Integration Techniques: Substitution, Parts & More

Why Techniques Are Needed

Differentiation follows mechanical rules: the derivative of any elementary function can be computed by applying chain, product, and quotient rules systematically.

Integration has no such universal algorithm.

THEOREM

Some elementary functions have no elementary antiderivative.

Others have antiderivatives that are difficult to find without insight.

Techniques bridge this gap. They transform integrands into forms matching known formulas. The transformation might involve changing variables, splitting products, or decomposing fractions—whatever reveals the underlying structure.

Substitution (u-Substitution)

Substitution reverses the chain rule. If the integrand contains a function and its derivative, substitution simplifies.

Method: Let

u = g(x)

, so

du = g'(x)\, dx

. Replace all

x

-expressions with

u

-expressions and integrate.

Substitution Rule

\int f(g(x))\, g'(x)\, dx = \int f(u)\, du \quad \text{with } u = g(x)

Learn more about this formula: Substitution Rule →

Browse all integral formulas →

Example

\int 2x \cos(x^2)\, dx

Let

u = x^2

, so

du = 2x\, dx

:

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

For definite integrals, convert the limits: when

x = a

,

u = g(a)

; when

x = b

,

u = g(b)

.

Pitfall: Keeping the x-bounds after substituting

Evaluating

\sin u

from

0

to

2

, the old

x

-bounds of

\int_0^2 2x\cos(x^2)\, dx

, gives

\sin 2 \approx 0.909

. In

u = x^2

the bounds are

0

and

4

, and the integral is

\sin 4 \approx -0.757

.

The bounds of a definite integral are values of the variable of integration. In

\int_0^2 2x\cos(x^2)\, dx

the numbers

0

and

2

are values of

x

; after the substitution

u = x^2

the integral runs over values of

u

, and the antiderivative

\sin u

is a function of

u

. Converting the bounds at the moment of substitution means the evaluation never has to return to

x

.

Notation Inside the Techniques

Notation

Notation Inside the Techniques

Substitution's paired equations, the four-slot ledger of parts, and the substitution that runs backwards — the bookkeeping is the technique.

The integral signs — indefinite and definite; the algebra of

du

— differentials.

u = g(x)

,

\; du = g'(x)\,dx

Let u equal g of x, so d u equals g prime of x, d x

A change of name recorded as two equations — and the second is real differential algebra, not ritual. Everything must convert: integrand,

dx

, all of it, as in Substitution above.

CasesIn a definite integral the bounds convert too, and the transition is often labelled on the sign itself:

\int_{x=a}^{x=b}

becomes

\int_{u=g(a)}^{u=g(b)}

— after which there is no going back to

x

. The alternative workflow keeps the

x

-bounds and back-substitutes

u = g(x)

before evaluating; both are legal, mixing them is not.

Also written

w

or

t

when

u

is spoken for; the final return to

x

is the “back-substitution” step.

Do not confuseA partial conversion.

\int \cos u \,dx

— one variable in the integrand, another in the differential — is meaningless; a leftover

x

under a

du

means the substitution was not completed.

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

The integral of u d v equals u v minus the integral of v d u

Parts factors the integrand into two slots:

u

, destined for differentiation, and

dv

, destined for integration — and

dv

swallows the

dx

. The four-slot ledger

u, du, dv, v

organises the whole computation in Integration by Parts below.

CasesChoosing

u

follows LIATE — logarithmic, inverse trig, algebraic, trigonometric, exponential — a classroom mnemonic, not a theorem; it fails often enough to keep judgment in the loop.

Also written

\int u\,v'\,dx = uv - \int u'\,v\,dx

— the prime form, standard in European texts, where the differential never appears bare.

Do not confuseSubstitution's

u

. Same letter, different job: substitution's

u

renames the variable; parts'

u

labels a factor and

x

remains the variable throughout. Writing

dv

without its

dx

is the commonest ledger slip.

x = a\sin\theta

Let x equal a sine theta

Substitution running backwards: the old variable is expressed through the new one, opposite in direction to

u = g(x)

. The radical collapses through a Pythagorean identity, as in Trigonometric Substitution below.

