Difficulty: Intermediate to Advanced | Prerequisites: Basic integration, u-substitution, trigonometric identities (Pythagorean, double-angle), partial derivative basics from Part 1 of these notes.
This material covers the main toolkit for evaluating integrals that do not yield to a straightforward antiderivative: trigonometric substitution, partial fractions, and algebraic u-substitutions. It also introduces improper integrals, where the interval of integration stretches to infinity or the integrand blows up. These techniques come up repeatedly in differential equations, physics, and probability, so they are worth internalising rather than just memorising for the exam.
When an integral has a square root of a quadratic, try a trig substitution. When the integrand is a rational function (polynomial over polynomial), decompose it with partial fractions. When the interval goes to infinity, set up a limit and check whether the answer converges to a finite number or diverges.
Trigonometric substitution (trig sub)
A technique for integrals involving √(a² − x²), √(a² + x²), or √(x² − a²). You replace x with a trig expression that eliminates the square root using a Pythagorean identity. Think of it as reshaping the integrand so the square root disappears.
Partial fraction decomposition
A method for integrating rational functions by splitting them into simpler fractions whose antiderivatives are known (usually logarithms or arctangent). In simple terms, you break one hard fraction into several easy ones.
Improper integral
An integral where the interval of integration is infinite (e.g. ∫ from 0 to ∞) or where the integrand is unbounded at some point in the interval. You evaluate it as a limit: replace ∞ with t, integrate, then take t → ∞.
Convergent (improper integral)
An improper integral whose limit exists and is finite. Think of it as: the area under the curve, even though the region stretches out forever, still adds up to a finite number.
Divergent (improper integral)
An improper integral whose limit does not exist or is infinite. The area grows without bound.
Completing the square
Rewriting a quadratic expression ax² + bx + c in the form a(x − h)² + k so that a trig or sec substitution becomes apparent. Essential for integrals like ∫ 1/√(x² − 4x − 12) dx.
The strategy depends on which form appears under the square root.
√(a² − x²): set x = a sin θ. Then √(a² − x²) = a cos θ and dx = a cos θ dθ. Example: ∫ 6/√(4 − 3x²) dx. Here 4 − 3x² = 4(1 − sin²θ) after setting √3 x = 2 sin θ, which gives 4 cos²θ. The integral simplifies to (6/√3) ∫ dθ, and evaluating from 0 to π/3 gives I = 2π/√3.
√(a² + x²): set x = a tan u. Then √(a² + x²) = a sec u and dx = a sec²u du. Example: ∫ 3/(x² + 1)^(3/2) dx from 0 to 1. Setting x = tan u, the denominator becomes sec³u. The integral reduces to 3 ∫ cos u du from 0 to π/4, giving I = (3/2)√2.
√(x² − a²): set x = a sec θ. Then √(x² − a²) = a tan θ and dx = a sec θ tan θ dθ. Example: ∫ 1/√(x² − 4x − 12) dx. First complete the square: x² − 4x − 12 = (x − 2)² − 16. Then set x − 2 = 4 sec θ. The integral becomes ∫ sec θ dθ = ln|sec θ + tan θ| + C, which back-substitutes to ln|x − 2 + √(x² − 4x − 12)| + C.
The pattern: identify which Pythagorean identity will kill the square root, make the substitution, convert the limits (for definite integrals), and simplify.
When the integrand is a ratio of polynomials with distinct linear factors, decompose it into a sum of simpler fractions.
Setting up the decomposition. Write 2/[(x − 2)(6 − x)] = A/(x − 2) + B/(6 − x). Multiply both sides by the denominator to get A(6 − x) + B(x − 2) = 2. Plug in convenient values of x: x = 2 gives A = 1/2, x = 6 gives B = 1/2.
Integrating the pieces. Each simple fraction integrates to a logarithm: ∫ [1/(x − 2) + 1/(6 − x)] dx = ln|x − 2| − ln|6 − x| + C. Evaluating from 3 to 5 gives (1/2) ln 9.
