Integration by Parts — Integrating Using Integration By Parts - Calculus 2

SECTION 1

Historical Context & Motivation

The technique of integration by parts arises from one of the most fundamental operations in calculus: reversing differentiation of a product. While the product rule for derivatives is straightforward—simply differentiate each factor and sum the results—undoing that process to find antiderivatives of products is considerably more subtle. Integration by parts provides a systematic framework for tackling integrals of the form ∫ f(x)g(x) dx when direct antidifferentiation or simpler substitution techniques fail. The method effectively redistributes complexity from one factor to another, ideally transforming a difficult integral into an easier one.

The intellectual origins of this technique trace back to the co-inventors of calculus, Isaac Newton and Gottfried Wilhelm Leibniz, who both recognized that the product rule for derivatives could be inverted to yield an integration technique. Leibniz, whose notation we still use today, was particularly attuned to the algebraic manipulation of differentials—writing d(uv) = u dv + v du—and immediately saw that rearranging this identity produced a tool for integration. Over the following centuries, mathematicians like Euler and Bernoulli refined the technique and extended it to handle increasingly sophisticated classes of functions, including logarithmic, inverse trigonometric, and exponential products.

1670s • Newton's Fluxions & Leibniz's Differentials

Newton and Leibniz independently develop calculus. Leibniz's differential notation d(uv) = u dv + v du implicitly contains the integration by parts formula, setting the stage for its formal articulation.

1690s • Bernoulli Brothers Apply the Technique

Johann and Jakob Bernoulli use integration by parts to solve problems in mechanics and the calculus of variations, demonstrating its power for evaluating integrals involving exponential and trigonometric products.

1730s • Euler's Systematic Treatment

Leonhard Euler codifies integration by parts as a standard technique in his textbooks, applying it to derive series expansions, evaluate definite integrals, and establish reduction formulas for powers of trigonometric functions.

1800s • Rigorization and the Definite Integral Form

Cauchy, Riemann, and others place integration on rigorous footing. The definite-integral version of integration by parts, complete with boundary terms, becomes central to analysis and mathematical physics.

Modern Era • Tabular Methods & Computational Extensions

The tabular (or 'DI') method streamlines repeated integration by parts. Computer algebra systems automate the technique, but understanding its logic remains essential for recognizing when and how to apply it.

The central question that integration by parts answers is: How can we evaluate the integral of a product of two functions when no elementary antiderivative is immediately apparent? By strategically choosing which factor to differentiate and which to integrate, we trade one integral for another—hopefully simpler—integral. Mastering this choice is the art at the heart of the technique.

SECTION 2

Core Principles & Definitions

Integration by parts rests on a single algebraic identity derived from the product rule of differentiation. If u = f(x) and v = g(x) are differentiable functions, then d(uv)/dx = u(dv/dx) + v(du/dx). Integrating both sides with respect to x and rearranging yields the integration by parts formula. The power of this result lies in its generality: it applies to any product of differentiable functions and converts one integral into a boundary term plus a new integral that may be easier to evaluate.

The Product Rule in Reverse

Integration by parts is the integral analogue of the product rule. Since d(uv) = u dv + v du, integrating both sides gives uv = ∫ u dv + ∫ v du, which rearranges to the integration by parts formula.

Strategic Decomposition (LIATE)

The effectiveness of the method depends on choosing u (the factor to differentiate) and dv (the factor to integrate). The LIATE mnemonic—Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential—guides which factor to designate as u.

Complexity Transfer

Differentiating u should simplify it (or at least not complicate it), while integrating dv should yield a manageable v. The new integral ∫ v du should be simpler than the original ∫ u dv.

Definite Integral Form

For definite integrals, integration by parts becomes ∫ₐᵇ u dv = [uv]ₐᵇ − ∫ₐᵇ v du. The boundary term [uv]ₐᵇ = u(b)v(b) − u(a)v(a) must be evaluated at the limits before subtracting the remaining integral.

