What Are The Methods Of Integration?
The methods of integration are the systematic techniques for finding the indefinite integral of a function when no single formula applies directly. Integration reverses differentiation, so most of these methods are a differentiation rule read backwards. The goal each time is the same: rewrite the integral until it matches something you already know how to integrate.
There are six techniques that cover almost every integral met in a first calculus course:
Standard formulas: the integral already matches a known antiderivative.
Substitution: the reverse of the chain rule, for a function paired with its own derivative.
Integration by parts: the reverse of the product rule, for a product of unlike functions.
Partial fractions: splits a rational function into simple fractions.
Trigonometric substitution: clears a square root of the form $\sqrt{a^2 - x^2}$, $\sqrt{a^2 + x^2}$, or $\sqrt{x^2 - a^2}$.
Trigonometric integrals: powers and products of sine, cosine, and their relatives, tamed with identities.
Every indefinite integral below carries a constant of integration $+C$, because differentiation destroys constants and integration cannot recover which one was there. For the wider map of the subject, see the calculus overview, and for the master list of results the standard formulas draw on, the list of integrals.
How Do You Integrate Using Standard Formulas?
The first method is to recognise the integral outright. A large share of integrals match a known antiderivative once you have memorised the basic table, so always check this before reaching for anything harder.
The power rule is the workhorse: for any $n \neq -1$,
$$\int x^n , dx = \frac{x^{n+1}}{n+1} + C$$
Example 1: A direct power-rule integral.
$$\int x^4 , dx = \frac{x^{5}}{5} + C$$
Check by differentiating the answer back: $\dfrac{d}{dx}\left(\dfrac{x^5}{5}\right) = \dfrac{5x^4}{5} = x^4$, the original integrand, so the antiderivative is correct.
Final answer: $\displaystyle\int x^4 , dx = \dfrac{x^{5}}{5} + C$.
Alongside the power rule sit $\int e^x,dx = e^x + C$, $\int \frac{1}{x},dx = \ln\lvert x\rvert + C$, $\int \cos x,dx = \sin x + C$, and $\int \sin x,dx = -\cos x + C$. When an integral does not fit one of these, the remaining methods exist to reshape it until it does. A fuller reference lives at basic integration formulas.
How Do You Integrate By Substitution?
Substitution is the chain rule run in reverse. Use it when the integrand contains a function $u = g(x)$ together with a multiple of its derivative $g'(x)$. You rename the inner function $u$, replace $g'(x),dx$ with $du$, integrate in $u$, then convert back to $x$.
$$\int f\big(g(x)\big),g'(x),dx = \int f(u),du, \qquad u = g(x)$$
Example 2: Substitution.
Evaluate $\displaystyle\int 2x\cos\left(x^2\right),dx$. The inner function is $u = x^2$, whose derivative $2x$ is sitting right there, so let $u = x^2$ and $du = 2x,dx$:
$$\int 2x\cos\left(x^2\right),dx = \int \cos u , du = \sin u + C = \sin\left(x^2\right) + C$$
Check: $\dfrac{d}{dx}\sin\left(x^2\right) = \cos\left(x^2\right)\cdot 2x = 2x\cos\left(x^2\right)$, the integrand, so the result holds.
Final answer: $\displaystyle\int 2x\cos\left(x^2\right),dx = \sin\left(x^2\right) + C$.
The deep version of this method, with more cases and definite-integral limits, is covered in integration by substitution.
How Do You Integrate By Parts?
Integration by parts is the product rule run in reverse, for an integrand that is a product of two unlike functions, such as a polynomial times an exponential, or a polynomial times a logarithm.
$$\int u , dv = uv - \int v , du$$
The trick is choosing which factor is $u$. The LIATE order, Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential, ranks candidates for $u$: pick whichever appears first in that list, and the remaining factor becomes $dv$.
Example 3: Integration by parts.
Evaluate $\displaystyle\int x,e^x,dx$. Here $x$ is algebraic and $e^x$ is exponential, so LIATE makes $u = x$ and $dv = e^x,dx$. Then $du = dx$ and $v = e^x$:
$$\int x,e^x,dx = x,e^x - \int e^x,dx = x,e^x - e^x + C = e^x(x - 1) + C$$
Check: $\dfrac{d}{dx}\big[e^x(x-1)\big] = e^x(x-1) + e^x = e^x,x = x,e^x$, the integrand, so the answer is verified.
