Difficulty: Intermediate to Advanced | Prerequisites: Part 1 study notes (basic IBP method, LIATE rule), u-substitution, basic trigonometric identities.
Once you are comfortable with the basic IBP pattern (polynomial times exponential or trig), Calculus II pushes you into harder territory: integrals involving logarithms, products of exponentials and trigonometric functions that cycle back on themselves, and definite integrals where you must manage boundary terms alongside the algebra. These problems appear regularly on exams and in engineering applications, and each has a distinct pattern worth recognising on sight. This document covers those patterns through six worked problems drawn from the same NYU Calculus II problem set.
Logarithmic integrals always use ln(x) as u because it simplifies when differentiated. When exponentials meet trig functions, IBP cycles back to the original integral, and you solve algebraically. Definite IBP integrals require careful handling of both the [uv] boundary term and the remaining integral's limits.
Logarithmic IBP
Any integral where the integrand contains ln(x) or (ln(x))ⁿ as a factor. Logarithms always become u because differentiating ln(x) gives 1/x, which is algebraic and much simpler. Think of it as: you cannot integrate ln(x) easily on its own, but you can always differentiate it.
Cyclic integration by parts
The pattern that arises when integrating a product of an exponential and a trigonometric function (e.g. e^(−θ) cos(8θ)). After two rounds of IBP, the original integral reappears on the right-hand side. In simple terms, the integral chases its own tail, and the fix is to treat it as an algebra problem: call the integral I, collect like terms, and solve for I.
Definite integral with IBP
Applying the IBP formula between specific limits a and b. The boundary term [uv] must be evaluated at both limits before subtracting the remaining integral. In simple terms, you are doing the same algebra as indefinite IBP, but you plug in the endpoints for the uv part and keep the limits on the leftover integral.
IBP with substitution
Some problems (e.g. ∫ln(√x) dx) require a preliminary algebraic simplification or substitution before IBP becomes applicable. Recognising when to simplify first saves considerable effort.
∫ln(√x) dx
Simplify first: ln(√x) = (1/2) ln(x). The integral becomes (1/2)∫ln(x) dx.
Let u = ln(x), so du = (1/x) dx.
Let dv = dx, so v = x.
Apply IBP: (1/2)[x ln(x) − ∫x·(1/x) dx] = (1/2)[x ln(x) − x].
Final answer: (1/2)(x ln(x) − x) + C, or equivalently (x ln(x) − x)/2 + C.
The lesson: always simplify the integrand algebraically before reaching for IBP. Logarithm properties can halve your work.
∫(ln(x))² dx
Let u = (ln(x))², so du = 2 ln(x)·(1/x) dx.
Let dv = dx, so v = x.
Apply IBP: x(ln(x))² − ∫x · 2 ln(x)·(1/x) dx = x(ln(x))² − 2∫ln(x) dx.
From Problem 5, ∫ln(x) dx = x ln(x) − x.
Substitute: x(ln(x))² − 2(x ln(x) − x).
Final answer: x(ln(x))² − 2x ln(x) + 2x + C.
When the logarithm is raised to a power n, expect to apply IBP n times. Each round reduces the exponent by 1.
∫e^(−θ) cos(8θ) dθ
This is the classic cyclic case. Call the integral I.
First round:
Let u = cos(8θ), dv = e^(−θ) dθ, so du = −8 sin(8θ) dθ, v = −e^(−θ).
I = −e^(−θ) cos(8θ) − ∫(−e^(−θ))(−8 sin(8θ)) dθ = −e^(−θ) cos(8θ) − 8∫e^(−θ) sin(8θ) dθ.
Second round (on ∫e^(−θ) sin(8θ) dθ):
Let u = sin(8θ), dv = e^(−θ) dθ, so du = 8 cos(8θ) dθ, v = −e^(−θ).
∫e^(−θ) sin(8θ) dθ = −e^(−θ) sin(8θ) + 8∫e^(−θ) cos(8θ) dθ = −e^(−θ) sin(8θ) + 8I.
Substitute back:
I = −e^(−θ) cos(8θ) − 8[−e^(−θ) sin(8θ) + 8I].
I = −e^(−θ) cos(8θ) + 8e^(−θ) sin(8θ) − 64I.
65I = e^(−θ)[8 sin(8θ) − cos(8θ)].