Repeated Application & Cyclic Integrals

Some integrals require applying integration by parts multiple times. Occasionally—as with ∫ eˣ sin x dx—the original integral reappears after two applications, allowing an algebraic solve rather than further iteration.

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KEY TAKEAWAY

Think of integration by parts as a negotiation. You have an integral that's too complex to evaluate directly, so you split it into two roles—one factor you differentiate (u) and one you integrate (dv). You're trading complexity from one factor to the other, much like bartering: you give up something (the boundary term uv) and receive a new integral (∫ v du) that is, ideally, a better deal than what you started with. The LIATE mnemonic acts like a market guide, telling you which trades tend to be favorable.

SECTION 3

Visual Explanation

Geometric Interpretation of Integration by Parts

Integration by parts has a beautiful geometric interpretation when the functions u and v are both non-negative and monotonic over the region of interest. Consider the integral ∫ₐᵇ u dv as the area of one region in the u–v plane, and ∫ᵤ₍ₐ₎ᵘ⁽ᵇ⁾ v du as the area of the complementary region. Together, these two areas plus the boundary rectangle terms compose the total rectangle u(b)v(b) − u(a)v(a). The following diagram illustrates this partition for a parametric curve (u(t), v(t)).

The violet region represents ∫ u dv (integrating along the v-axis), while the cyan region represents ∫ v du (integrating along the u-axis). Together with the boundary rectangle corners, these two areas reconstruct the total rectangle u(b)v(b) − u(a)v(a), visually confirming the integration by parts identity.

This geometric perspective reveals why integration by parts works: it partitions the rectangle in the u–v plane into two complementary regions separated by the parametric curve. The formula ∫ₐᵇ u dv = [uv]ₐᵇ − ∫ₐᵇ v du is simply the statement that the total rectangle area (u(b)v(b)) minus the small corner rectangle (u(a)v(a)) equals the sum of the violet and cyan regions. When we rearrange to isolate one integral, we are computing one region's area by subtracting the other from the known total. This viewpoint also explains why choosing u and dv wisely matters: a poor choice may produce a complementary region whose area is even harder to compute than the original.

SECTION 4

Mathematical Framework

Derivation from the Product Rule

The derivation begins with the product rule for differentiation. If u(x) and v(x) are differentiable functions, then the derivative of their product is d(uv)/dx = u(dv/dx) + v(du/dx). Integrating both sides with respect to x from a to b yields uv|ₐᵇ = ∫ₐᵇ u(dv/dx) dx + ∫ₐᵇ v(du/dx) dx. Rearranging isolates the desired integral and produces the definite-integral form of integration by parts. The indefinite form follows by omitting limits and including the constant of integration.

PRODUCT RULE (DIFFERENTIAL FORM)

d(uv) = u dv + v du

This is the starting point. Here u and v are differentiable functions of x, and du, dv denote their differentials.

INTEGRATION BY PARTS — INDEFINITE FORM

∫ u dv = uv − ∫ v du

Obtained by integrating both sides of the product rule and solving for ∫ u dv. The term uv is the boundary term, and ∫ v du is the new integral that must be simpler than the original.

INTEGRATION BY PARTS — DEFINITE FORM

∫ₐᵇ u dv = [uv]ₐᵇ − ∫ₐᵇ v du

The definite form evaluates the boundary term at the limits: [uv]ₐᵇ = u(b)v(b) − u(a)v(a). This is the version used for computing numerical answers to definite integrals.

The LIATE Selection Heuristic

The most critical step in applying integration by parts is the selection of u and dv. The LIATE rule is a widely used mnemonic that ranks function types by priority for designation as u. Functions higher on the list are better candidates for u because differentiating them tends to simplify the integrand, while functions lower on the list integrate cleanly. The acronym stands for: Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. When facing an integral of a product, assign u to whichever factor appears earliest in the LIATE ordering, and let dv be the remainder.

