Integration By Parts

Definition and Formula

Basic Statement

Integration by parts is a technique for integrating products of functions. Formula: ∫u dv = uv − ∫v du. Converts integral of product into simpler integrals.

Components

u = function chosen for differentiation. dv = remaining part chosen for integration. du = derivative of u. v = integral of dv.

Purpose

Simplifies integrals involving products where direct integration is difficult or impossible.

∫ u(x) dv(x) = u(x)v(x) − ∫ v(x) du(x)

Derivation from Product Rule

Product Rule Recap

Differentiation: d(uv)/dx = u dv/dx + v du/dx. Basis for integration by parts.

Rearrangement

Rearranged: u dv = d(uv) − v du. Integrate both sides to form integration by parts formula.

Integral Form

∫ u dv = uv − ∫ v du. Shows direct link between differentiation and integration operators.

d(uv) = u dv + v du∫ u dv = uv − ∫ v du

Indefinite Integrals

Usage

Applies to antiderivatives without limits. Result includes constant of integration.

Formula

∫ u dv = uv − ∫ v du + C. C represents arbitrary constant.

Example

∫ x e^x dx. Choose u = x, dv = e^x dx. Then du = dx, v = e^x.

∫ x e^x dx = x e^x − ∫ e^x dx = x e^x − e^x + C = e^x (x − 1) + C

Definite Integrals

Formula with Limits

∫_a^b u dv = [uv]_a^b − ∫_a^b v du. Evaluated at boundaries a and b.

Boundary Evaluation

Compute uv at limits, subtract integral of v du over same interval.

Example

∫_0^1 x e^x dx. Use u = x, dv = e^x dx.

∫_0^1 x e^x dx = [x e^x]_0^1 − ∫_0^1 e^x dx = (1)(e^1) − (0)(e^0) − (e^1 − e^0) = e − 0 − (e − 1) = 1

Choosing u and dv

LIATE Rule

Priority order: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential functions. Choose u highest in LIATE.

Rationale

Choosing u to simplify upon differentiation; dv easy to integrate.

Examples

For ∫ x ln(x) dx: u = ln(x) (logarithmic), dv = x dx (algebraic). Simplifies integral.

Common Strategies and Examples

Polynomial × Exponential

Example: ∫ x e^x dx. Polynomial reduces degree; exponential easy integration.

Logarithmic Functions

Example: ∫ ln(x) dx. Set u = ln(x), dv = dx. Simplifies integral.

Inverse Trigonometric Functions

Example: ∫ arctan(x) dx. Choose u = arctan(x), dv = dx.

Example: ∫ ln(x) dxu = ln(x), dv = dxdu = (1/x) dx, v = x∫ ln(x) dx = x ln(x) − ∫ x (1/x) dx = x ln(x) − x + C

Tabular Integration Method

Overview

Systematic tabular approach for repeated integration by parts. Useful for polynomials × exponentials/trigonometric.

Procedure

1. Differentiate u repeatedly until zero.
2. Integrate dv repeatedly.
3. Multiply entries diagonally with alternating signs.
4. Sum products to find integral.

Example

∫ x^3 e^x dx using tabular method.

Integral = + x^3 * e^x − 3x^2 * e^x + 6x * e^x − 6 * e^x + C= e^x (x^3 − 3x^2 + 6x − 6) + C

Reduction Formulas

Definition

Recursive relations reducing integral powers or orders step-by-step via integration by parts.

Example: ∫ sin^n x dx

Reduction formula relates ∫ sin^n x dx to ∫ sin^(n-2) x dx.

∫ sin^n x dx = − (1/n) sin^(n−1) x cos x + ((n−1)/n) ∫ sin^(n−2) x dx + C

Utility

Breaks down complex powers into simpler integrals iteratively.

Integration By Parts for Improper Integrals

Definition

Improper integrals have infinite limits or integrand singularities. Integration by parts applies with convergence considerations.

Conditions

Boundary terms uv|_a^b must converge or vanish. Integral ∫ v du must converge.

Example

∫_1^∞ (ln x)/x^2 dx. Use integration by parts to evaluate convergence and value.

u = ln x, dv = x^−2 dxdu = (1/x) dx, v = −x^−1∫_1^∞ (ln x)/x^2 dx = [− (ln x)/x]_1^∞ + ∫_1^∞ (1/x)(1/x) dx= 0 + ∫_1^∞ 1/x^2 dx = 1

Applications in Calculus and Beyond

Definite and Indefinite Integration

Solves integrals involving product functions, logarithmic, inverse trig, exponential, and polynomial terms.

Differential Equations

Used in solving linear differential equations with variable coefficients.

Physics and Engineering

Calculates work, energy, and solves integrals in signal processing and quantum mechanics.

Limitations and Pitfalls

Incorrect Choice of u and dv

Poor choices lead to more complicated integrals or infinite loops.

Non-Applicable Cases

Not useful when neither function simplifies upon differentiation/integration.

Convergence Issues

In improper integrals, boundary terms may diverge, invalidating method.

Practice Problems and Solutions

Problem 1

Evaluate ∫ x cos x dx.

Solution 1

u = x, dv = cos x dxdu = dx, v = sin x∫ x cos x dx = x sin x − ∫ sin x dx = x sin x + cos x + C

Problem 2

Evaluate ∫ e^x sin x dx.

Solution 2

Use integration by parts twice:Let I = ∫ e^x sin x dxu = sin x, dv = e^x dxdu = cos x dx, v = e^xI = e^x sin x − ∫ e^x cos x dxLet J = ∫ e^x cos x dxu = cos x, dv = e^x dxdu = −sin x dx, v = e^xJ = e^x cos x + ∫ e^x sin x dx = e^x cos x + ITherefore,I = e^x sin x − J = e^x sin x − e^x cos x − I2I = e^x (sin x − cos x)I = (e^x / 2) (sin x − cos x) + C

References

• Stewart, J. "Calculus: Early Transcendentals." Brooks Cole, 8th Edition, 2015, pp. 564-590.
• Thomas, G.B., Weir, M.D., Hass, J. "Thomas’ Calculus." Pearson, 14th Edition, 2017, pp. 520-545.
• Apostol, T.M. "Calculus, Volume 1." Wiley, 2nd Edition, 1967, pp. 180-210.
• Spivak, M. "Calculus." Publish or Perish, 3rd Edition, 1994, pp. 230-260.
• Strang, G. "Calculus." Wellesley-Cambridge Press, 2016, pp. 310-335.