CasesThree radical shapes, three choices:

\sqrt{a^2 - x^2} \to x = a\sin\theta

;

\sqrt{a^2 + x^2} \to x = a\tan\theta

;

\sqrt{x^2 - a^2} \to x = a\sec\theta

. The answer returns to

x

through

\theta = \arcsin\frac{x}{a}

and a reference triangle.

Also writtenHyperbolic versions —

x = a\sinh t

— in analysis and European texts, trading the identity

1 - \sin^2 = \cos^2

for

\cosh^2 - \sinh^2 = 1

.

Do not confuseThe

u

-direction. Here the new variable is not a function of

x

you can read off — inverting requires the inverse trig functions, and the domain restriction on

\theta

is what makes the inversion legal.

All calculus symbols →↑ Integrals

Integration by Parts

Integration by parts reverses the product rule:

Integration by Parts

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

Learn more about this formula: Integration by Parts →

Browse all integral formulas →

Method: Identify factors

u

and

dv

in the integrand. Differentiate

u

to get

du

; integrate

dv

to get

v

. Apply the formula.

Example

\int x e^x\, dx

Let

u = x

and

dv = e^x\, dx

. Then

du = dx

and

v = e^x

:

= xe^x - \int e^x\, dx = xe^x - e^x + C = e^x(x - 1) + C

Pitfall: Dropping the minus sign

Adding the second integral instead of subtracting it gives

\int x e^x\, dx = xe^x + e^x + C

, whose derivative is

xe^x + 2e^x

. Subtracting gives

xe^x - e^x + C

, whose derivative is

xe^x

.

The formula comes from integrating the product rule

(uv)' = u'v + uv'

and solving for

\int u\, dv

, which is why the second integral is subtracted. In the example,

u = x

and

v = e^x

give

uv = xe^x

, and the integral left over,

\int v\, du = \int e^x\, dx

, is simple enough to finish at once.

The formula accepts any split of the integrand, even an odd one. For

\int \frac{1}{x}\, dx

, take

u = \frac{1}{x}

and

dv = dx

; then

du = -\frac{1}{x^2}\, dx

and

v = x

, and the formula gives

\int \frac{1}{x}\, dx = x \cdot \frac{1}{x} - \int x \left(-\frac{1}{x^2}\right) dx = 1 + \int \frac{1}{x}\, dx

. The split is legal, but it only returns the integral it started from.

Fact: Parts seems to prove

0 = 1

Subtracting

\int \frac{1}{x}\, dx

from both sides of

\int \frac{1}{x}\, dx = 1 + \int \frac{1}{x}\, dx

leaves

0 = 1

. Nothing is broken: each indefinite integral stands for the whole family

\ln|x| + C

, and two members of that family may differ by

1

.

Choosing the first factor: LIATE guides selection—Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential. Earlier types typically make better choices for

u

.

The two splits above show what the choice decides. In both, the formula moves the derivative from

v

to

u

: with

u = x

in

\int x e^x\, dx

the integral left behind is

\int e^x\, dx

, simpler than the original, while

u = \frac{1}{x}

in

\int \frac{1}{x}\, dx

leaves the original integral unchanged. A choice is good when the factor chosen as

u

becomes simpler after differentiation and

dv

can still be integrated.

Integration by parts moves the derivative from one factor to the other, and it pays off only when the integral left behind is simpler.

Trigonometric Integrals

Integrals involving powers of sine and cosine require strategic use of identities.

Odd power of sine: Save one

\sin x

, convert remaining

\sin^2 x = 1 - \cos^2 x

, substitute

u = \cos x

.

Odd power of cosine: Save one

\cos x

, convert remaining

\cos^2 x = 1 - \sin^2 x

, substitute

u = \sin x

.

