Advanced Integration Techniques

Once an MSTE integral stops yielding to the power rule, the board is really testing whether you can recognize which special method unlocks it. This lesson works through the four techniques that cover almost every hard antiderivative on the paper: integration by parts, the trigonometric integrals and their partner trigonometric substitution, integration by partial fractions, and improper integrals over infinite or unbounded regions. Each method is reduced to a short recipe and then carried all the way to a number so you can grade your own habits against it.

Integration by parts

Integration by parts is the product rule read backwards. From d(uv) = u dv + v du, integrate both sides and rearrange to get the working formula:

integral of u dv = uv - integral of v du.

The whole game is choosing u and dv so that the leftover integral of v du is easier than the one you started with. The LIATE rule picks u by taking whichever factor appears first in this order, because that choice tends to simplify when differentiated.

Priority for u	Type	Example factor
L	Logarithmic	ln(x)
I	Inverse trigonometric	arctan(x)
A	Algebraic	x, x^2
T	Trigonometric	sin(x), cos(x)
E	Exponential	e^x

Worked example (indefinite): Evaluate integral of x e^x dx. By LIATE the algebraic factor x is picked as u, so u = x and dv = e^x dx, giving du = dx and v = e^x. Then integral of x e^x dx = x e^x - integral of e^x dx = x e^x - e^x + C = e^x(x - 1) + C. Worked example (definite): Evaluate integral from 0 to 1 of x e^x dx. Using the antiderivative e^x(x - 1), at x = 1 it is e(1 - 1) = 0 and at x = 0 it is 1 times (0 - 1) = -1, so the definite integral is 0 - (-1) = 1.

Worked example (the lone-logarithm trick): Evaluate integral of ln(x) dx. There is no obvious dv, so take u = ln(x) and dv = dx, which gives du = dx/x and v = x. Then integral of ln(x) dx = x ln(x) - integral of x times (1/x) dx = x ln(x) - integral of dx = x ln(x) - x + C. The same trick handles arcsin and arctan.

Trigonometric integrals

Powers of sine and cosine follow a two-case rule. If either power is odd, peel off one factor and convert the rest with sin^2(x) + cos^2(x) = 1, then substitute. If both powers are even, knock them down with the half-angle identities sin^2(x) = (1 - cos(2x))/2 and cos^2(x) = (1 + cos(2x))/2.

Worked example (even power): Evaluate integral of sin^2(x) dx. Replace sin^2(x) by (1 - cos(2x))/2, so integral of sin^2(x) dx = (1/2) integral of dx - (1/2) integral of cos(2x) dx = x/2 - sin(2x)/4 + C. As a definite check, integral from 0 to pi of sin^2(x) dx = [x/2 - sin(2x)/4] from 0 to pi = (pi/2 - 0) - (0 - 0) = pi/2, about 1.571.

Worked example (odd power): Evaluate integral of sin^3(x) dx. Write sin^3(x) = sin(x)(1 - cos^2(x)) and let u = cos(x), so du = -sin(x) dx. Then integral of sin^3(x) dx = -integral of (1 - u^2) du = -(u - u^3/3) + C = -cos(x) + (1/3) cos^3(x) + C.

Trigonometric substitution

When a radical of the form sqrt(a^2 - x^2), sqrt(a^2 + x^2), or sqrt(x^2 - a^2) blocks an integral, a trigonometric substitution turns the radical into a single trig function using the Pythagorean identities. Memorize the three pairings.

Radical in the integrand	Substitution	Radical becomes
sqrt(a^2 - x^2)	x = a sin(theta)	a cos(theta)
sqrt(a^2 + x^2)	x = a tan(theta)	a sec(theta)
sqrt(x^2 - a^2)	x = a sec(theta)	a tan(theta)

After integrating in theta you must translate every trig ratio back into x, and the cleanest way is to draw a reference right triangle built directly from the substitution.

Worked example: Evaluate integral of dx / (x^2 sqrt(4 - x^2)). Here a = 2, so let x = 2 sin(theta), dx = 2 cos(theta) d(theta), sqrt(4 - x^2) = 2 cos(theta), and x^2 = 4 sin^2(theta). Substituting, integral of (2 cos(theta) d(theta)) / (4 sin^2(theta) times 2 cos(theta)) = (1/4) integral of csc^2(theta) d(theta) = -(1/4) cot(theta) + C. Reading cot(theta) = sqrt(4 - x^2) / x off the reference triangle gives the final answer -(1/4) times sqrt(4 - x^2) / x + C, that is -sqrt(4 - x^2) / (4x) + C. The most common standard result, integral of dx / sqrt(a^2 - x^2) = arcsin(x/a) + C, comes from the same substitution.

