Techniques of Integration

Differentiation follows rules that always work. Integration has no single procedure, so it is learned as a set of techniques, and the skill is reading the shape of the integrand and recognizing which technique it calls for.

What is integration?

Integrating asks the reverse of differentiating: which function has this derivative? That function is the antiderivative, and every answer can be checked by differentiating it. See The antiderivative.

To integrate a power of x, add one to the power and divide by the new power: $x^3$x³ integrates to $\frac{x^4}{4}$x⁴/4. The rule fails only for $\frac{1}{x}$1/x, where the new power would be 0. The natural logarithm fills that gap. See Integrating powers.

Why is there always a + C?

Differentiating removes any constant term. $x^2$x², $x^2+7$x² + 7 and $x^2-100$x² − 100 all differentiate to 2x, so integrating 2x gives the whole family $x^2+C$x² + C, one curve for each value of C.

One extra fact pins down C. If the curve passes through (1, 4), substitute the point in: $4=1^2+C$4 = 1² + C, so C = 3 and the function is $x^2+3$x² + 3. In a differential equation that fact is the initial condition. See Finding the constant of integration.

What does the integral sign actually say?

The sign $\int$∫ is a letter S stretched tall, and S stands for sum. $\int f\left(x\right)dx$∫ f(x) dx means "add up slivers of height f(x) and width dx". The function f(x) is the integrand, and the numbers at the top and bottom of the sign are the limits of integration. See Reading the integral sign.

dx has a meaning of its own. $\frac{dy}{dx}$dy/dx began as a rise over a run, and the pieces can be separated: if $y=x^2$y = x², then dy = 2x dx. Substitution depends on this. See Differentials.

Why does an integral measure area?

A definite integral is the limit of a sum of rectangles. Slice the region under the curve into strips of width $\Delta x$Δx, treat each strip as a rectangle of height f(x), and add the areas $f\left(x\right)\Delta x$f(x) Δx. Make the strips thinner and the total approaches the exact area. See The definite integral.

Each height can be read at the left edge, the right edge or the middle of its strip. On a rising curve the left sum underestimates and the right sum overestimates. In sigma notation the sum is $\sum _{i=1}^{n}f\left(x_i\right)\Delta x$Σ f(xᵢ) Δx from i = 1 to n; let n run to infinity and $\sum$Σ becomes $\int$∫ and $\Delta x$Δx becomes dx. See Left, right and midpoint Riemann sums and The Riemann sum in sigma notation.

The fundamental theorem of calculus finds the area without strips. Let A(x) be the area under the curve up to x. Increase x a little and the area grows by a thin strip of height f(x), so A differentiates to f. To find the area from a to b, take any antiderivative F and work out F(b) − F(a). See The fundamental theorem of calculus.

Area below the axis counts as negative. The arch of sin x from 0 to $\pi$π has integral 2, from $\pi$π to $2\pi$2π the integral is −2, and over the whole cycle it is 0. For the total area, split at the crossings and add the sizes: 2 + 2 = 4. See Area below the axis.

Swapping the limits changes the sign, a constant factor comes outside, and a to b plus b to c joins into a to c. See Properties of the definite integral.

How do you integrate a function inside another function?

Substitution is the chain rule read backwards. Differentiating a composite function leaves the derivative of the inside multiplying the result. When that factor is present in an integral, name the inside u. See Integration by substitution.

1. Choose u, usually the expression inside the bracket or under the root.
2. Find $\frac{du}{dx}$du/dx, and write dx in terms of du.
3. Replace every x. If one is left over, u was the wrong choice.
4. Integrate in u, then write the x expression back. For a definite integral, convert the limits instead.

$\int 2x{\left(x^2+1\right)}^{5}dx$∫ 2x(x² + 1)⁵ dx: 2x is the derivative of $x^2+1$x² + 1, so let $u=x^2+1$u = x² + 1 and du = 2x dx. The integral becomes $\int u^{5}du=\frac{u^6}{6}+C=\frac{{\left(x^2+1\right)}^6}{6}+C$∫ u⁵ du = u⁶/6 + C = (x² + 1)⁶/6 + C.

When the inside is linear, ax + b, its derivative is the constant a and there is a shortcut: integrate as if the inside were plain x, then divide by a. $\int \cos 3xdx=\left(\frac{1}{3}\right)\sin 3x+C$∫ cos 3x dx = (1/3) sin 3x + C. A curved inside has a derivative that changes with x, so the shortcut does not apply. See Integrating f(ax + b).

How do you integrate a product?

Integration by parts is the product rule read backwards: $\int udv=uv-\int vdu$∫ u dv = uv − ∫ v du. It trades one integral for another, so choose u as the factor that simplifies when differentiated. A polynomial differentiates down to a constant, so against $e^x$eˣ or sin x the polynomial is u. See Integration by parts.

$\int x{e}^{x}dx$∫ x eˣ dx: let u = x and $dv=e^{x}dx$dv = eˣ dx, so du = dx and $v=e^x$v = eˣ. Then $\int x{e}^{x}dx=x{e}^x-\int e^{x}dx=x{e}^x-e^x+C$∫ x eˣ dx = x eˣ − ∫ eˣ dx = x eˣ − eˣ + C. If the new integral is harder than the old one, swap the choice of u.