Both powers even: Use half-angle identities:

Trigonometric Integrals

\sin^2 x = \frac{1 - \cos 2x}{2} \qquad \cos^2 x = \frac{1 + \cos 2x}{2}

Example

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

Let

u = \cos x

:

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

Pitfall: Losing the sign of du

Taking

du = \sin x\, dx

for

u = \cos x

gives

\int \sin^3 x\, dx = \cos x - \frac{\cos^3 x}{3} + C

, whose derivative is

-\sin^3 x

. With

du = -\sin x\, dx

the answer is

-\cos x + \frac{\cos^3 x}{3} + C

.

The saved factor

\sin x

is what makes the substitution possible: together with

dx

it is matched against

du

for

u = \cos x

, and the remaining even power becomes a polynomial in

u

. For an odd power of cosine the roles swap, and

\cos x\, dx

is matched against

du

for

u = \sin x

. In both cases the answer can be checked by differentiating it back to the integrand.

Power pattern	Strategy	Substitution
Odd power of sin x	save one sin x, rewrite the rest using sin²x = 1 − cos²x	u = cos x, du = −sin x dx
Odd power of cos x	save one cos x, rewrite the rest using cos²x = 1 − sin²x	u = sin x, du = cos x dx
Both powers even	apply half-angle identities to lower the powers	none directly — reduce, then integrate term by term

Trigonometric Substitution

Square roots of quadratics suggest trigonometric substitutions.

For

\sqrt{a^2 - x^2}

: Let

x = a\sin\theta

, so

\sqrt{a^2 - x^2} = a\cos\theta

For

\sqrt{a^2 + x^2}

: Let

x = a\tan\theta

, so

\sqrt{a^2 + x^2} = a\sec\theta

For

\sqrt{x^2 - a^2}

: Let

x = a\sec\theta

, so

\sqrt{x^2 - a^2} = a\tan\theta

Example

\int \frac{1}{\sqrt{1 - x^2}}\, dx

Let

x = \sin\theta

, so

dx = \cos\theta\, d\theta

and

\sqrt{1 - x^2} = \cos\theta

:

= \int \frac{\cos\theta}{\cos\theta}\, d\theta = \int d\theta = \theta + C = \arcsin x + C

Pitfall: Replacing dx with dθ

Swapping

dx

for

d\theta

without the factor

\cos\theta

leaves

\int \frac{d\theta}{\cos\theta} = \ln|\sec\theta + \tan\theta| + C

, a different function. With

dx = \cos\theta\, d\theta

the cosines cancel and the answer is

\arcsin x + C

.

The substitution

x = \sin\theta

is applied to the whole integral. It rewrites the radical

\sqrt{1 - x^2}

as

\cos\theta

, and differentiating it gives the differential

dx

in terms of

d\theta

; the integrand in

\theta

is built from both. At the end,

\theta = \arcsin x

returns the answer to the original variable.

Radical in integrand	Substitution	Radical simplifies to
√(a² − x²)	x = a sin θ	a cos θ
√(a² + x²)	x = a tan θ	a sec θ
√(x² − a²)	x = a sec θ	a tan θ

A trigonometric substitution trades a square root for a Pythagorean identity, and the radical collapses to a single trigonometric factor.

Partial Fractions

Rational functions—polynomials divided by polynomials—decompose into simpler fractions.

Method: Factor the denominator. Write the fraction as a sum of terms with linear or irreducible quadratic denominators. Solve for coefficients. Integrate each term.

Example

\int \frac{1}{x^2 - 1}\, dx = \int \frac{1}{(x-1)(x+1)}\, dx

Decompose:

\frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}

Solving gives

A = 1/2

,

B = -1/2

:

= \frac{1}{2}\int \frac{1}{x-1}\, dx - \frac{1}{2}\int \frac{1}{x+1}\, dx = \frac{1}{2}\ln|x-1| - \frac{1}{2}\ln|x+1| + C

Pitfall: Taking the log of the whole denominator

Writing

\ln|x^2 - 1| + C

for

\int \frac{1}{x^2 - 1}\, dx

gives a function whose derivative is

\frac{2x}{x^2 - 1}

, not

\frac{1}{x^2 - 1}

. The decomposition gives

\frac{1}{2}\ln|x - 1| - \frac{1}{2}\ln|x + 1| + C

.