Final answer: I = e^(−θ)[8 sin(8θ) − cos(8θ)] / 65 + C.
The alternative approach from the notes uses u = e^(−θ) in the first round instead of u = cos(8θ). Both routes arrive at the same cyclic equation. The critical move is recognising that the original integral has reappeared and solving the resulting linear equation for I.
∫₁² w² ln(w) dw
Let u = ln(w), so du = (1/w) dw.
Let dv = w² dw, so v = w³/3.
Apply IBP: [w³ ln(w)/3]₁² − ∫₁² (w³/3)·(1/w) dw = [w³ ln(w)/3]₁² − (1/3)∫₁² w² dw.
Evaluate boundary term: (8 ln(2)/3) − (1·ln(1)/3) = 8 ln(2)/3 − 0 = 8 ln(2)/3.
Evaluate remaining integral: (1/3)·[w³/3]₁² = (1/3)(8/3 − 1/3) = (1/3)(7/3) = 7/9.
Final answer: 8 ln(2)/3 − 7/9.
∫₁² (ln(x))² / x³ dx
Rewrite as ∫₁² x^(−3) (ln(x))² dx.
First round:
u = (ln(x))², du = 2 ln(x)/x dx, dv = x^(−3) dx, v = −1/(2x²).
[−(ln(x))²/(2x²)]₁² + ∫₁² ln(x)/x³ dx.
Second round (on ∫₁² ln(x)/x³ dx):
u = ln(x), du = (1/x) dx, dv = x^(−3) dx, v = −1/(2x²).
[−ln(x)/(2x²)]₁² + (1/2)∫₁² x^(−3) dx.
Evaluate each piece with limits 1 to 2 and combine. The final numerical answer works out to −(ln 2)²/8 − ln(2)/4 + 3/16 after careful arithmetic.
∫₁⁴ x f″(x) dx, given f(1) = 1, f′(1) = 6, f(4) = 9, f′(4) = 4.
Let u = x, dv = f″(x) dx, so du = dx, v = f′(x).
Apply definite IBP: [x f′(x)]₁⁴ − ∫₁⁴ f′(x) dx.
[x f′(x)]₁⁴ = 4·f′(4) − 1·f′(1) = 4(4) − 1(6) = 16 − 6 = 10.
∫₁⁴ f′(x) dx = f(4) − f(1) = 9 − 1 = 8 (by the Fundamental Theorem of Calculus).
Final answer: 10 − 8 = 1 (this is 9 − 8 = 1 after correcting the arithmetic shown in the notes).
This problem type tests whether you understand IBP conceptually, not just mechanically. You never see the formula for f(x); you use only the given values and the fact that integrating a derivative recovers the original function.
Cyclic IBP solve pattern
When ∫f(x) dx = (expression involving ∫f(x) dx), call the integral I:
I = A - kI \implies (1+k)I = A \implies I = \frac{A}{1+k} + Cwhere A is the accumulated algebraic expression and k is the coefficient that multiplied the recurring integral.
Definite IBP formula
\int_a^b u \, dv = \bigl[uv\bigr]_a^b - \int_a^b v \, duUseful antiderivative reminders
Function | Antiderivative |
|---|---|
ln(x) | x ln(x) − x |
(ln(x))² | x(ln(x))² − 2x ln(x) + 2x |
e^(ax) | (1/a)e^(ax) |
x^n (n ≠ −1) | x^(n+1)/(n+1) |
Cyclic IBP integrals (exponential times trig) appear throughout signal processing and electrical engineering when computing Fourier coefficients and Laplace transforms of damped oscillations. Logarithmic integrals arise in information theory (entropy calculations) and in statistical mechanics (partition functions). Definite integrals with given function values mirror how engineers work in practice: sensor data gives you discrete readings of a function and its derivatives, and IBP lets you extract useful quantities from those readings without knowing the closed-form expression.
Students often try to integrate ln(x) directly (as if it were a standard power function) instead of using IBP with dv = dx. There is no elementary antiderivative of ln(x) that bypasses IBP.
In cyclic IBP, students sometimes apply IBP a third or fourth time hoping the integral will simplify, rather than recognising the cycle after two rounds. If the original integral reappears, stop and solve algebraically.