Partial fractions after a substitution. Sometimes a substitution produces a rational function. For ∫ 3/(e^x + 2) dx from 0 to ln 3, setting u = e^x + 2 gives ∫ 3/[u(u − 2)] du. Then 1/[u(u − 2)] = (1/2)[1/(u − 2) − 1/u], and the integral works out to (3/2) ln(9/5).
Handling irreducible quadratics. When a factor like u² + 4 appears in the denominator, it does not split into linear factors. Instead, use ∫ 1/(u² + a²) du = (1/a) arctan(u/a) + C. For ∫ √(x − 7)/[4(x − 3)] dx from 7 to 11, after substituting u² = x − 7 and reducing, you get a term with 1/(u² + 4) that evaluates via arctan. The result is (1/2)(2 − π/2).
Inverse-tangent pattern. For ∫ 1/[9 + (x − 1)²] dx from 1 to 4, recognise the form ∫ 1/(a² + u²) du = (1/a) arctan(u/a). Set 3 tan u = x − 1. Then 9 + (x − 1)² = 9 sec²u and dx = 3 sec²u du. The integral becomes (1/3) ∫ du from 0 to π/4, giving I = π/12.
Substitution to handle sin and cos. For ∫ 8 sin x/(1 + 4 cos²x) dx from π/3 to π/2, set u = 2 cos x so du = −2 sin x dx. The integral becomes ∫ 4/(1 + u²) du from 0 to 1 = 4 arctan(1) = π.
Eliminating a square root with u² = expression. For ∫ √(x − 7)/[4(x − 3)] dx, set u² = x − 7 to remove the square root. Then x − 3 = u² + 4 and dx = 2u du. The integrand becomes a rational function in u, which you can then handle with partial fractions or direct integration.
An integral is improper when the interval of integration is infinite or the integrand is unbounded. The procedure is the same in both cases: replace the problematic bound with a finite parameter t, evaluate the integral, then take the limit.
Divergent example. ∫ from 0 to ∞ of f(x) dx where f(x) = x² for x ≤ 1 and f(x) = 1/x for x > 1. For t > 1 the integral equals 1/3 + ln t. Since ln t → ∞ as t → ∞, the integral diverges. The 1/x piece grows without bound.
Convergent example (power decay). ∫ from 0 to ∞ of 3/(x + 2)² dx. Set u = x + 2, giving ∫ from 2 to t+2 of 3/u² du = [−3/u] from 2 to t+2 = 3/2 − 3/(t + 2). As t → ∞ the second term vanishes, so I = 3/2.
Convergent example (exponential decay). ∫ from 0 to ∞ of e^(−6x) dx. The antiderivative is (−1/6)e^(−6x). Evaluating from 0 to t gives (−1/6)e^(−6t) + 1/6. Since e^(−6t) → 0, we get I = 1/6.
The intuition: functions that decay fast enough (like 1/x² or e^(−x)) produce convergent integrals. Functions that decay too slowly (like 1/x) do not.
Trig substitution reference:
√(a² − x²) → set x = a sin θ, dx = a cos θ dθ
√(a² + x²) → set x = a tan u, dx = a sec²u du
√(x² − a²) → set x = a sec θ, dx = a sec θ tan θ dθ
Partial fractions (distinct linear factors): A/[(x − r₁)(x − r₂)] = C₁/(x − r₁) + C₂/(x − r₂). Solve for C₁, C₂ by plugging in x = r₁ and x = r₂.
Key antiderivatives:
∫ 1/(a² + u²) du = (1/a) arctan(u/a) + C
∫ sec θ dθ = ln|sec θ + tan θ| + C
∫ 1/u du = ln|u| + C
∫ e^(ku) du = (1/k) e^(ku) + C
Improper integral test: ∫ from 0 to ∞ of f(x) dx = lim (t → ∞) ∫ from 0 to t of f(x) dx. Convergent if the limit is finite, divergent otherwise.
Completing the square: x² + bx + c = (x + b/2)² + (c − b²/4).
