MATH 104, Sections 8.2, 8.3, and 8.4 | Source: Rimmer lecture notes
Difficulty: Intermediate to Advanced | Prerequisites: Integration by parts (Section 8.1), trig identities (Pythagorean, double-angle, half-angle), polynomial long division
Chapter 8 is about expanding your integration toolkit. Section 8.2 teaches you systematic strategies for integrals involving products and powers of trig functions. Section 8.3 handles integrals containing square roots of quadratic expressions by substituting a trig function for x. Section 8.4 breaks rational functions (polynomial over polynomial) into simpler fractions that you can integrate term by term. Together, these three sections cover most of the "hard" integrals you will meet on exams. If you are behind, make sure you are solid on trig identities and polynomial factoring before diving in.
Powers of trig functions are handled by pulling out one factor and converting the rest using Pythagorean identities (odd power) or half-angle identities (both even). Trig substitution replaces x with a sin theta, a tan theta, or a sec theta to eliminate a square root. Partial fractions decompose a rational function into a sum of simpler fractions with linear or irreducible quadratic denominators, each of which integrates to a logarithm or arctangent.
Pythagorean identity
cos^2 x + sin^2 x = 1, and its rearrangements. Also: 1 + tan^2 x = sec^2 x. These let you convert between trig functions inside an integral.
Half-angle identities
sin^2 x = (1 minus cos 2x) / 2 and cos^2 x = (1 + cos 2x) / 2. Used to reduce even powers of sine and cosine to first-power expressions.
Trigonometric substitution
A technique where you replace x with a trig expression to eliminate a square root of a quadratic. Three standard templates exist, one for each form of the quadratic under the root.
Reference triangle
A right triangle drawn from the trig substitution that lets you convert back from theta to x after integrating. Each side is labelled with the original variable, the constant, and the square root expression.
Partial fraction decomposition
The process of writing a rational function p(x)/q(x) as a sum of simpler fractions. Requires that the degree of p is less than the degree of q (use polynomial long division first if it is not).
Irreducible quadratic
A quadratic factor ax^2 + bx + c that cannot be factored over the real numbers. Test: b^2 minus 4ac is less than 0. These contribute an (Ax + B) / (ax^2 + bx + c) term to the decomposition.
Type 1: integral of cos^m x sin^n x dx
Case A (one or both powers odd, greater than 1): Factor out one power from the trig function with the odd exponent. Convert the remaining even power using cos^2 x + sin^2 x = 1. Finish with u-substitution, letting u equal the other trig function.
Case B (both powers even): Replace all even powers using half-angle identities: sin^2 x = (1 minus cos 2x)/2 and cos^2 x = (1 + cos 2x)/2. Expand and integrate.
Case C (one or both powers equal to 1): Straightforward u-substitution. Let u = the trig function with the higher power (or either, if both are 1).
Type 2: integral of tan^m x sec^n x dx
If the power of sec is even (n >= 2): Factor out sec^2 x, convert the remaining sec powers to tan using sec^2 x = 1 + tan^2 x, let u = tan x.
If the power of tan is odd: Factor out sec x tan x, convert the remaining tan powers to sec using tan^2 x = sec^2 x minus 1, let u = sec x.
Both conditions fail (tan is even and sec is odd): Convert everything to sin and cos, or use reduction formulas. These are the hardest cases.
Type 3: Products of different arguments
Integrals of the form sin(mx) sin(nx), cos(mx) cos(nx), or sin(mx) cos(nx) where m is not equal to n. Use the product-to-sum identities to convert the product into a sum of first-power trig functions, then integrate term by term.
Use trig substitution when the integrand contains one of three square-root expressions. Each form has its own substitution and reference triangle.
Case 1: sqrt(a^2 minus x^2)
Let x = a sin theta, so dx = a cos theta d(theta).
Then sqrt(a^2 minus x^2) = a cos theta.
Valid for -pi/2 <= theta <= pi/2.
Reference triangle: hypotenuse = a, opposite = x, adjacent = sqrt(a^2 minus x^2).
Case 2: sqrt(a^2 + x^2)
Let x = a tan theta, so dx = a sec^2 theta d(theta).
Then sqrt(a^2 + x^2) = a sec theta.
Valid for -pi/2 < theta < pi/2.
Reference triangle: adjacent = a, opposite = x, hypotenuse = sqrt(a^2 + x^2).
Case 3: sqrt(x^2 minus a^2)
Let x = a sec theta, so dx = a sec theta tan theta d(theta).
Then sqrt(x^2 minus a^2) = a tan theta.
Valid for 0 <= theta < pi/2 (when x > 0) or pi <= theta < 3pi/2 (when x < 0).
Reference triangle: hypotenuse = x, adjacent = a, opposite = sqrt(x^2 minus a^2).
After integrating in theta, use the reference triangle to convert back to x. Every side of the triangle is expressed in terms of x and a, so you can read off any trig function of theta directly.
Goal: Integrate a rational function p(x)/q(x) by breaking it into simpler fractions.
Step 0: If degree of p >= degree of q, perform polynomial long division first to get a polynomial plus a proper fraction.
Step 1: Factor q(x) completely into linear factors and irreducible quadratic factors.
To check whether a quadratic ax^2 + bx + c is irreducible: compute b^2 minus 4ac. If b^2 minus 4ac < 0, the quadratic is irreducible.
Step 2: Write the decomposition.
Each distinct linear factor (x minus r) contributes a term A / (x minus r).
A repeated linear factor (x minus r)^k contributes k terms: A_1 / (x minus r) + A_2 / (x minus r)^2 + ... + A_k / (x minus r)^k.