Final answer: $\displaystyle\int x,e^x,dx = e^x(x - 1) + C$.
The full treatment, including the $\int u,dv$ setup with logarithms and repeated parts, is in integration by parts.
How Do You Integrate Using Partial Fractions?
Partial fractions handle a proper rational function, a ratio of polynomials where the numerator has lower degree than the denominator. You factor the denominator, split the fraction into a sum of simpler pieces with unknown numerators, solve for those numerators, then integrate each piece with a logarithm or power rule.
Example 4: Partial fractions.
Evaluate $\displaystyle\int \dfrac{1}{x^2 - 1},dx$. Factor the denominator as $(x-1)(x+1)$ and split:
$$\frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$$
Clearing denominators gives $A(x+1) + B(x-1) = 1$. Setting $x = 1$ gives $2A = 1$, so $A = \tfrac{1}{2}$; setting $x = -1$ gives $-2B = 1$, so $B = -\tfrac{1}{2}$. Integrate each simple fraction:
$$\int \frac{1}{x^2 - 1},dx = \frac{1}{2}\ln\lvert x - 1\rvert - \frac{1}{2}\ln\lvert x + 1\rvert + C = \frac{1}{2}\ln\left\lvert \frac{x-1}{x+1} \right\rvert + C$$
Check: differentiating $\tfrac{1}{2}\big(\ln\lvert x-1\rvert - \ln\lvert x+1\rvert\big)$ gives $\tfrac{1}{2}\left(\tfrac{1}{x-1} - \tfrac{1}{x+1}\right) = \tfrac{1}{2}\cdot\tfrac{2}{x^2-1} = \tfrac{1}{x^2-1}$, the integrand.
Final answer: $\displaystyle\int \dfrac{1}{x^2 - 1},dx = \dfrac{1}{2}\ln\left\lvert \dfrac{x-1}{x+1} \right\rvert + C$.
For the full decomposition method, including repeated and quadratic factors, see integration by partial fractions and the broader integration of rational functions.
How Do You Integrate Using Trigonometric Substitution?
Trigonometric substitution clears an awkward square root by turning it into a trig identity. A root of the form $\sqrt{a^2 - x^2}$ calls for $x = a\sin\theta$, because $a^2 - a^2\sin^2\theta = a^2\cos^2\theta$ collapses the root. Roots $\sqrt{a^2 + x^2}$ and $\sqrt{x^2 - a^2}$ use $x = a\tan\theta$ and $x = a\sec\theta$.
Example 5: Trigonometric substitution.
Evaluate $\displaystyle\int \dfrac{dx}{\sqrt{9 - x^2}}$. With $a = 3$, let $x = 3\sin\theta$, so $dx = 3\cos\theta,d\theta$ and $\sqrt{9 - x^2} = 3\cos\theta$:
$$\int \frac{dx}{\sqrt{9 - x^2}} = \int \frac{3\cos\theta}{3\cos\theta},d\theta = \int d\theta = \theta + C = \arcsin\left(\frac{x}{3}\right) + C$$
Check: $\dfrac{d}{dx}\arcsin\left(\dfrac{x}{3}\right) = \dfrac{1}{\sqrt{1 - (x/3)^2}}\cdot\dfrac{1}{3} = \dfrac{1}{3}\cdot\dfrac{3}{\sqrt{9 - x^2}} = \dfrac{1}{\sqrt{9 - x^2}}$, the integrand.
Final answer: $\displaystyle\int \dfrac{dx}{\sqrt{9 - x^2}} = \arcsin\left(\dfrac{x}{3}\right) + C$.
The method with all three root shapes is worked in full at trigonometric substitution.
How Do You Integrate Powers Of Trigonometric Functions?
Integrals of powers and products of sine, cosine, tangent, and secant rarely match a table row directly. The move is to rewrite them with a trig identity until they do. The most common tool is the power-reducing identity $\sin^2 x = \dfrac{1 - \cos 2x}{2}$.
Example 6: A trigonometric integral.