$\int \ln xdx$∫ ln x dx is done by parts with u = ln x and dv = 1 dx. $\int e^x\sin xdx$∫ eˣ sin x dx returns to itself after two rounds; treat the original integral as an unknown and solve for it.

How do you integrate trigonometric functions?

Sine differentiates to cosine, so cosine integrates to sine. Cosine differentiates to −sine, so sine integrates to −cosine. See Integrating trigonometric functions.

${\sin }^{2}x$sin²x is not the derivative of any standard function, so rewrite it with the double angle formula: ${\sin }^{2}x=\frac{\left(1-\cos 2x\right)}{2}$sin²x = (1 − cos 2x)/2, and $\int {\sin }^{2}xdx=\frac{x}{2}-\frac{\left(\sin 2x\right)}{4}+C$∫ sin²x dx = x/2 − (sin 2x)/4 + C. ${\cos }^{2}x=\frac{\left(1+\cos 2x\right)}{2}$cos²x = (1 + cos 2x)/2 integrates the same way. The identities are in radians and trigonometric identities; see Integrating ${\sin }^2$sin² and ${\cos }^2$cos².

${\sec }^{2}x$sec²x integrates to tan x, and sec x tan x integrates to sec x. tan x is sin x over cos x, and the numerator is minus the derivative of the denominator, so $\int \tan xdx=-\ln |\cos x|+C=\ln |\sec x|+C$∫ tan x dx = −ln|cos x| + C = ln|sec x| + C. See Integrating ${\sec }^{2}x$sec²x, sec x tan x and tan x.

How do you integrate exponentials, logarithms and fractions?

$e^x$eˣ is its own derivative, so it is its own integral. ln x differentiates to $\frac{1}{x}$1/x, so $\int \frac{1}{x}dx=\ln |x|+C$∫ 1/x dx = ln|x| + C. The modulus keeps the result defined for negative x. See Integrating exponentials and logarithms.

Differentiating ln f(x) by the chain rule gives $\frac{f^{\prime }\left(x\right)}{f\left(x\right)}$f'(x)/f(x). Read backwards: when the numerator is the derivative of the denominator, the integral is ln of the denominator. $\int \frac{2x}{\left(x^2+1\right)}dx=\ln \left(x^2+1\right)+C$∫ 2x/(x² + 1) dx = ln(x² + 1) + C, because 2x is the derivative of $x^2+1$x² + 1. Differentiate the denominator first and check whether the numerator matches. See Integrals of f' over f.

When the denominator factors instead, split into partial fractions. Write $\frac{1}{\left(\left(x-1\right)\left(x+1\right)\right)}$1/((x − 1)(x + 1)) as $\frac{A}{\left(x-1\right)}+\frac{B}{\left(x+1\right)}$A/(x − 1) + B/(x + 1), clear the denominators, then substitute x = 1 to find A and x = −1 to find B. Each piece is a constant over a linear bracket, so each integrates to a logarithm. A repeated factor needs one fraction for each power; a quadratic that will not factor keeps a numerator Ax + B. See Integration by partial fractions.

Which integrals give inverse functions?

${\tan }^{-1}x$tan⁻¹x differentiates to $\frac{1}{\left(1+x^2\right)}$1/(1 + x²), so $\frac{1}{\left(1+x^2\right)}$1/(1 + x²) integrates to ${\tan }^{-1}x$tan⁻¹x, and $\frac{1}{\sqrt{1-x^2}}$1/√(1 − x²) integrates to ${\sin }^{-1}x$sin⁻¹x. With $a^2$a² in place of the 1, the answers are $\left(\frac{1}{a}\right){\tan }^{-1}\left(\frac{x}{a}\right)$(1/a) tan⁻¹(x/a) and ${\sin }^{-1}\left(\frac{x}{a}\right)$sin⁻¹(x/a). These are derivatives from the rules of differentiation read in reverse; see Integrals that give ${\tan }^{-1}$tan⁻¹ and ${\sin }^{-1}$sin⁻¹.

When $\sqrt{a^2-x^2}$√(a² − x²) appears with no factor of x beside it, substitute $x=a\sin \theta$x = a sin θ. Then $a^2-x^2=a^2{\cos }^2\theta$a² − x² = a² cos²θ, the root becomes $a\cos \theta$a cos θ, and $dx=a\cos \theta d\theta$dx = a cos θ dθ. A minus inside takes sine; a plus takes tangent. See Choosing a trigonometric substitution.

A plus under the root gives a hyperbolic answer: $\frac{1}{\sqrt{x^2+1}}$1/√(x² + 1) integrates to arsinh x, and $\frac{1}{\sqrt{x^2-1}}$1/√(x² − 1) to arcosh x. Both are logarithms, so either form is correct. See Integrals that give arsinh and arcosh.

How do you choose?

Read the shape of the integrand before starting.

The mistakes worth naming

• Dropping the + C. An indefinite integral is a family of curves.
• Leaving an x behind after substituting. Every x must go, including the one in dx.
• Reporting a signed area as a total. Split at the crossings first.
• Choosing the wrong factor in parts. If the new integral is harder, swap.
• Forgetting the modulus in ln|x|. Without it the result fails for negative x.

Learn this properly in the app

Every lesson linked above is in Math Challenge, with a diagram and practice questions. The rules being reversed are in the rules of differentiation, their uses in applications of integration and polar curves, and the limits behind the definition in limits and continuity.