Difficulty: Intermediate | Prerequisites: U-substitution, product rule for derivatives.

Big Picture

Integration by parts handles integrals of products that u-substitution cannot simplify. It is the integration equivalent of the product rule for derivatives. In Calculus II, it sits directly after u-substitution in the toolkit and is used throughout the rest of the course, including for trigonometric integrals, improper integrals, and series convergence tests. If you can differentiate products confidently and you have u-substitution down, you are ready for this.

TL;DR

When an integral is the product of two different types of function (say, a polynomial times an exponential, or x times a trig function), you split the integrand into a part you differentiate (u) and a part you integrate (dv). The formula trades the original integral for a simpler one. Sometimes you need to apply it more than once.

Key Terms

Integration by parts

A technique that rewrites an integral of a product as a boundary term minus a new (ideally simpler) integral. It comes from reversing the product rule. In simple terms, you trade the integral you have for one that is easier to solve.

LIATE rule (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential)

A mnemonic for choosing u. Whichever function type appears earlier in the list LIATE is usually the better choice for u (the part you differentiate). In simple terms, pick the piece that gets simpler when you differentiate it.

Tabular integration (repeated integration by parts)

A shortcut for integrals where one factor is a polynomial. You repeatedly differentiate u and integrate dv in a table, applying alternating signs. In simple terms, it automates the bookkeeping when you need integration by parts more than once in a row.

Cyclic (looping) integration by parts

A situation where applying integration by parts twice returns the original integral on the right-hand side. You solve algebraically for the integral by collecting it on one side. Think of it as the integral "coming back around" so you can isolate it.

Core Content: The Formula and Choosing u/dv

Derivation From the Product Rule

The product rule says (u·v)' = u'v + uv'. Integrating both sides:

uv = \int u'v\, dx + \int uv'\, dx

Rearranging gives the integration by parts formula:

\int u\, dv = uv - \int v\, du

You pick part of the integrand to call u (which you will differentiate) and the rest to call dv (which you will integrate).

How to Choose u and dv

The goal is to make the new integral ∫ v du simpler than the original. The LIATE mnemonic helps:

1. L - Logarithmic functions (ln x, log x)

1. I - Inverse trigonometric functions (arctan x, arcsin x)

1. A - Algebraic functions (x, x², polynomials)

1. T - Trigonometric functions (sin x, cos x)

1. E - Exponential functions (eˣ, 2ˣ)

Functions higher in the list make better choices for u, because differentiating them tends to simplify things. Functions lower in the list are easier to integrate, making them better choices for dv.

Choosing Wrongly

The lecture explicitly demonstrates a wrong choice: for ∫ x sin(x) dx, if you set u = sin x and dv = x dx, you get v = x²/2, and the new integral ∫ (x²/2) cos x dx is harder than what you started with. The correct choice is u = x, dv = sin x dx.

The test is simple: after one application, is the new integral easier? If not, reverse your choice.

Worked Examples From the Lecture

Example 1: Polynomial Times Trig – ∫ x sin(x) dx

Set u = x, so du = dx. Set dv = sin(x) dx, so v = -cos(x).

\int x\sin(x)\, dx = -x\cos(x) - \int -\cos(x)\, dx = -x\cos(x) + \sin(x) + C

This is the textbook first example. The polynomial x drops its degree from 1 to 0 after one differentiation, leaving you with a pure trig integral.

Example 2: Polynomial Times Exponential – ∫ x²eˣ dx (Repeated Application)

Here u = x² and dv = eˣ dx, giving du = 2x dx and v = eˣ.

x^2 e^x - \int 2x e^x\, dx

The new integral still has x times eˣ, so apply by parts again: u = 2x, dv = eˣ dx, du = 2 dx, v = eˣ.

x^2 e^x - 2xe^x + 2e^x + C = e^x(x^2 - 2x + 2) + C

Each application lowers the polynomial degree by one. A degree-n polynomial requires n rounds of integration by parts.

Example 3: Cyclic Case – ∫ eˣ sin(x) dx

Neither eˣ nor sin(x) simplifies when differentiated, so integration by parts cycles.

First pass: u = eˣ, dv = sin(x) dx, v = -cos(x).

I = -e^x\cos(x) + \int e^x\cos(x)\, dx

Second pass on the new integral: u = eˣ, dv = cos(x) dx, v = sin(x).