Evaluate $\displaystyle\int \sin^2 x , dx$. Replace $\sin^2 x$ using the identity:
$$\int \sin^2 x , dx = \int \frac{1 - \cos 2x}{2},dx = \frac{1}{2}\left(x - \frac{1}{2}\sin 2x\right) + C = \frac{x}{2} - \frac{1}{4}\sin 2x + C$$
Check: $\dfrac{d}{dx}\left(\dfrac{x}{2} - \dfrac{1}{4}\sin 2x\right) = \dfrac{1}{2} - \dfrac{1}{4}\cdot 2\cos 2x = \dfrac{1}{2} - \dfrac{1}{2}\cos 2x = \dfrac{1 - \cos 2x}{2} = \sin^2 x$, the integrand.
Final answer: $\displaystyle\int \sin^2 x , dx = \dfrac{x}{2} - \dfrac{1}{4}\sin 2x + C$.
The identities and patterns for higher powers and mixed products are collected in integration of trigonometric functions.
Which Method Of Integration Should You Choose?
Most of the difficulty in integration is not the algebra, it is reading the integrand and picking the right tool. This table settles the choice by shape.
Table: Which method of integration to try, based on what the integrand looks like.
The integrand looks like | Try this method | First move |
|---|---|---|
A known antiderivative ($x^n$, $e^x$, $\sin x$, $\tfrac{1}{x}$) | Standard formula | Match the table row |
A function and (a multiple of) its own derivative | Substitution | Let $u = $ the inner function |
A product of unlike functions (polynomial $\times\ e^x$, $\ln x$, or trig) | Integration by parts | Choose $u$ by LIATE |
A proper rational function $\dfrac{P(x)}{Q(x)}$ with $Q$ factorable | Partial fractions | Factor $Q$, then split |
A square root $\sqrt{a^2 - x^2}$, $\sqrt{a^2 + x^2}$, or $\sqrt{x^2 - a^2}$ | Trig substitution | Set $x = a\sin\theta$, $a\tan\theta$, or $a\sec\theta$ |
Powers or products of $\sin$, $\cos$, $\tan$, $\sec$ | Trigonometric integrals | Rewrite with an identity |
Read the table top to bottom. Check for a standard formula first, then substitution, since it is the fastest reshaping tool, and only reach for parts, partial fractions, or a trig substitution when the simpler moves fail. Methods also combine: a substitution can turn an integral into a rational function, which then needs partial fractions.
Why Do The Methods Of Integration Work?
The methods are not a bag of unrelated tricks. Each standard technique is a differentiation rule turned around, which is why the same short list keeps reappearing.
Substitution is the chain rule backwards. The chain rule differentiates $F\big(g(x)\big)$ into $F'\big(g(x)\big)g'(x)$. Integration by substitution reads that equation right to left, recovering $F\big(g(x)\big)$ from the product of the outer derivative and the inner derivative.
Integration by parts is the product rule backwards. Differentiating $uv$ gives $u,dv + v,du$. Rearranging and integrating both sides leaves $\int u,dv = uv - \int v,du$, which trades a hard integral for an easier one.
Partial fractions and trig substitution are reshaping moves. Neither reverses a single rule; each rewrites the integrand into pieces the table above can finish. Partial fractions turn one rational function into a sum of logarithms, and trig substitution turns a square root into a clean trig integral.
There is also a geometric reading. A definite integral is the signed area under a curve, and every method here is a way of changing variables or splitting a region so that the area becomes computable. Substitution rescales the horizontal axis, parts swaps which piece accumulates, and the fundamental theorem of calculus then turns any of these antiderivatives into an exact area by a single subtraction.
Who Shaped The Methods Of Integration?
The techniques were assembled over a century, from the notation that made substitution natural to the first textbook that organised the methods into a subject.
Two figures anchor the timeline:
Gottfried Wilhelm Leibniz (1646 to 1716, Germany) gave integration its notation and framed substitution as a manipulation of differentials.
Leonhard Euler (1707 to 1783, Switzerland) wrote the systematic integral-calculus textbook that grouped the techniques into the method set taught today.
Where Are The Methods Of Integration Used In The Real World?
Every field that measures a total from a rate ends up choosing an integration method to evaluate it.
Physics and engineering: finding displacement from velocity, work from a variable force, or the centre of mass of a shape usually needs substitution or parts before the fundamental theorem of calculus finishes the job.
Electronics and signals: the average power of an alternating current comes from integrating $\sin^2$ or $\cos^2$ over a cycle, the exact trig-integral move shown above.