Integration by partial fractions

A proper rational function (numerator degree lower than denominator degree) splits into a sum of simpler fractions that each integrate to a logarithm or an arctangent. Factor the denominator first, then match the fraction type. If the numerator degree is not smaller, divide long-hand first to get a polynomial plus a proper remainder.

Denominator factor	Terms it contributes
Distinct linear (x - r)	A / (x - r)
Repeated linear (x - r)^2	A / (x - r) + B / (x - r)^2
Irreducible quadratic (x^2 + bx + c)	(Ax + B) / (x^2 + bx + c)

Worked example: Evaluate integral of (5x - 4) / (x^2 - x - 2) dx. Factor the denominator as (x - 2)(x + 1), then write (5x - 4) / ((x - 2)(x + 1)) = A / (x - 2) + B / (x + 1). Clearing denominators, 5x - 4 = A(x + 1) + B(x - 2). The cover-up (Heaviside) shortcut sets x = 2 to get 6 = 3A, so A = 2, and sets x = -1 to get -9 = -3B, so B = 3. Therefore integral of (5x - 4) / (x^2 - x - 2) dx = 2 ln|x - 2| + 3 ln|x + 1| + C. As a definite check, integral from 3 to 4 of the same integrand is [2 ln|x - 2| + 3 ln|x + 1|] from 3 to 4 = (2 ln 2 + 3 ln 5) - (0 + 3 ln 4), which is about 2.056.

Improper integrals

An integral is improper when a limit is infinite (Type 1) or when the integrand blows up somewhere on the interval (Type 2). You never plug infinity in directly; you replace the bad endpoint with a variable and take a limit. If the limit is a finite number the integral converges to it, and if the limit is infinite or does not exist the integral diverges.

integral from 1 to infinity of f(x) dx = lim as b approaches infinity of integral from 1 to b of f(x) dx.

Worked example (converges): Evaluate integral from 1 to infinity of 1/x^2 dx. Integrate to b: integral from 1 to b of x^(-2) dx = [-1/x] from 1 to b = -1/b + 1. As b approaches infinity, -1/b approaches 0, so the value is 1. Worked example (diverges): Evaluate integral from 1 to infinity of 1/x dx. Now integral from 1 to b of 1/x dx = ln(b) - ln(1) = ln(b), and ln(b) grows without bound, so this integral diverges. The two share the same shape but only the faster-decaying 1/x^2 encloses a finite area.

Worked example (exponential tail): Evaluate integral from 0 to infinity of e^(-x) dx = lim as b approaches infinity of [-e^(-x)] from 0 to b = lim of (1 - e^(-b)) = 1.

A Type 2 integral is improper because the integrand has a vertical asymptote inside or at an endpoint of the interval, so you take the limit as you approach the bad point.

Worked example (Type 2): Evaluate integral from 0 to 1 of 1/sqrt(x) dx. The integrand is undefined at x = 0, so integral from a to 1 of x^(-1/2) dx = [2 sqrt(x)] from a to 1 = 2 - 2 sqrt(a). As a approaches 0 from the right, 2 sqrt(a) approaches 0, so the value is 2. The single most useful convergence fact ties these together in the p-test: integral from 1 to infinity of 1/x^p dx converges only when p > 1, while integral from 0 to 1 of 1/x^p dx converges only when p < 1.

Improper integral	Converges when	Diverges when
integral from 1 to infinity of 1/x^p dx	p > 1	p <= 1
integral from 0 to 1 of 1/x^p dx	p < 1	p >= 1

Exam-day strategy

• Read the integrand and name the method before writing anything: a product of unlike functions signals parts, a lone radical of the sqrt(a^2 +/- x^2) family signals a trig substitution, a factorable denominator signals partial fractions, and an infinite limit or an interior blow-up signals an improper integral.
• For integration by parts, pick u by LIATE and, for a stubborn single function like ln(x) or arctan(x), set dv = dx so v = x.
• On powers of sine and cosine, check for an odd power first (peel one factor and use sin^2 + cos^2 = 1); reach for the half-angle identities only when both powers are even.
• After a trig substitution, always sketch the reference triangle so you can translate cot, sec, and the rest back into x instead of leaving a stray theta in the answer.
• Before applying partial fractions, confirm the fraction is proper; if the numerator degree is not smaller than the denominator degree, do the long division first, then decompose the remainder.
• Never substitute infinity or a point of discontinuity directly; rewrite the improper integral as a limit, integrate to the variable endpoint, and read off whether the limit is a finite number (converges) or blows up (diverges).
• When a convergence question offers no time to integrate, fall back on the p-test: 1/x^p over 1 to infinity converges only for p > 1, and over 0 to 1 converges only for p < 1.