Techniques of Integration

Differentiation is mechanical: give me a formula and a set of rules, and I can differentiate almost anything on autopilot. Integration is the opposite. There is no single algorithm that cracks every integral, so integration is less like following a recipe and more like a craft — you learn a toolbox of techniques and, crucially, how to recognize which tool a given problem is asking for. This page builds that toolbox: u-substitution, integration by parts, trigonometric integrals, and partial fractions, and then ties them together with a strategy for choosing among them.

Think of each technique as a way of rewriting an integral you cannot do into one you can. That reframing — "which move turns this unknown integral into a known one?" — is the entire game.

Learning Objectives

• Reverse the chain rule using u-substitution, including how to change limits for definite integrals.
• Apply integration by parts and use the LIATE guide to pick $u$ and $dv$.
• Evaluate common trigonometric integrals using power reductions and Pythagorean identities.
• Decompose rational functions with partial fractions and integrate the pieces.
• Diagnose an unfamiliar integral and choose the most efficient technique.

Quick Answer

The four workhorse techniques each undo a specific differentiation rule. U-substitution reverses the chain rule: spot an inner function $u$ whose derivative also appears, and the integral collapses. Integration by parts, $\int u\,dv = uv - \int v\,du$, reverses the product rule and is the go-to for products like $x e^x$ or $x\ln x$. Trigonometric integrals use identities to rewrite powers of sine, cosine, and their relatives into integrable forms. Partial fractions splits a complicated rational function into simple fractions you can integrate to logs and arctangents. Choosing well means reading the structure of the integrand: a composition suggests substitution, a product suggests parts, a fraction of polynomials suggests partial fractions.

Where It Came From

When Newton and Leibniz created calculus in the 1660s–1680s, they had the Fundamental Theorem — the insight that integration is the inverse of differentiation — but they did not yet have a systematic toolkit for actually computing antiderivatives. The 1700s were the century in which that toolkit was assembled, driven by a very practical need: the differential equations coming out of physics (planetary motion, vibrating strings, fluid flow, heat) produced integrals that nobody could evaluate by inspection, and progress in science stalled on them.

Integration by parts is the integral form of Leibniz's own product rule, made explicit as calculus was formalized. Substitution is the chain rule read backwards, and became routine as Euler standardized notation and pushed calculus into a computational discipline. Leonhard Euler (1707–1783) was the great systematizer here: his textbooks treated integration as a collection of learnable strategies rather than a bag of clever tricks. Partial fraction decomposition grew out of the algebra of rational functions — Johann Bernoulli and others recognized that every rational function integrates to a combination of logarithms and arctangents, and the decomposition made that guaranteed. By the early 1800s these methods were consolidated into essentially the form students learn today. The motivation throughout was never abstract elegance; it was the relentless supply of real integrals that applied mathematics kept generating.

U-Substitution: Reversing the Chain Rule

The chain rule says $\frac{d}{dx}f(g(x)) = f'(g(x))\,g'(x)$. Integration by substitution reads this backwards. If you can write the integrand as a function of some inner quantity $u$, multiplied by that quantity's derivative, you can substitute and simplify.

The mechanics: choose $u = g(x)$, compute $du = g'(x)\,dx$, and rewrite the entire integral in terms of $u$ until no $x$ remains.

Worked example. Evaluate $\int 2x\cos(x^2)\,dx$.

The inner function is $x^2$, and its derivative $2x$ is sitting right there. Let $u = x^2$, so $du = 2x\,dx$. The integral becomes

$\int \cos(u)\,du = \sin(u) + C = \sin(x^2) + C.$

Check by differentiating: $\frac{d}{dx}\sin(x^2) = 2x\cos(x^2)$. Correct.

Definite integrals — change the limits. For $\int_{0}^{2} x\,e^{x^2}\,dx$, let $u = x^2$, $du = 2x\,dx$, so $x\,dx = \tfrac12\,du$. When $x=0$, $u=0$; when $x=2$, $u=4$. Then

$\int_{0}^{2} x\,e^{x^2}\,dx = \frac12\int_{0}^{4} e^{u}\,du = \frac12\left(e^{4} - e^{0}\right) = \frac{e^{4}-1}{2}.$

Converting the limits saves you from having to back-substitute at the end.

Integration by Parts: Reversing the Product Rule

From the product rule $(uv)' = u'v + uv'$, integrating both sides and rearranging gives the parts formula:

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

The art is choosing which factor to call $u$ (which you will differentiate) and which to call $dv$ (which you will integrate). A reliable guide is LIATE — pick $u$ as whichever type appears first: Logarithmic, Inverse trig, Algebraic (polynomials), Trigonometric, Exponential. The goal is for $\int v\,du$ to be simpler than what you started with.

