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Integration Techniques



Why Techniques Are Needed
Substitution (u-Substitution)
Integration by Parts
Trigonometric Integrals
Trigonometric Substitution
Partial Fractions
Choosing the Right Technique
Summary: The Techniques at a Glance



Beyond Direct Formulas


Most functions resist direct antidifferentiation. The integral of ex2e^{x^2} has no elementary formula. Even innocuous-looking expressions like 1+x3\sqrt{1 + x^3} lack closed-form antiderivatives.

Integration techniques transform difficult integrals into tractable ones. Substitution reverses the chain rule. Integration by parts reverses the product rule. Partial fractions decompose rational functions. Trigonometric methods handle roots and powers.

No single algorithm covers all cases—unlike differentiation, which follows systematic rules. Success requires recognizing which technique applies and executing it correctly. This pattern-matching skill develops through practice.

Key Terms

Integrand— the expression being transformed into a recognizable form
Indefinite Integral— the target: find the antiderivative family
Definite Integral— substitution in definite integrals requires converting bounds

See All Calculus Definitions →


Substitution (u-Substitution)


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

Method: Let u=g(x)u = g(x), so du=g′(x) dxdu = g'(x)\, dx. Replace all xx-expressions with uu-expressions and integrate.

Substitution Rule
∫f(g(x)) g′(x) dx=∫f(u) duwith u=g(x)\int f(g(x))\, g'(x)\, dx = \int f(u)\, du \quad \text{with } u = g(x)
Learn more about this formula: Substitution Rule →


Example:

∫2xcos⁡(x2) dx\int 2x \cos(x^2)\, dx


Let u=x2u = x^2, so du=2x dxdu = 2x\, dx:

=∫cos⁡u du=sin⁡u+C=sin⁡(x2)+C= \int \cos u\, du = \sin u + C = \sin(x^2) + C


For definite integrals, convert the limits: when x=ax = a, u=g(a)u = g(a); when x=bx = b, u=g(b)u = g(b).

Integration by Parts


Integration by parts reverses the product rule:

Integration by Parts
∫u dv=uv−∫v du\int u\, dv = uv - \int v\, du
Learn more about this formula: Integration by Parts →


Method: Identify factors uu and dvdv in the integrand. Differentiate uu to get dudu; integrate dvdv to get vv. Apply the formula.

Example:

∫xex dx\int x e^x\, dx


Let u=xu = x and dv=ex dxdv = e^x\, dx. Then du=dxdu = dx and v=exv = e^x:

=xex−∫ex dx=xex−ex+C=ex(x−1)+C= xe^x - \int e^x\, dx = xe^x - e^x + C = e^x(x - 1) + C


Choosing $u$: LIATE guides selection—Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential. Earlier types typically make better choices for uu.

Trigonometric Integrals


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

Odd power of sine: Save one sin⁡x\sin x, convert remaining sin⁡2x=1−cos⁡2x\sin^2 x = 1 - \cos^2 x, substitute u=cos⁡xu = \cos x.

Odd power of cosine: Save one cos⁡x\cos x, convert remaining cos⁡2x=1−sin⁡2x\cos^2 x = 1 - \sin^2 x, substitute u=sin⁡xu = \sin x.

Both powers even: Use half-angle identities:

sin⁡2x=1−cos⁡2x2cos⁡2x=1+cos⁡2x2\sin^2 x = \frac{1 - \cos 2x}{2} \qquad \cos^2 x = \frac{1 + \cos 2x}{2}


Example:

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


Let u=cos⁡xu = \cos x:

=−∫(1−u2) du=−u+u33+C=−cos⁡x+cos⁡3x3+C= -\int (1 - u^2)\, du = -u + \frac{u^3}{3} + C = -\cos x + \frac{\cos^3 x}{3} + C

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=asin⁡θx = a\sin\theta, so a2−x2=acos⁡θ\sqrt{a^2 - x^2} = a\cos\theta

For $\sqrt{a^2 + x^2}$: Let x=atan⁡θx = a\tan\theta, so a2+x2=asec⁡θ\sqrt{a^2 + x^2} = a\sec\theta

For $\sqrt{x^2 - a^2}$: Let x=asec⁡θx = a\sec\theta, so x2−a2=atan⁡θ\sqrt{x^2 - a^2} = a\tan\theta

Example:

∫11−x2 dx\int \frac{1}{\sqrt{1 - x^2}}\, dx


Let x=sin⁡θx = \sin\theta, so dx=cos⁡θ dθdx = \cos\theta\, d\theta and 1−x2=cos⁡θ\sqrt{1 - x^2} = \cos\theta:

=∫cos⁡θcos⁡θ dθ=∫dθ=θ+C=arcsin⁡x+C= \int \frac{\cos\theta}{\cos\theta}\, d\theta = \int d\theta = \theta + C = \arcsin x + C

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

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:

∫1x2−1 dx=∫1(x−1)(x+1) dx\int \frac{1}{x^2 - 1}\, dx = \int \frac{1}{(x-1)(x+1)}\, dx


Decompose:

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


Solving gives A=1/2A = 1/2, B=−1/2B = -1/2:

=12∫1x−1 dx−12∫1x+1 dx=12ln⁡∣x−1∣−12ln⁡∣x+1∣+C= \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


Choosing the Right Technique


Pattern recognition guides technique selection.

Substitution: Look for a function paired with its derivative. Expressions like f(g(x))⋅g′(x)f(g(x)) \cdot g'(x) signal substitution with u=g(x)u = g(x).

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

Trigonometric integrals: Powers of sin⁡x\sin x and cos⁡x\cos x, products of trigonometric functions.

Trigonometric substitution: Square roots of a2−x2a^2 - x^2, a2+x2a^2 + x^2, or x2−a2x^2 - a^2.

Partial fractions: Rational functions with factorable denominators.

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

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