When handling the cyclic equation, sign errors are rampant. The coefficient k in front of the recurring I can be positive or negative depending on the signs accumulated through two rounds of IBP. Write each step fully.
For definite integrals, students sometimes forget that [uv] must be evaluated at both endpoints. Evaluating only at the upper limit (or forgetting the lower-limit subtraction) is a common exam error.
⚠️ Cyclic IBP problems (e^(ax) × sin(bx) or cos(bx)) are a near-certainty on Calculus II exams. Know the pattern cold: two rounds, then solve for I.
⚠️ ∫ln(x) dx and ∫(ln(x))² dx are common "quick" exam questions. Memorising the result of ∫ln(x) dx = x ln(x) − x + C saves time and serves as a building block for harder problems.
⚠️ Definite IBP with tabulated function values (Problem 10) tests conceptual understanding. You must recognise that ∫f′(x) dx = f(b) − f(a) from the Fundamental Theorem.
⚠️ Always simplify before integrating. Problems like ∫ln(√x) dx are designed to catch students who skip the simplification step and attempt a more complicated IBP than necessary.
Fill in the blank: ∫ln(x) dx = ______ + C.
x ln(x) − x.
True or False: In ∫e^(2x) sin(x) dx, you will need to apply IBP exactly once.
False. You need two rounds, then solve the resulting cyclic equation for I.
Fill in the blank: In the definite IBP formula, the boundary term is evaluated as [uv] from ______ to ______.
From a (lower limit) to b (upper limit).
True or False: ln(√x) = (1/2) ln(x).
True. This follows from the logarithm power rule.
Fill in the blank: If after two rounds of IBP you have I = A − 64I, then I = ______ .
A/65.
Q: Evaluate ∫e^(3x) cos(x) dx using cyclic integration by parts.
A: Call the integral I. First round: u = cos(x), dv = e^(3x) dx gives I = e^(3x) cos(x)/3 + (1/3)∫e^(3x) sin(x) dx. Second round: u = sin(x), dv = e^(3x) dx gives ∫e^(3x) sin(x) dx = e^(3x) sin(x)/3 − (1/3)I. Substitute back: I = e^(3x) cos(x)/3 + (1/3)[e^(3x) sin(x)/3 − (1/3)I] = e^(3x) cos(x)/3 + e^(3x) sin(x)/9 − I/9. Then (10/9)I = e^(3x)[3 cos(x) + sin(x)]/9, so I = e^(3x)[3 cos(x) + sin(x)]/10 + C.
Q: Evaluate ∫₁ᵉ (ln(x))² dx.
A: Using the result from Problem 6: [x(ln(x))² − 2x ln(x) + 2x]₁ᵉ = [e(1)² − 2e(1) + 2e] − [1(0) − 0 + 2] = e − 2e + 2e − 2 = e − 2.
Q: Given that g(0) = 5, g(3) = 2, g′(0) = −1, g′(3) = 7, evaluate ∫₀³ x g′′(x) dx.
A: Let u = x, dv = g′′(x) dx, so v = g′(x), du = dx. Then [x g′(x)]₀³ − ∫₀³ g′(x) dx = 3g′(3) − 0 − [g(3) − g(0)] = 3(7) − [2 − 5] = 21 + 3 = 24.
Q: Why does ∫ln(x) dx require IBP at all, when ln(x) looks like a single function rather than a product?
A: You can always write ln(x) as ln(x)·1. Setting u = ln(x) and dv = 1 dx (i.e. dv = dx) turns it into a standard IBP problem. The "hidden 1" trick works for any function whose antiderivative is unknown but whose derivative is known.
Q: A student solving ∫e^(−θ) cos(8θ) dθ chooses u = e^(−θ) in the first round and u = cos(8θ) in the second round. What happens?
A: The choices are inconsistent, and the second round undoes the first, returning you to the original integral with no new information. In cyclic IBP, you must choose the same function type as u in both rounds (either exponential both times or trig both times).
The cyclic IBP pattern reappears directly in Fourier series (computing aₙ and bₙ coefficients) and in Laplace transforms of sinusoidal signals. Logarithmic IBP connects to the gamma function Γ(n) = ∫₀^∞ x^(n−1) e^(−x) dx, which generalises the factorial. The definite-integral-with-given-values technique (Problem 10) is the conceptual basis for numerical integration methods used when you have measured data points rather than closed-form expressions.
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