Each simple fraction integrates on its own: a constant over a linear factor

x - r

gives that constant times

\ln|x - r|

, because the derivative of

x - r

is

1

. The two logarithms can be combined at the end into

\frac{1}{2}\ln\left|\frac{x - 1}{x + 1}\right| + C

, a single term with the same derivative.

A fraction taken apart

The section’s 1/(x² − 1) (blue) has two vertical asymptotes, at x = −1 and x = 1. The decomposition writes it as (1/2)/(x − 1) (amber dashes), which carries only the asymptote at 1, plus −(1/2)/(x + 1) (navy dashes), which carries only the one at −1. Adding the two dashed curves point by point gives the blue curve, and each simple piece integrates to a logarithm.

Each factor of the denominator contributes one simple fraction with its own asymptote.

Choosing the Right Technique

Pattern recognition guides technique selection.

Substitution: Look for a function paired with its derivative. Expressions like

f(g(x)) \cdot g'(x)

signal substitution with

u = g(x)

.

Parts: Products of different function types—polynomial times exponential, polynomial times trigonometric, logarithm times polynomial.

Trigonometric integrals: Powers of

\sin x

and

\cos x

, products of trigonometric functions.

Trigonometric substitution: Square roots of

a^2 - x^2

,

a^2 + x^2

, or

x^2 - a^2

.

Partial fractions: Rational functions with factorable denominators.

Fact: One substitution for every trig fraction

With

t = \tan\frac{x}{2}

,

\sin x = \frac{2t}{1 + t^2}

,

\cos x = \frac{1 - t^2}{1 + t^2}

and

dx = \frac{2\, dt}{1 + t^2}

. Every rational expression in

\sin x

and

\cos x

becomes a rational function of

t

, so partial fractions can always finish it.

Multiple techniques often combine. A problem might require substitution followed by parts, or partial fractions followed by a trigonometric integral.

Some combinations are planned from the start. An integrand such as

\frac{1}{2 + \cos x}

is a fraction built from

\cos x

, with no odd power to save and no radical to replace, so none of the trigonometric patterns above fits it directly. A substitution can still turn it into a rational function of a new variable, and partial fractions take over from there.

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Before you read

Read the integrand first

The shape of the integrand points to the technique. Each card names the pattern to look for.

Integration techniques5 methods

Substitution

Look for

f(g(x)) \cdot g'(x)

Movelet

u = g(x)

Integration by parts

Look forproducts of different function types

Examplespolynomial × exponential, polynomial × trig, log × polynomial

Trigonometric integrals

Look forpowers or products of

\sin

and

\cos

Trigonometric substitution

Look for

\sqrt{a^2 - x^2}

,

\sqrt{a^2 + x^2}

,

\sqrt{x^2 - a^2}

Partial fractions

Look forrational functions with a factorable denominator

Takeaway

Match the pattern, then commit

Substitution and parts cover most integrals; trigonometric methods and partial fractions handle the specialised shapes.

Choosing an integration technique·/calculus/integrals/techniquesLearn Math Class

Summary: The Techniques at a Glance

The techniques covered above all share the same goal—reshape the integrand until it matches a known formula—but each one reverses a different differentiation rule and is triggered by a different signal in the integrand. The table below collects all five in one place, pairing each technique with what it reverses, the pattern that signals it, the standard setup, and a canonical example. Read the "Recognize when…" column first when scanning an unfamiliar integral; read across the row once a match is found.