I = -e^x\cos(x) + e^x\sin(x) - \int e^x\sin(x)\, dx

The original integral I appears on the right. Collect it:

2I = e^x(\sin(x) - \cos(x))

I = \frac{e^x(\sin(x) - \cos(x))}{2} + C

This algebraic trick, solving for I, is a favourite exam question.

Formulas to Memorise

Integration by parts (indefinite):

\int u\, dv = uv - \int v\, du

Integration by parts (definite):

\int_a^b u\, dv = [uv]_a^b - \int_a^b v\, du

LIATE priority for choosing u: Logarithmic > Inverse trig > Algebraic > Trigonometric > Exponential.

Common Misconceptions

• Students sometimes set dv to the piece that is harder to integrate. The whole point is that you must be able to integrate dv to get v. If you cannot integrate your chosen dv, swap your assignments.

• A frequent error is forgetting the minus sign in the formula. The formula is uv minus ∫ v du, and dropping that sign propagates through the rest of the calculation.

• Students occasionally apply integration by parts when u-substitution would have been simpler and faster. Always check for a substitution first.

• In the cyclic case (eˣ sin x or eˣ cos x), students sometimes switch which function they call u between the first and second application. You must be consistent, or the algebra will not produce the original integral on the right-hand side.

Why It Matters / Exam Flags

⚠️ Integration by parts is tested directly on every Calculus II midterm and final. Expect at least one straightforward problem (polynomial times eˣ or trig) and possibly one cyclic problem.

⚠️ The cyclic case (∫ eˣ sin x dx or ∫ eˣ cos x dx) is a classic exam question. Being able to recognise it and solve for I algebraically is worth practising until it feels automatic.

⚠️ Problems requiring repeated application (like ∫ x² eˣ dx) test patience and sign tracking. Tabular integration can speed these up, though you should understand the full method first.

⚠️ Definite integrals using by parts will appear. Remember to evaluate the uv boundary term at both limits before subtracting the remaining integral.

Quick Self-Test

1. True or False: In integration by parts, u should always be the function that is hardest to differentiate.
   Answer: False. u should be the function that becomes simpler when differentiated.

1. Fill in the blank: The integration by parts formula is ∫ u dv = ______.
   Answer: uv - ∫ v du

1. True or False: If applying integration by parts once gives you a harder integral, your choice of u and dv was probably wrong.
   Answer: True.

1. Fill in the blank: For ∫ x eˣ dx, a good choice of u is ______ and dv is ______.
   Answer: u = x, dv = eˣ dx

1. True or False: In the cyclic case, you can switch which function is u between the first and second application of integration by parts.
   Answer: False. Switching prevents the original integral from reappearing, and the method fails.

Practice Q&A

Q: Evaluate ∫ x cos(x) dx.

A: u = x, dv = cos x dx, du = dx, v = sin x. Result: x sin(x) - ∫ sin(x) dx = x sin(x) + cos(x) + C.

Q: Evaluate ∫ ln(x) dx.

A: u = ln x, dv = dx, du = (1/x) dx, v = x. Result: x ln(x) - ∫ 1 dx = x ln(x) - x + C.

Q: Evaluate ∫ x² sin(x) dx.

A: Apply by parts twice. First: u = x², dv = sin x dx gives x²(-cos x) + ∫ 2x cos x dx. Second: u = 2x, dv = cos x dx gives 2x sin x - ∫ 2 sin x dx = 2x sin x + 2 cos x. Final: -x² cos x + 2x sin x + 2 cos x + C.

Q: How do you recognise a cyclic integration by parts problem?

A: The integrand is a product of eˣ and a trig function (sin or cos). Neither factor simplifies under differentiation, and after two rounds of by parts the original integral reappears. Solve for it algebraically.

Q: What happens if you pick u and dv backwards?

A: The resulting integral ∫ v du becomes more complicated than the original. The fix is to swap your assignments and try again.

Connections to Other Topics

Integration by parts builds directly on u-substitution: many by-parts problems contain a sub-integral that itself requires substitution.

The technique is essential for deriving reduction formulas for powers of trig functions (∫ sinⁿ x dx, ∫ cosⁿ x dx), which appear later in the course.

It also surfaces in improper integrals and in the integral test for series convergence, both of which are central topics in the second half of Calculus II.

Related Terms / Search Tags

integration by parts, IBP, product rule, LIATE, tabular integration, cyclic integral, reduction formula, ∫ x eˣ, ∫ eˣ sin x, ∫ ln x, repeated integration by parts, Calculus II, NYU, Calculus II for Engineers, Sections 5.5-7.1, by parts definite integral