Each distinct irreducible quadratic (ax^2 + bx + c) contributes (Ax + B) / (ax^2 + bx + c).
A repeated irreducible quadratic contributes the same pattern with increasing powers.
Step 3: Solve for the unknown constants (A, B, C, ...) using one of two methods: multiply both sides by q(x) and either plug in convenient x-values or compare coefficients.
Step 4: Integrate each simpler fraction.
A / (x minus r) integrates to A ln|x minus r| + C.
(Ax + B) / (x^2 + bx + c) may require completing the square and splitting into a logarithm part and an arctangent part:
\int \frac{1}{x^2 + a^2}\, dx = \frac{1}{a}\arctan\!\left(\frac{x}{a}\right) + CTechnique | Key identity or substitution | Use when |
|---|---|---|
Trig powers (odd case) | cos^2 x + sin^2 x = 1 | One power is odd: peel off one factor, convert the rest |
Trig powers (even case) | sin^2 x = (1 minus cos 2x)/2, cos^2 x = (1 + cos 2x)/2 | Both powers are even |
tan/sec (sec even) | sec^2 x = 1 + tan^2 x | Factor out sec^2 x, let u = tan x |
tan/sec (tan odd) | tan^2 x = sec^2 x minus 1 | Factor out sec x tan x, let u = sec x |
Trig sub: sqrt(a^2 minus x^2) | x = a sin theta | Eliminates the root, leaves a cos theta |
Trig sub: sqrt(a^2 + x^2) | x = a tan theta | Eliminates the root, leaves a sec theta |
Trig sub: sqrt(x^2 minus a^2) | x = a sec theta | Eliminates the root, leaves a tan theta |
Partial fractions: linear factor | A / (x minus r) | Integrates to A ln abs(x minus r) |
Partial fractions: irreducible quadratic | (Ax + B) / (x^2 + bx + c) | Complete the square, split into ln and arctan |
Trig substitution is how engineers evaluate integrals that arise in electrical engineering (impedance calculations) and physics (gravitational and electrostatic fields near spherical bodies). Partial fractions are essential in control systems engineering, where transfer functions are decomposed to determine the behaviour of a system via inverse Laplace transforms.
Students confuse which trig substitution goes with which root form. Memorise: "minus" goes with sine, "plus" goes with tangent, "x^2 minus" goes with secant.
In partial fractions, students forget to do long division first when the numerator's degree is greater than or equal to the denominator's. If you skip this, the decomposition will fail.
When a linear factor is repeated, students often write only one term instead of one for each power up to the multiplicity.
In trig integrals, students attempt to use half-angle identities when one power is odd. The odd-power strategy (peel-and-convert) is both faster and less error-prone.
Students forget to convert back to x after doing a trig substitution. The reference triangle is the tool for this step.
The decision tree for trig integrals (odd vs. even powers) is testable on its own. Be ready to state which strategy applies before computing.
Trig substitution problems are typically set up so the integral simplifies cleanly. If you are getting a mess, re-check your substitution and reference triangle.
Partial fractions problems with irreducible quadratics are a reliable exam topic. Know how to complete the square and split the integral.
Many exam problems combine techniques: a partial fraction whose terms require integration by parts or a trig substitution that produces a trig integral.
Fill in the blank: To integrate cos^3 x sin^4 x dx, you would factor out one ______ and convert the remaining cos^2 x using the Pythagorean identity.
cos x
True or False: For sqrt(9 minus x^2), you substitute x = 3 tan theta.
False. You substitute x = 3 sin theta.
Fill in the blank: Before doing partial fractions, if degree of numerator >= degree of denominator, you must first ______.
perform polynomial long division.
True or False: An irreducible quadratic factor contributes a term of the form A / (ax^2 + bx + c).
False. It contributes (Ax + B) / (ax^2 + bx + c), with a linear numerator.
Q: Evaluate the integral of sin^5 x cos^2 x dx.
A: Sin has the odd power. Factor out sin x: integral of sin^4 x cos^2 x sin x dx. Convert sin^4 x = (1 minus cos^2 x)^2. Let u = cos x, du = minus sin x dx. The integral becomes minus integral of (1 minus u^2)^2 u^2 du. Expand and integrate term by term.
Q: Which trig substitution would you use for integral of dx / (x^2 sqrt(x^2 + 4))?
A: The expression under the root is x^2 + 4 = x^2 + a^2, so let x = 2 tan theta.
Q: Decompose (x minus 12) / (x^2 + 3x minus 18) into partial fractions.
A: Factor the denominator: x^2 + 3x minus 18 = (x + 6)(x minus 3). Set (x minus 12) / [(x + 6)(x minus 3)] = A / (x + 6) + B / (x minus 3). Multiply through: x minus 12 = A(x minus 3) + B(x + 6). Plug in x = 3: -9 = 9B, so B = -1. Plug in x = -6: -18 = -9A, so A = 2. Result: 2 / (x + 6) minus 1 / (x minus 3).
These three techniques, together with integration by parts from Section 8.1, form the complete toolkit for evaluating the integrals that appear in Sections 8.7 (improper integrals) and 8.8 (probability). Trig substitution in particular reappears in multivariable calculus when converting to polar or spherical coordinates. Partial fractions are essential for inverse Laplace transforms in differential equations courses.
Trig integrals, powers of sine and cosine, powers of tangent and secant, half-angle identity, product-to-sum identity, trig substitution, reference triangle, completing the square, partial fraction decomposition, irreducible quadratic, reducible polynomial, discriminant, long division, rational function integration, MATH 104, Calculus II techniques, Rimmer, Penn State Math 104