Technique	What it reverses	Recognize when…	Setup	Example trigger
u-substitution	chain rule	integrand contains a function and its derivative	u = g(x), du = g'(x) dx; rewrite and integrate in u	∫ 2x cos(x²) dx
Integration by parts	product rule	product of unrelated factors (poly · exp, poly · trig, log · poly)	∫ u dv = uv − ∫ v du; LIATE picks u	∫ x eˣ dx
Trigonometric integrals	Pythagorean & half-angle identities	powers and products of sin x, cos x (and tan, sec variants)	save a factor; convert via identity; substitute u = sin x or cos x	∫ sin³ x dx
Trigonometric substitution	Pythagorean identity	integrand contains √(a² − x²), √(a² + x²), or √(x² − a²)	x = a sin θ, a tan θ, or a sec θ depending on the radical	∫ 1 / √(1 − x²) dx
Partial fractions	addition of rational fractions	rational function with a factorable denominator	decompose into A/(x − r) + B/(x − s) + … and integrate each piece	∫ 1 / (x² − 1) dx

Common Pitfalls

Five mistakes cause many of the wrong answers when these techniques are carried out. Each is marked where it happens in the sections above; here each is worked through in full, with the wrong and the right working side by side.

Keeping the x-bounds after substituting

In a definite integral, the integrand and

dx

are rewritten in

u

, but the bounds are left as they were. The antiderivative in

u

is then evaluated at numbers that are values of

x

.

Wrong working: with

u = x^2

,

\int_0^2 2x\cos(x^2)\, dx

becomes

\int_0^2 \cos u\, du = \sin 2 - \sin 0 \approx 0.909

.

Right working: when

x = 0

,

u = 0

; when

x = 2

,

u = 4

. So the integral is

\int_0^4 \cos u\, du = \sin 4 - \sin 0 \approx -0.757

.

Why it happens: the bounds are written once, at the start, and are read as part of the integral sign rather than as values of a particular variable.

How to check: take the other route. Back-substitute to get the antiderivative

\sin(x^2)

and evaluate it at the

x

-bounds:

\sin(2^2) - \sin(0^2) = \sin 4

. The two routes must agree.

Dropping the minus sign

In integration by parts,

uv - \int v\, du

is written with a plus sign, so the second integral is added instead of subtracted.

Wrong working:

\int x e^x\, dx = xe^x + \int e^x\, dx = xe^x + e^x + C

.

Right working:

\int x e^x\, dx = xe^x - \int e^x\, dx = xe^x - e^x + C = e^x(x - 1) + C

.

Why it happens: the product rule has a plus sign, and the parts formula is remembered by its shape. The minus only appears when

\int v\, du

is moved to the other side of the integrated product rule.

How to check: differentiate the answer.

(xe^x + e^x)' = xe^x + 2e^x

, which is not the integrand;

(xe^x - e^x)' = e^x + xe^x - e^x = xe^x

, which is.

Losing the sign of du

With

u = \cos x

, the differential is taken as

du = \sin x\, dx

instead of

du = -\sin x\, dx

, and the whole answer comes out with the wrong sign.

Wrong working:

\int \sin x (1 - \cos^2 x)\, dx = \int (1 - u^2)\, du = u - \frac{u^3}{3} + C = \cos x - \frac{\cos^3 x}{3} + C

.

Right working:

\sin x\, dx = -du

, so the integral is

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

.

Why it happens: the derivative of

\sin x

is

\cos x

with no sign, and the same is expected the other way round.

How to check: on

[0, \pi]

the integrand

\sin^3 x

is never negative, so

\int_0^\pi \sin^3 x\, dx

must be positive. The right answer gives

\left(1 - \frac{1}{3}\right) - \left(-1 + \frac{1}{3}\right) = \frac{4}{3}

; the wrong one gives

-\frac{4}{3}

.

Replacing dx with dθ

In a trigonometric substitution, the radical is rewritten in

\theta

, but

dx

is replaced by

d\theta

alone, without the factor that comes from differentiating

x = \sin\theta

.

Wrong working:

\int \frac{1}{\sqrt{1 - x^2}}\, dx = \int \frac{1}{\cos\theta}\, d\theta = \ln|\sec\theta + \tan\theta| + C = \ln\left|\frac{1 + x}{\sqrt{1 - x^2}}\right| + C

.