Students often choose the wrong trig substitution. The choice depends on the sign pattern under the square root: a² − x² uses sin, a² + x² uses tan, x² − a² uses sec. Mixing these up produces an identity that does not simplify anything.
Forgetting to convert the limits of integration after a substitution. If you change from x to u, the limits must change too. Alternatively, back-substitute into x before evaluating, but do not mix the two approaches.
Assuming every integral over an infinite interval diverges. Exponential decay (e^(−kx) with k > 0) and power decay faster than 1/x (such as 1/x²) both converge.
When decomposing into partial fractions, students sometimes forget to check that the degree of the numerator is strictly less than the degree of the denominator. If it is not, you must perform polynomial long division first.
⚠️ Trig substitution problems almost always appear on exams. Know the three standard forms cold: sin for a² − x², tan for a² + x², sec for x² − a².
⚠️ Partial fractions combined with a prior substitution (e.g. u = e^x + 2 turning the integral into a rational function) is a common multi-step problem.
⚠️ Improper integral convergence/divergence questions are reliable exam material. You must show the limit setup explicitly, not just evaluate the antiderivative.
⚠️ Completing the square before a trig substitution is a standard exam technique. Look for quadratics like x² − 4x − 12 that need rewriting as (x − 2)² − 16.
⚠️ Knowing ∫ sec θ dθ = ln|sec θ + tan θ| + C by heart saves significant time.
Fill in the blank: To integrate √(a² − x²), substitute x = ______. (a sin θ.)
True or false: ∫ from 0 to ∞ of 1/x dx converges. (False, it diverges.)
Fill in the blank: ∫ 1/(a² + u²) du = ______. ((1/a) arctan(u/a) + C.)
True or false: Before applying partial fractions, the numerator's degree must be less than the denominator's degree. (True.)
Fill in the blank: x² − 4x − 12, after completing the square, equals ______. ((x − 2)² − 16.)
Q: Evaluate ∫ from 1 to 4 of 1/[9 + (x − 1)²] dx.
A: Set 3 tan u = x − 1. The integral becomes (1/3) ∫ du from 0 to π/4 = π/12.
Q: Evaluate ∫ from 0 to 1 of 6/√(4 − 3x²) dx.
A: Set √3 x = 2 sin θ. The integral reduces to (6/√3) ∫ dθ from 0 to π/3 = 2π/√3.
Q: Evaluate ∫ from 0 to 1 of 3/(x² + 1)^(3/2) dx.
A: Set x = tan u. The integral becomes 3 ∫ cos u du from 0 to π/4 = (3/2)√2.
Q: Evaluate ∫ from 3 to 5 of 2/[(x − 2)(6 − x)] dx.
A: Partial fractions give (1/2)[1/(x − 2) + 1/(6 − x)]. Integrating yields (1/2) ln 9.
Q: Is ∫ from 0 to ∞ of 3/(x + 2)² dx convergent? If so, find its value.
A: Set u = x + 2. The integral equals [−3/u] from 2 to ∞ = 3/2. Convergent.
Q: Is ∫ from 0 to ∞ of e^(−6x) dx convergent? If so, find its value.
A: The antiderivative is (−1/6)e^(−6x). As t → ∞, e^(−6t) → 0, so I = 1/6. Convergent.
Q: Find the indefinite integral ∫ 1/√(x² − 4x − 12) dx.
A: Complete the square: (x − 2)² − 16. Set x − 2 = 4 sec θ. The integral equals ln|x − 2 + √(x² − 4x − 12)| + C.
Trigonometric substitution reappears in arc-length and surface-area integrals later in the course. Partial fractions are essential for inverse Laplace transforms in differential equations. Improper integrals are the foundation of probability theory: every continuous probability distribution has a PDF whose integral over all of R must converge to 1.
Trigonometric substitution, trig sub, partial fractions, partial fraction decomposition, improper integral, convergent integral, divergent integral, u-substitution, completing the square, arctan integral, secant integral, inverse trig integration, infinite interval, limit of integration, 1/x² convergence, exponential decay integral, rational function integration, polynomial long division, Pythagorean identity, sec substitution, tan substitution, sin substitution