Difficulty: Intermediate | Prerequisites: Calculus I (differentiation rules, basic antiderivatives), familiarity with u-substitution.
Integration by parts is one of the central techniques in Calculus II and the first method you reach for when a straightforward u-substitution does not work. It reverses the product rule of differentiation, letting you break an integral of two multiplied functions into simpler pieces. You will use it constantly in later topics (differential equations, Laplace transforms, Fourier series), so building fluency here pays off across the rest of the course. If you can differentiate products and do basic antiderivatives, you have what you need to start.
Integration by parts converts a hard integral into an easier one by splitting the integrand into two pieces (u and dv), differentiating one and integrating the other. The LIATE mnemonic tells you which piece to call u. Sometimes you need to apply the method more than once, or use a cyclic trick when the original integral reappears.
Integration by parts (IBP)
A technique that rewrites ∫u dv as uv − ∫v du, effectively trading one integral for another that is (hopefully) simpler. Think of it as running the product rule for derivatives in reverse.
u and dv
The two pieces you split the integrand into. You differentiate u (to get du) and integrate dv (to get v). The whole method hinges on choosing these well.
LIATE rule
A mnemonic for choosing u. The function type that appears earliest in the list Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential should usually be picked as u. In simple terms, this is a priority ranking: whatever simplifies most when you differentiate it should be u.
Cyclic integration by parts
What happens when you apply IBP twice and the original integral shows up again on the right-hand side. You treat it as an equation, solve for the integral algebraically. This typically occurs with products of exponentials and trig functions.
Tabular integration (rapid repeated IBP)
A shortcut table for cases where one factor eventually differentiates to zero (e.g. polynomials times exponentials or trig). You list successive derivatives and antiderivatives in columns and alternate signs.
Identify a product of two different types of function in the integrand.
Choose u and dv using the LIATE priority (see Formulas below).
Differentiate u to find du.
Integrate dv to find v.
Substitute into the formula: ∫u dv = uv − ∫v du.
Evaluate the new integral ∫v du. If it is still not elementary, apply IBP again or use another technique.
Add the constant of integration (+C) for indefinite integrals.
∫x e^(9x) dx
Let u = x, so du = dx.
Let dv = e^(9x) dx, so v = (1/9)e^(9x).
Apply the formula: x · (1/9)e^(9x) − ∫(1/9)e^(9x) dx.
The remaining integral is a plain exponential: (1/9) · (1/81)e^(9x) = (1/81)e^(9x).
Combine, watching the subtraction sign: (1/9)x·e^(9x) − (1/81)e^(9x) + C.
Why u = x? It is algebraic (A in LIATE), while e^(9x) is exponential (E), so x has higher priority. Differentiating x simplifies it to 1, which is exactly the goal.
∫5x cos(4x) dx
Let u = 5x, so du = 5 dx.
Let dv = cos(4x) dx, so v = sin(4x)/4.
Apply: 5x · sin(4x)/4 − ∫(sin(4x)/4)·5 dx = 5x sin(4x)/4 − (5/4)∫sin(4x) dx.
∫sin(4x) dx = −cos(4x)/4.
Final answer: 5x sin(4x)/4 + 5 cos(4x)/16 + C.
Sign management is the main hazard here. The minus from the IBP formula and the minus from integrating sin combine to produce a positive cosine term.
∫(x − 4) sin(πx) dx
Split or treat (x − 4) as a single linear term.
Let u = x − 4, so du = dx.
Let dv = sin(πx) dx, so v = −cos(πx)/π.
Apply: (x − 4)·(−cos(πx)/π) − ∫(−cos(πx)/π) dx.
∫cos(πx)/π dx = sin(πx)/π².
Final answer: −(x − 4)cos(πx)/π + sin(πx)/π² + C.
When the trig argument contains π (or any constant), every antiderivative picks up a factor of 1/π. Missing that divisor is one of the most common errors on exams.
∫(x² + 6x) cos(x) dx
Split into ∫x² cos(x) dx + 6∫x cos(x) dx and handle each separately.
For ∫x² cos(x) dx, apply IBP twice:
First round: u = x², dv = cos(x) dx, v = sin(x), du = 2x dx.
x² sin(x) − 2∫x sin(x) dx.
Second round on ∫x sin(x) dx: u = x, dv = sin(x) dx, v = −cos(x), du = dx.
x(−cos(x)) − ∫(−cos(x)) dx = −x cos(x) + sin(x).
Combine: x² sin(x) − 2(−x cos(x) + sin(x)) = x² sin(x) + 2x cos(x) − 2 sin(x).
For 6∫x cos(x) dx (one round of IBP):
u = x, dv = cos(x) dx, v = sin(x), du = dx.
6[x sin(x) − ∫sin(x) dx] = 6x sin(x) + 6 cos(x).
Final answer: x² sin(x) + 2x cos(x) − 2 sin(x) + 6x sin(x) + 6 cos(x) + C.
The key takeaway: when the polynomial factor has degree n, expect to apply IBP n times until it differentiates away to a constant.
