx.
Check yourself
10 questions. Every correct answer earns XP.