Probability and statistics: normalising a probability density, or finding an expected value, routinely calls for substitution and integration by parts.
Economics: recovering total cost or total revenue from a marginal-rate function is an integration problem, and rational marginal models bring in partial fractions.
Computer graphics and geometry: arc length, surface area, and volumes of revolution generate square-root integrands that trigonometric substitution is built to clear.
One small toolkit of techniques serves motion, circuits, risk, markets, and rendering, which is why every quantitative degree teaches it early.
What Are The Most Common Mistakes With The Methods Of Integration?
These four errors account for most lost marks on integration, and each matches a question real students ask on r/calculus, r/learnmath, and university calculus handouts.
Forgetting to change $dx$ when you substitute.
Where it slips in:
A student sets $u = x^2$ but keeps writing $dx$, integrating a mix of $u$ and $x$ that means nothing.
Don't do this:
Do not swap the variable halfway. Once $u$ enters, every $x$ and the $dx$ must be gone.
The correct way:
Compute $du = g'(x),dx$ and use it to replace the whole $g'(x),dx$ block, so the integral is written purely in $u$ before you integrate.
Not substituting back to the original variable.
Where it slips in:
A student integrates in $u$, reaches $\sin u + C$, and stops, leaving the answer in a variable the question never used.
Don't do this:
Do not leave $u$ in a final indefinite answer. The result must be a function of $x$.
The correct way:
Replace $u$ with its definition at the end: for $u = x^2$, write $\sin u + C = \sin\left(x^2\right) + C$.
Choosing $u$ and $dv$ the wrong way round in parts.
Where it slips in:
A student sets $u = e^x$ and $dv = x,dx$, so the new integral $\int v,du$ is harder than the original.
Don't do this:
Do not pick $u$ at random. A poor choice makes parts loop or grow.
The correct way:
Use LIATE: choose $u$ as the function appearing earliest in Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential, so its derivative simplifies and $dv$ stays integrable.
Dropping the constant of integration.
Where it slips in:
A student finishes an indefinite integral and writes the antiderivative with no $+C$.
Don't do this:
Do not omit $+C$ on an indefinite integral. Infinitely many antiderivatives differ by a constant.
The correct way:
Add $+C$ to every indefinite result. The reason is spelled out under constant of integration; it only disappears when you evaluate a definite integral between two limits.
Practice Problems On The Methods Of Integration
Work each one, name the method, then check by differentiating your answer. Answers are verified.
$\displaystyle\int x^5 , dx$.
(Standard formula. Answer: $\dfrac{x^6}{6} + C$.)$\displaystyle\int 3x^2\cos\left(x^3\right),dx$.
(Substitution, $u = x^3$. Answer: $\sin\left(x^3\right) + C$.)$\displaystyle\int x,\ln x , dx$.
(By parts, $u = \ln x$. Answer: $\dfrac{x^2}{2}\ln x - \dfrac{x^2}{4} + C$.)$\displaystyle\int \dfrac{1}{x^2 - 4},dx$.
(Partial fractions. Answer: $\dfrac{1}{4}\ln\left\lvert\dfrac{x-2}{x+2}\right\rvert + C$.)$\displaystyle\int \dfrac{dx}{\sqrt{4 - x^2}}$.
(Trig substitution, $x = 2\sin\theta$. Answer: $\arcsin\left(\dfrac{x}{2}\right) + C$.)$\displaystyle\int \cos^2 x , dx$.
(Trig integral, $\cos^2 x = \tfrac{1+\cos 2x}{2}$. Answer: $\dfrac{x}{2} + \dfrac{1}{4}\sin 2x + C$.)
Where Should You Go Next After The Methods Of Integration?
Each method has a page that goes deeper than this overview, and the natural next steps are the two you will use most.
Integration by substitution. The fastest and most-used method, with definite-integral limits and harder inner functions.
Integration by parts. The reverse product rule in full, including repeated parts and the $\int u,dv$ setup for logarithms.
Integration by partial fractions. The decomposition method for rational functions with repeated and quadratic factors.
If your child is meeting the methods of integration for the first time, a live Bhanzu trainer teaches them from the decision table up, so choosing a technique becomes a habit rather than a guess, in the Bhanzu math program.
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