CYCLIC INTEGRAL TECHNIQUE

If I = ∫ f(x) dx and after two applications I reappears: I = g(x) − h(x) + I, then 2I = g(x) − h(x), so I = ½[g(x) − h(x)] + C

This algebraic trick handles cyclic integrals such as ∫ eˣ sin x dx. After two rounds of integration by parts, the original integral I appears on both sides, allowing you to solve for I without further integration.

SECTION 5

Detailed Breakdown: Strategy Selection & The Tabular Method

LIATE Priority Table

The following table summarizes the LIATE hierarchy with representative functions for each category. When the integrand is a product of two functions from different categories, the function from the higher row should generally be chosen as u. This heuristic is not infallible—there are exceptions—but it provides an effective starting point for the vast majority of integrals encountered in a standard Calculus 2 course.

Priority	Category	Typical Functions	Why Choose as u?
1 (Highest)	L — Logarithmic	ln x, log₂ x, ln(x² + 1)	Differentiating removes the logarithm; integrating logs directly is hard.
2	I — Inverse Trig	arctan x, arcsin x, arcsec x	Differentiation yields algebraic expressions (e.g., 1/(1 + x²)).
3	A — Algebraic	x², x³, √x, x⁻¹ (polynomials, radicals)	Differentiating lowers degree; polynomials eventually vanish.
4	T — Trigonometric	sin x, cos x, tan x, sec x	Trig functions cycle under differentiation; integrating them is straightforward.
5 (Lowest)	E — Exponential	eˣ, 2ˣ, e⁻³ˣ	Exponentials are essentially unchanged by differentiation or integration.

LIATE priority for choosing u in integration by parts

The Tabular (DI) Method for Repeated Integration by Parts

When integration by parts must be applied multiple times—typically when u is a polynomial of degree n—the tabular method provides a compact, organized shortcut. Create two columns: the left column lists successive derivatives of u (the 'D' column), and the right column lists successive antiderivatives of dv (the 'I' column). Alternating signs (+, −, +, −, …) are applied to the diagonal products from left to right. The process terminates when a row in the D column reaches zero (for polynomials) or when the product of a D-entry and the corresponding I-entry is directly integrable. The diagram below illustrates this method for ∫ x³ eˣ dx.

The tabular method organizes repeated integration by parts into a compact table. The diagonal arrows connect each D-entry with the I-entry one row below, and the alternating + / − signs are applied to each product. The process stops when the D-column reaches 0.

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When Does the Tabular Method Apply?

The tabular method works best when one factor is a polynomial (which eventually differentiates to zero) and the other is repeatedly integrable (such as eˣ, sin x, or cos x). For cyclic integrals where the D-column never reaches zero (e.g., ∫ eˣ sin x dx), you stop after two rows and use the cyclic technique: isolate the original integral algebraically from the resulting equation.

SECTION 6

Worked Example

Example 1: ∫ x ln x dx

Evaluate ∫ x ln x dx

1
Step 1 — Identify the Integrand Type

The integrand is a product of two function types: ln x (logarithmic) and x (algebraic). This product cannot be integrated by basic antidifferentiation or u-substitution alone, so integration by parts is the appropriate technique.
2
Step 2 — Choose u and dv Using LIATE

By LIATE, logarithmic functions have higher priority than algebraic functions, so we set u = ln x and dv = x dx. This is strategically advantageous because differentiating ln x produces 1/x (simpler), and integrating x dx is straightforward.

u = ln x, du = (1/x) dx; dv = x dx, v = x²/2
3
Step 3 — Apply the Integration by Parts Formula

Substituting into ∫ u dv = uv − ∫ v du: ∫ x ln x dx = (ln x)(x²/2) − ∫ (x²/2)(1/x) dx = (x²/2) ln x − ∫ x/2 dx

(x²/2) ln x − (1/2) ∫ x dx
4
Step 4 — Evaluate the Remaining Integral

The remaining integral ∫ x dx is elementary: ∫ x dx = x²/2 So (1/2) ∫ x dx = x²/4.