The IBP formula (indefinite)
\int u \, dv = uv - \int v \, duThe IBP formula (definite)
\int_a^b u \, dv = \bigl[uv\bigr]_a^b - \int_a^b v \, duLIATE priority table for choosing u
Priority | Function type | Example | Why it goes first |
|---|---|---|---|
1 (highest) | Logarithmic | ln(x), log₂(x) | Differentiates to a simpler algebraic form |
2 | Inverse trigonometric | arctan(x), arcsin(x) | Differentiates to an algebraic fraction |
3 | Algebraic | x², x, polynomials | Degree drops by 1 each time |
4 | Trigonometric | sin(x), cos(x) | Cycles, does not simplify |
5 (lowest) | Exponential | eˣ, e^(9x) | Stays the same when differentiated |
The function appearing earliest in LIATE should be your u. Everything else becomes part of dv.
Integration by parts appears whenever engineers model systems with decaying or oscillating signals. Electrical engineers use it to derive the Laplace transforms that govern circuit analysis. Mechanical engineers encounter it when computing moments of inertia and centres of mass for non-uniform bodies, and it underpins the derivation of many probability distributions (e.g. the expected value of continuous random variables).
Students often swap u and dv, picking the exponential as u and the polynomial as dv. This makes the integral harder, not easier, because differentiating the exponential leaves it unchanged while integrating the polynomial raises its degree.
Forgetting to include the 1/a factor when integrating e^(ax) or sin(ax). Integrating e^(9x) gives (1/9)e^(9x), not e^(9x). Missing this cascades through the entire solution.
Dropping the minus sign from the IBP formula. The formula is uv minus ∫v du. A flipped sign propagates through every subsequent step and gives the wrong answer.
Assuming IBP is always the right technique. If the integrand is a simple composition (e.g. e^(x²)·2x), u-substitution is faster and cleaner. IBP is for products of different function types that substitution cannot untangle.
⚠️ IBP problems appear on nearly every Calculus II exam. You will see at least one, typically involving a polynomial times an exponential or trig function.
⚠️ Sign errors are the number-one point-loser. Write out every step of the formula explicitly; do not try to combine steps in your head.
⚠️ The LIATE mnemonic is not a law, it is a reliable heuristic. On rare occasions (especially with inverse trig) you may need to override it. If your choice of u makes ∫v du worse than the original, try swapping.
⚠️ For definite integrals, do not forget to evaluate the uv boundary term [uv] from a to b before subtracting the remaining integral. Students sometimes evaluate only the integral part and lose easy marks.
True or False: In ∫x eˣ dx, you should let u = eˣ.
False. u = x (algebraic beats exponential in LIATE).
Fill in the blank: The IBP formula is ∫u dv = uv − ______.
∫v du.
True or False: Applying IBP to ∫x³ sin(x) dx requires three rounds of IBP.
True. Each round reduces the polynomial degree by 1.
Fill in the blank: ∫e^(9x) dx = ______.
(1/9)e^(9x) + C.
True or False: If you get the original integral back after two rounds of IBP, you are stuck.
False. You set the expression equal to the original integral and solve algebraically (the cyclic method).
Q: Evaluate ∫x sin(3x) dx using integration by parts.
A: Let u = x, dv = sin(3x) dx. Then du = dx, v = −cos(3x)/3. Applying IBP: −x cos(3x)/3 + (1/3)∫cos(3x) dx = −x cos(3x)/3 + sin(3x)/9 + C.
Q: In ∫x² eˣ dx, how many times must you apply IBP, and why?
A: Twice. Each round reduces the polynomial degree by 1 (x² becomes 2x, then 2), until the polynomial factor differentiates away entirely.
Q: You are integrating ∫eˣ sin(x) dx and after two rounds of IBP you get the original integral on the right-hand side. What do you do next?
A: Call the original integral I. You now have an equation of the form I = (expression) − I (or + I depending on signs). Move the I term to the left, solve: 2I = (expression), so I = (expression)/2. Add +C.
Q: What is the result of ∫ln(x) dx?
A: Let u = ln(x), dv = dx. Then du = (1/x) dx, v = x. Applying IBP: x ln(x) − ∫x·(1/x) dx = x ln(x) − x + C.
Q: A student sets u = cos(4x) and dv = 5x dx for ∫5x cos(4x) dx. What goes wrong?
A: Differentiating cos(4x) gives −4 sin(4x), which is no simpler, and integrating 5x gives (5/2)x², which is more complex. The new integral ∫(5/2)x²·(−4 sin(4x)) dx is harder than the original. Swapping the choices (u = 5x, dv = cos(4x) dx) reduces the polynomial and produces a solvable integral.
Integration by parts is the foundation for reduction formulas (e.g. ∫sinⁿ(x) dx), which you will meet later in this course. It also underpins the derivation of Taylor and Maclaurin series remainders in the integral form, and is essential for solving second-order linear differential equations with non-constant forcing functions.
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