Worked example. Evaluate $\int x e^{x}\,dx$.

Here $x$ is algebraic and $e^x$ is exponential, so LIATE picks $u = x$, $dv = e^x\,dx$. Then $du = dx$ and $v = e^x$:

$\int x e^{x}\,dx = x e^{x} - \int e^{x}\,dx = x e^{x} - e^{x} + C = e^{x}(x-1) + C.$

A subtler case — the logarithm. Evaluate $\int \ln x\,dx$. There seems to be no product, but write it as $\ln x \cdot 1$. LIATE says $u = \ln x$, $dv = dx$, giving $du = \tfrac{1}{x}\,dx$ and $v = x$:

$\int \ln x\,dx = x\ln x - \int x\cdot\frac{1}{x}\,dx = x\ln x - \int 1\,dx = x\ln x - x + C.$

Sometimes parts must be applied twice (e.g. $\int x^2 e^x\,dx$), and occasionally the original integral reappears and you solve for it algebraically (a "boomerang," as in $\int e^x\sin x\,dx$).

Trigonometric Integrals: Identities Do the Work

Powers of trig functions rarely integrate directly, but the right identity converts them into a substitution problem. Two situations dominate.

Odd power of sine or cosine. For $\int \sin^3 x\,dx$, peel off one factor and convert the rest using $\sin^2 x = 1 - \cos^2 x$:

$\int \sin^3 x\,dx = \int (1-\cos^2 x)\sin x\,dx.$

Now let $u = \cos x$, $du = -\sin x\,dx$:

$= -\int (1 - u^2)\,du = -\left(u - \frac{u^3}{3}\right) + C = -\cos x + \frac{\cos^3 x}{3} + C.$

Even powers — use the half-angle formula. For $\int \cos^2 x\,dx$, there is no single factor to peel, so apply $\cos^2 x = \tfrac{1+\cos 2x}{2}$:

$\int \cos^2 x\,dx = \frac12\int (1 + \cos 2x)\,dx = \frac12\left(x + \frac{\sin 2x}{2}\right) + C = \frac{x}{2} + \frac{\sin 2x}{4} + C.$

The same philosophy handles powers of $\tan x$ and $\sec x$ using $\sec^2 x = 1 + \tan^2 x$ and the fact that $\frac{d}{dx}\tan x = \sec^2 x$.

Partial Fractions: Splitting Rational Functions

When you integrate a ratio of polynomials $\frac{P(x)}{Q(x)}$ (with the degree of $P$ less than that of $Q$), factor the denominator and break the fraction into a sum of simpler pieces — each of which integrates to a logarithm or arctangent.

Worked example. Evaluate $\int \frac{5x - 4}{x^2 - x - 2}\,dx$.

Factor: $x^2 - x - 2 = (x-2)(x+1)$. Set up the decomposition

$\frac{5x-4}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1}.$

Multiply through: $5x - 4 = A(x+1) + B(x-2)$. Substitute $x = 2$: $6 = 3A$, so $A = 2$. Substitute $x = -1$: $-9 = -3B$, so $B = 3$. Therefore

$\int \frac{5x-4}{x^2-x-2}\,dx = \int\left(\frac{2}{x-2} + \frac{3}{x+1}\right)dx = 2\ln|x-2| + 3\ln|x+1| + C.$

For repeated factors you include terms like $\frac{B}{(x-2)^2}$, and for an irreducible quadratic $x^2 + a^2$ you use a numerator $Bx + C$, which leads to an arctangent. If the numerator's degree is not smaller than the denominator's, do polynomial long division first.

Real-World Applications

• Physics and engineering: Solving differential equations for circuits, springs, and heat flow constantly requires integration by parts and partial fractions; the $\frac{1}{s^2+a^2}$ terms in Laplace transforms invert via partial fractions.
• Probability and statistics: Expected values and moments of continuous distributions (e.g. $\int x e^{-\lambda x}\,dx$ for the exponential distribution) are textbook integration-by-parts problems.
• Signal processing: Fourier analysis leans on trigonometric integrals to compute the coefficients that decompose signals into frequencies.
• Chemical kinetics and biology: Rate equations produce rational integrands that partial fractions untangle into logarithmic growth or decay solutions.

Common Mistakes

1. Forgetting to convert $dx$ in a substitution. Students replace $x^2$ with $u$ but leave a stray $dx$ or an extra factor of $x$. Every $x$ — including the differential — must be expressed in terms of $u$. Correction: always compute $du$ explicitly and substitute for it before integrating.