(1/2)(x²/2) = x²/4
5
Step 5 — Combine and Simplify

Putting it all together: ∫ x ln x dx = (x²/2) ln x − x²/4 + C This can be factored as (x²/4)(2 ln x − 1) + C.

∫ x ln x dx = (x²/2) ln x − x²/4 + C
6
Step 6 — Verify by Differentiating

Differentiate the result to confirm: d/dx[(x²/2) ln x − x²/4] = x ln x + (x²/2)(1/x) − x/2 = x ln x + x/2 − x/2 = x ln x ✓. The derivative matches the original integrand, confirming our answer.

Example 2: ∫ eˣ cos x dx (Cyclic Integral)

Evaluate ∫ eˣ cos x dx

1
Step 1 — Recognize the Cyclic Structure

The integrand is a product of eˣ (exponential) and cos x (trigonometric). Neither factor simplifies under repeated differentiation—exponentials reproduce themselves, and trig functions cycle. This signals a cyclic integral that will require two applications of integration by parts followed by algebraic manipulation.
2
Step 2 — First Application of IBP

Let u = eˣ, dv = cos x dx. Then du = eˣ dx, v = sin x. I = ∫ eˣ cos x dx = eˣ sin x − ∫ eˣ sin x dx

I = eˣ sin x − ∫ eˣ sin x dx
3
Step 3 — Second Application of IBP

For the new integral ∫ eˣ sin x dx, let u = eˣ, dv = sin x dx. Then du = eˣ dx, v = −cos x. ∫ eˣ sin x dx = −eˣ cos x + ∫ eˣ cos x dx = −eˣ cos x + I Substituting back: I = eˣ sin x − (−eˣ cos x + I) = eˣ sin x + eˣ cos x − I

I = eˣ sin x + eˣ cos x − I
4
Step 4 — Solve Algebraically for I

Adding I to both sides: 2I = eˣ sin x + eˣ cos x. Dividing by 2: I = (eˣ/2)(sin x + cos x) + C

∫ eˣ cos x dx = (eˣ/2)(sin x + cos x) + C

SECTION 7

Integration by Parts vs. Other Techniques

Integration by parts is one of several major techniques in the integration toolkit. Understanding when to use it—and when a different method would be more efficient—is an essential skill. The table below compares integration by parts with other standard techniques, highlighting their respective strengths, typical applications, and limitations.

Technique	Best For	Strengths	Limitations
Integration by Parts	Products of functions from different categories (e.g., x·eˣ, x²·sin x, ln x·xⁿ)	Handles products of dissimilar functions; works for ln x and arctan x integrals; enables reduction formulas	Requires good u/dv choice; may need multiple iterations; can produce more complex integrals if applied carelessly
u-Substitution	Composite functions where the derivative of the inner function appears as a factor (e.g., ∫ 2x·eˣ² dx)	Simple and direct; usually the first technique to try; reverses the chain rule	Only works when the integrand has a clear composite structure with the inner derivative present
Partial Fractions	Rational functions P(x)/Q(x) where deg P < deg Q	Systematically decomposes any proper rational function; results in elementary antiderivatives	Only applies to rational functions; factoring the denominator can be algebraically intensive
Trigonometric Substitution	Integrands involving √(a² − x²), √(a² + x²), or √(x² − a²)	Eliminates radicals; converts to trig integrals that are often manageable	Can be cumbersome; requires back-substitution; limited to specific radical forms

Comparison of major integration techniques

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WHEN TO REACH FOR INTEGRATION BY PARTS

Integration by parts is your method of choice when the integrand is a product of functions from two different LIATE categories, and u-substitution does not apply (because the derivative of one factor does not appear as a multiplier). Think of your integration toolkit like a mechanic's wrench set: u-substitution is your everyday adjustable wrench, but integration by parts is the torque wrench you bring out for products that resist simpler approaches. If you see ln x, arctan x, or a polynomial multiplied by eˣ or sin x, integration by parts should be your first instinct.