Right working:

dx = \cos\theta\, d\theta

, so

\int \frac{\cos\theta}{\cos\theta}\, d\theta = \theta + C = \arcsin x + C

.

Why it happens: in

u

-substitution the new differential is often just

du

, and the habit carries over to a substitution that runs the other way, where

x

is the function of

\theta

.

How to check: differentiate the answer in

x

. The wrong answer has derivative

\frac{1}{1 - x^2}

, not

\frac{1}{\sqrt{1 - x^2}}

; at

x = 0.6

these are

1.5625

and

1.25

.

Taking the log of the whole denominator

The rule

\int \frac{1}{x}\, dx = \ln|x| + C

is stretched to any denominator, so

\frac{1}{x^2 - 1}

is integrated straight to a logarithm of

x^2 - 1

without being decomposed.

Wrong working:

\int \frac{1}{x^2 - 1}\, dx = \ln|x^2 - 1| + C

.

Right working: decompose first.

\frac{1}{x^2 - 1} = \frac{1/2}{x - 1} - \frac{1/2}{x + 1}

, so the integral is

\frac{1}{2}\ln|x - 1| - \frac{1}{2}\ln|x + 1| + C

.

Why it happens: the rule

\int \frac{f'(x)}{f(x)}\, dx = \ln|f(x)| + C

is remembered without its condition. It needs the derivative of the denominator in the numerator, here

2x

, and the numerator is

1

.

How to check: differentiate.

\frac{d}{dx}\ln|x^2 - 1| = \frac{2x}{x^2 - 1}

, which at

x = 2

is

\frac{4}{3}

, while the integrand at

x = 2

is

\frac{1}{3}

.

Integration Techniques FAQ

Do you have to change the limits when you use u-substitution?

+

You have two legal options, and the only mistake is mixing them. Either convert the bounds — when x = a, u = g(a) — and finish entirely in u, never returning to x; or keep the original x-bounds, back-substitute u = g(x) to get the answer in x, and only then evaluate. Evaluating u-bounds against an x-expression is the error.Read more →

What does it mean if there's a leftover x after substituting?

+

It means the substitution is incomplete, and the expression is meaningless as written. Something like ∫cos u dx mixes one variable in the integrand with another in the differential. Every x must convert, the integrand and the dx alike. A stray x usually signals that the chosen u does not have its derivative present.Read more →

Is the u in integration by parts the same as the u in u-substitution?

+

No — same letter, different job. In substitution, u renames the variable: everything converts and x disappears. In integration by parts, u labels one factor of the integrand and x remains the variable throughout; the ledger is u, du, dv, v. Reading a parts problem as a substitution, or the reverse, is a common source of stuck work.Read more →

Does LIATE always work?

+

No. LIATE — logarithmic, inverse trig, algebraic, trigonometric, exponential — is a classroom mnemonic for choosing u in integration by parts, not a theorem. It picks well most of the time, but it fails often enough that judgment stays in the loop. If the resulting ∫v du is harder than what you started with, swap the roles.Read more →

Why is trigonometric substitution written x = a sin θ instead of u = g(x)?

+

Because it runs backwards. Ordinary substitution names the new variable as a function of the old one; trigonometric substitution expresses the old variable through the new one, which is what lets a Pythagorean identity collapse the radical. The price is that returning to x needs an inverse trig function, a reference triangle, and a domain restriction on θ.Read more →

Integration Techniques

Beyond Direct Formulas

Key Terms

Why Techniques Are Needed

Substitution (u-Substitution)

Notation Inside the Techniques

Notation Inside the Techniques

Integration by Parts

Trigonometric Integrals

Trigonometric Substitution

Partial Fractions

Choosing the Right Technique

Summary: The Techniques at a Glance

Common Pitfalls

Integration Techniques FAQ

Do you have to change the limits when you use u-substitution?

What does it mean if there's a leftover x after substituting?

Is the u in integration by parts the same as the u in u-substitution?

Does LIATE always work?

Why is trigonometric substitution written x = a sin θ instead of u = g(x)?