Integration by Parts

Integration by parts is a powerful technique for evaluating integrals of products of functions. Just as the product rule tells us how to differentiate a product of two functions, integration by parts gives us a systematic way to integrate products. It is one of the most important tools in a calculus student's toolkit, applicable whenever a direct antiderivative or substitution is not obvious. The technique transforms a difficult integral into a (hopefully simpler) one by strategically splitting the integrand into two parts: one to differentiate and one to integrate. The classic formula states that ∫u dv = uv − ∫v du. The art of the method lies in choosing which part of the integrand plays the role of u and which plays the role of dv — a decision that can make the problem straightforward or painfully circular if done poorly. Integration by parts appears throughout mathematics and physics: it is used to derive reduction formulas for powers of trigonometric functions, to evaluate integrals involving logarithms and inverse trigonometric functions, and even to define integration in an abstract sense in functional analysis. Mastering it at the high school level opens doors to solving a wide class of problems in calculus and beyond.

Key takeaways
  • Start with the Product Rule
  • Choose u and dv Using LIATE
  • Compute du and v
  • Apply the Formula and Simplify
  • Repeated Application and Circular Cases

Start with the Product Rule

Recall the product rule for differentiation: if u and v are functions of x, then d(uv)/dx = u(dv/dx) + v(du/dx). Rearranging gives u(dv/dx) = d(uv)/dx − v(du/dx). Integrating both sides with respect to x yields the integration by parts formula: ∫u dv = uv − ∫v du.

Choose u and dv Using LIATE

The hardest part is choosing which factor becomes u and which becomes dv. A reliable mnemonic is LIATE: Logarithmic, Inverse trigonometric, Algebraic (polynomials), Trigonometric, Exponential. Choose u to be the type that appears earliest in LIATE, because that type gets simpler when differentiated. For example, in ∫x·eˣ dx, x is Algebraic and eˣ is Exponential, so set u = x and dv = eˣ dx.

Compute du and v

Once you have chosen u and dv, differentiate u to get du and integrate dv to get v. You do not need a constant of integration when finding v at this stage — any antiderivative will do. For u = x: du = dx. For dv = eˣ dx: v = eˣ.

Apply the Formula and Simplify

Substitute into ∫u dv = uv − ∫v du. In the example: ∫x eˣ dx = x·eˣ − ∫eˣ dx = x eˣ − eˣ + C = eˣ(x − 1) + C. Always verify by differentiating your answer using the product rule to confirm it equals the original integrand.

Repeated Application and Circular Cases

Sometimes ∫v du is just as hard as the original integral, requiring integration by parts again (tabular method is useful here for polynomials times exponentials or trigonometric functions). In other cases, such as ∫eˣ sin x dx, applying the formula twice produces a copy of the original integral on both sides, allowing you to solve algebraically for the unknown integral.

The Key Equation

∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du
  • u — First factor (chosen function) (dimensionless (or same units as integrand context)): The part of the integrand selected to be differentiated; chosen to become simpler after differentiation.
  • dv — Differential of second factor (same units as the remaining factor times dx): The part of the integrand (including dx) that is integrated to produce v.
  • v — Antiderivative of dv (same units as dv divided by dx): The antiderivative of the chosen dv, found by integrating dv with respect to x.
  • du — Differential of u (same units as u divided by x): The derivative of u multiplied by dx; appears in the new, simpler integral on the right-hand side.

Worked Example

Problem. Evaluate ∫ x cos x dx.

xsin⁡x+cos⁡x+Cx \sin x + \cos x + C

Answer. x sin x + cos x + C

Figure
Diagram
Integration by Parts: LIATE Selection GuideLIATE OrderL — Logarithmice.g., ln xI — Inverse Trige.g., arctan xA — Algebraice.g., x², xT — Trige.g., sin xE — Exponentiale.g., eˣExample Choices∫x eˣ dxu = x (Algebraic)dv = eˣ dx (Exp)∫x cos x dxu = x (Algebraic)dv = cos x dx (Trig)∫ln x dxu = ln x (Log)dv = dx (Algebraic)Formula∫u dv =uv − ∫v duChoose u so thatdu is simplerthan u.Choose dv sothat v is easyto find.
Origins · 1 minBrook Taylor and how this was discovered
BT
Brook Taylor
1685–1731 · British

Brook Taylor developed the method of integration by parts as a consequence of his work on the summation of series and differences, publishing it in his 1715 treatise Methodus Incrementorum. He showed how the technique follows naturally from reversing the product rule of differentiation, a conceptual bridge that remains central to how the method is taught today. Taylor is best remembered for 'Taylor series,' but he was also an accomplished musician and painter, writing a treatise on the mathematics of perspective in art called 'Linear Perspective' in 1715 — the same year he published his integration work.

  • U-Substitution — U-substitution is often the first integration technique to try before integration by parts; both are strategies for transforming integrals into simpler forms.
  • Product Rule (Differentiation) — Integration by parts is derived directly by integrating both sides of the product rule, so understanding the product rule is essential.
  • Trigonometric Integrals — Integration by parts is frequently used alongside trigonometric integrals, especially when evaluating powers of sin and cos via reduction formulas.
  • Definite Integrals — Integration by parts applies to definite integrals too, with the formula ∫[a to b] u dv = [uv] from a to b minus ∫[a to b] v du, requiring careful evaluation of boundary terms.