SECTION 8

Connections to Advanced Theory

Integration by parts is not merely a computational trick—it is a structural result that permeates higher mathematics. In analysis, it leads to the theory of weak derivatives and Sobolev spaces, where functions that are not classically differentiable can still be handled rigorously by transferring derivatives onto smooth test functions via integration by parts. In differential equations, the technique underlies the method of variation of parameters and the construction of Green's functions. The multi-variable generalization—Green's theorem, the divergence theorem, and Stokes' theorem—can all be viewed as higher-dimensional versions of integration by parts.

Calculus 2 Concept	Advanced Extension	Key Idea
∫ u dv = uv − ∫ v du	Weak Derivatives / Distribution Theory	Transfer differentiation from non-smooth functions to smooth test functions, defining derivatives in a generalized sense.
Reduction formulas via IBP	Recursion in special functions (Gamma, Beta)	Γ(n+1) = nΓ(n) is derived by integration by parts; the factorial function extends to non-integers.
Boundary term [uv]ₐᵇ	Boundary value problems in PDEs	Boundary conditions in physics (heat equation, wave equation) emerge naturally from the boundary terms in higher-dimensional IBP.
1D integration by parts	Green's / Divergence / Stokes' theorems	Multi-variable generalizations relate volume integrals of derivatives to surface integrals, fundamental in electromagnetism and fluid dynamics.

From Calculus 2 to advanced mathematics

Perhaps the most elegant connection is to the Gamma function, defined by Γ(n) = ∫₀^∞ $t^{n−1}$ $e^{−t}$ dt. Applying integration by parts once yields Γ(n+1) = nΓ(n), which for positive integers gives Γ(n+1) = n!, thereby extending the factorial to all complex numbers with positive real part. This single application of integration by parts opens the door to analytic number theory, quantum mechanics, and statistical mechanics—a testament to the far-reaching consequences of what begins as a simple algebraic rearrangement of the product rule.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain in your own words why the LIATE rule suggests choosing u = ln x rather than u = x³ when evaluating ∫ x³ ln x dx. What would go wrong if you made the opposite choice?

Show answer

By LIATE, logarithmic functions have higher priority than algebraic functions for the role of u. Setting u = ln x gives du = (1/x) dx, which simplifies the integrand because the logarithm is eliminated from the new integral ∫ v du. If instead we chose u = x³ and dv = ln x dx, we would need to find ∫ ln x dx (which itself requires integration by parts) just to determine v. Moreover, du = 3x² dx does not simplify the problem—the new integral ∫ v du would involve the product of x² with an antiderivative of ln x, creating a more complicated expression than the original. The LIATE heuristic steers us toward the choice that transfers complexity away from the integrand.

Question 2: BASIC CALCULATION

Evaluate ∫ x eˣ dx using integration by parts.

Show answer

Set u = x (algebraic) and dv = eˣ dx (exponential). Then du = dx and v = eˣ. Applying the formula: ∫ x eˣ dx = xeˣ − ∫ eˣ dx = xeˣ − eˣ + C = eˣ(x − 1) + C. This can be verified by differentiating: d/dx[eˣ(x − 1)] = eˣ(x − 1) + eˣ = xeˣ − eˣ + eˣ = xeˣ, confirming the result.

Question 3: INTERMEDIATE

Evaluate ∫ x² sin x dx. (Hint: you will need to apply integration by parts twice, or use the tabular method.)

Show answer

Using the tabular method with D-column (derivatives of x²): x², 2x, 2, 0 and I-column (antiderivatives of sin x): sin x, −cos x, −sin x, cos x. Applying alternating signs: (+)x²(−cos x) + (−)(2x)(−sin x) + (+)(2)(cos x) = −x² cos x + 2x sin x + 2 cos x + C. This factors to (2 − x²) cos x + 2x sin x + C. Verification: differentiating gives −2x cos x + (x² − 2) sin x · (−1) · (−1)... which, after careful application of the product rule to each term, yields x² sin x ✓.