2. Choosing $u$ and $dv$ backwards in integration by parts. Picking $u = e^x$ in $\int x e^x\,dx$ makes $\int v\,du$ harder, not easier. Correction: use LIATE, and sanity-check that the new integral is genuinely simpler.

3. Not adjusting the limits of a definite integral after substituting. Writing $\int_0^2 \dots = \tfrac12\int_0^2 e^u\,du$ uses the old limits with the new variable — a common and costly slip. Correction: either change the limits to match $u$, or convert back to $x$ before plugging in.

Comparison and Connections

Each technique targets a recognizable structure. Reading that structure is how you choose quickly.

Technique	Best clue in the integrand	Reverses	Typical result
U-substitution	A composition where the inner function's derivative also appears	Chain rule	Same form, simpler
Integration by parts	A product of unlike types (e.g. polynomial $\times$ exponential)	Product rule	$uv - \int v\,du$
Trig integrals	Powers of $\sin, \cos, \tan, \sec$	— (identity-driven)	Trig antiderivative
Partial fractions	A ratio of polynomials	— (algebra-driven)	Logs and arctangents

Decision strategy: First try to simplify or spot a substitution — it is the fastest and most common. If the integrand is a product of different function types, reach for parts. If it is all trig, use identities. If it is a rational function, factor and decompose. Many hard integrals chain these: substitution often sets up an integral you then finish by parts.

Practice Questions

Recall

State the integration by parts formula and identify what LIATE stands for.

Answer: $\int u\,dv = uv - \int v\,du$. LIATE = Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential — the priority order for choosing $u$.

Understanding

Explain why $\int \cos^2 x\,dx$ needs the half-angle identity but $\int \cos^3 x\,dx$ does not.

Guidance: An odd power leaves a single $\cos x$ (or $\sin x$) factor to serve as $du$ after converting the rest with a Pythagorean identity. An even power has no leftover factor, so you must reduce the power first using $\cos^2 x = \tfrac{1+\cos 2x}{2}$.

Application

Evaluate $\int x\cos x\,dx$.

Answer: Parts with $u = x$, $dv = \cos x\,dx$ gives $x\sin x - \int \sin x\,dx = x\sin x + \cos x + C$.

Analysis

Which technique(s) does $\int \frac{x}{x^2+1}\,dx$ call for, and what is the answer?

Answer: Substitution, not partial fractions: let $u = x^2+1$, $du = 2x\,dx$, so the integral is $\tfrac12\ln(x^2+1) + C$. Recognizing that the numerator is (half) the denominator's derivative avoids unnecessary decomposition.

FAQ

How do I know which technique to use? Read the structure. A composition with its derivative present means substitution; a product of unlike functions means parts; pure trig means identities; a polynomial fraction means partial fractions. When unsure, try substitution first — it is the most frequently useful.

What if two techniques seem to apply? They often both work, and one is usually shorter. $\int \frac{x}{x^2+1}\,dx$ can be forced through other routes, but substitution is instant. Efficiency comes with practice; there is rarely a single "legal" method.

When does integration by parts loop back to the start? With integrands like $e^x\sin x$, applying parts twice reproduces the original integral. That is not failure — set the result equal to the original and solve for the integral algebraically.

Do I always need partial fractions for a fraction? No. If the numerator is the derivative of the denominator (or a constant multiple), a simple substitution to a logarithm is faster. Use partial fractions when the denominator factors and no direct substitution presents itself.

Why does my answer look different from the textbook's? Antiderivatives are unique only up to a constant, and trig identities can disguise equivalent expressions. Differentiate your answer — if you recover the integrand, you are correct.

Quick Revision

• U-substitution: $u = g(x)$, $du = g'(x)\,dx$; reverses the chain rule; change limits for definite integrals.
• Parts: $\int u\,dv = uv - \int v\,du$; choose $u$ by LIATE.
• Odd trig power: peel one factor, convert the rest with $\sin^2 + \cos^2 = 1$.
• Even trig power: reduce with $\cos^2 x = \tfrac{1+\cos 2x}{2}$, $\sin^2 x = \tfrac{1-\cos 2x}{2}$.
• Partial fractions: factor denominator, solve for constants, integrate to logs/arctangents; long-divide first if degree of top $\geq$ degree of bottom.
• Always verify by differentiating.

Related Topics

Prerequisites

• Calculus overview
• The Fundamental Theorem of Calculus and basic antiderivatives
• Trigonometric and Pythagorean identities

Related Topics

• Applications of integration (area, volume, arc length)
• Differential equations, where these techniques are indispensable

Next Topics

• Improper integrals and convergence
• Numerical integration for integrals with no elementary antiderivative
• Sequences and series, including Taylor series