Question 4: APPLIED

In signal processing, the Laplace transform of t·sin(ωt) requires evaluating ∫₀^∞ t sin(ωt)

e^{−st}

dt (for s > 0). Set up and evaluate this integral using integration by parts (treat ω and s as positive constants). Express your answer in terms of s and ω.

Show answer

This integral can be approached by first noting that ∫₀^∞ t sin(ωt)

e^{−st}

dt = −d/ds [∫₀^∞ sin(ωt)

e^{−st}

dt] by differentiation under the integral sign. The inner integral is the Laplace transform of sin(ωt), which equals ω/(s² + ω²). Differentiating with respect to s: −d/ds[ω/(s² + ω²)] = 2ωs/(s² + ω²)². Alternatively, one can apply integration by parts directly with u = t, dv = sin(ωt)

e^{−st}

dt, then evaluate the resulting integral using the cyclic technique for ∫

e^{−st}

sin(ωt) dt = ω/(s² + ω²). Either way, the answer is 2ωs/(s² + ω²)², valid for s > 0. This result is fundamental in analyzing second-order systems in electrical engineering.

Question 5: CRITICAL THINKING

Derive the reduction formula ∫ sinⁿ x dx = −(1/n) sinⁿ⁻¹ x cos x + ((n−1)/n) ∫ sinⁿ⁻² x dx using integration by parts. Then use your formula to explain why ∫₀^(π/2) sinⁿ x dx can be expressed as a ratio of factorials (Wallis's formula).

Show answer

Write sinⁿ x = sinⁿ⁻¹ x · sin x and set u = sinⁿ⁻¹ x, dv = sin x dx. Then du = (n−1) sinⁿ⁻² x cos x dx and v = −cos x. Integration by parts gives: ∫ sinⁿ x dx = −sinⁿ⁻¹ x cos x + (n−1) ∫ sinⁿ⁻² x cos² x dx. Replacing cos² x = 1 − sin² x: ∫ sinⁿ x dx = −sinⁿ⁻¹ x cos x + (n−1) ∫ sinⁿ⁻² x dx − (n−1) ∫ sinⁿ x dx. Moving the last term to the left side: n ∫ sinⁿ x dx = −sinⁿ⁻¹ x cos x + (n−1) ∫ sinⁿ⁻² x dx, which gives the reduction formula upon dividing by n. For the definite integral over [0, π/2], the boundary term vanishes (sin 0 = 0 and cos(π/2) = 0), so Iₙ = ((n−1)/n) Iₙ₋₂. Iterating from I₀ = π/2 or I₁ = 1, one obtains Wallis's formulas: for even n, Iₙ = [(n−1)!!/(n!!)] · (π/2); for odd n, Iₙ = (n−1)!!/(n!!). These can be unified using the Gamma function as Iₙ = √π · Γ((n+1)/2) / (2 · Γ(n/2 + 1)), connecting integration by parts to the factorial and its generalization.

SUMMARY

Summary & Key Concepts

Summary

Integration by parts is the integral analogue of the product rule, expressed as ∫ u dv = uv − ∫ v du. The technique transforms a difficult integral into a boundary term (uv) plus a new integral (∫ v du) that is ideally simpler. The LIATE mnemonic (Logarithmic → Inverse trig → Algebraic → Trigonometric → Exponential) guides the strategic choice of u—the factor to differentiate—ensuring that the resulting integral decreases in complexity. For integrands involving polynomials multiplied by exponentials or trigonometric functions, the tabular (DI) method organizes repeated applications efficiently.

Special situations include cyclic integrals (such as ∫ eˣ sin x dx), where the original integral reappears after two applications and can be solved algebraically, and reduction formulas (such as ∫ sinⁿ x dx), where integration by parts expresses an integral in terms of a simpler version of itself. Beyond Calculus 2, integration by parts underpins weak derivatives, Green's and Stokes' theorems, and the Gamma function, making it one of the most far-reaching single identities in all of mathematics.