Difficulty: Intermediate to Advanced | Prerequisites: Ch. 5 review (definite/indefinite integrals, FTC), basic trig identities.

TL;DR

These three techniques handle the majority of integrals you will meet in Calc II. U-substitution reverses the chain rule, integration by parts reverses the product rule, and trigonometric integral strategies use identities to reshape products of sin, cos, tan and sec into forms you can integrate with substitution. Learning when to reach for each one is at least as important as the mechanics.

Key Terms

U-substitution (the reverse chain rule)

A method for evaluating integrals of the form integral of f(g(x)) g'(x) dx by setting u = g(x), so du = g'(x) dx, reducing it to integral of f(u) du.

Think of it as: spotting a function and its derivative sitting together inside the integral, then renaming to simplify.

Complete substitution

When substitution replaces every part of the integrand including leftover x terms, sometimes by solving u = g(x) back for x. Used when a simple u-sub leaves x terms behind.

Integration by parts

The integration counterpart of the product rule: integral of u dv = uv - integral of v du. Used when the integrand is a product of two different types of function.

In simple terms, this means: you trade one integral for a (hopefully simpler) one by differentiating one factor and integrating the other.

LIATE priority

A mnemonic for choosing u in integration by parts. In order of priority for u: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential. Pick u from whichever type appears earliest in the list.

Trigonometric integrals

Integrals involving products and powers of trigonometric functions (sin^m x cos^n x, tan^m x sec^n x) that are solved by strategically applying trig identities and substitution.

Half-angle identities

cos^2 x = (1 + cos 2x)/2 and sin^2 x = (1 - cos 2x)/2. Used when both powers of sine and cosine are even, so you cannot peel off a single factor for substitution.

Product-to-sum formulas

Identities that convert products like sin A cos B into sums: sin A cos B = (1/2)[sin(A+B) + sin(A-B)]. Useful for integrating products of sines and cosines with different arguments.

Symmetric integrals (even and odd functions)

If f is even (f(-x) = f(x)), then integral from -a to a of f(x) dx = 2 times integral from 0 to a. If f is odd (f(-x) = -f(x)), the integral from -a to a is 0. Saves work and catches traps.

Core Content, Part 1: U-Substitution

The Substitution Rule

If u = g(x) is differentiable and f is continuous on the range of g, then:

\int f(g(x))\,g'(x)\,dx = \int f(u)\,du

The practical steps:

1. Identify a composite structure f(g(x)) times g'(x) dx in the integrand.

1. Set u = g(x), compute du = g'(x) dx.

1. Rewrite the entire integral in terms of u and du.

1. Integrate with respect to u.

1. Substitute back to x (for indefinite integrals). For definite integrals, convert the limits to u-values instead.

Substitution for Definite Integrals

When using u-sub on a definite integral, change the limits: if u = g(x), the new lower limit is g(a) and the new upper limit is g(b). No need to substitute back to x.

Complete Substitution

Sometimes after setting u = g(x), leftover x terms remain. Solve the substitution equation for x in terms of u and replace those terms. For example, to evaluate integral of k sqrt(3 - k) dk, set u = 3 - k, so k = 3 - u and dk = -du, and the whole integrand converts.

Symmetric Integrals

For f continuous on [-a, a]:

• Even function (f(-x) = f(x)): integral from -a to a = 2 times integral from 0 to a.

• Odd function (f(-x) = -f(x)): integral from -a to a = 0.

Before grinding through a computation, check symmetry. If x^2 sin(x) / (1 + x^6) is odd (it is), the integral from -7 to 7 is immediately 0.

Core Content, Part 2: Integration by Parts

The Formula

Two equivalent forms:

\int u\,dv = uv - \int v\,du

or in function notation: integral of f(x) g'(x) dx = f(x) g(x) - integral of g(x) f'(x) dx.

Choosing u and dv

Two guiding principles (they usually agree):

• Let dv be the most complicated part of the integrand that you can easily integrate. Whatever remains is u.

• Let u be the factor whose derivative is simpler than itself (so the new integral is easier). dv is the rest.

The LIATE mnemonic ranks u candidates: Logarithmic > Inverse trig > Algebraic > Trig > Exponential. Pick u from whichever category appears first.

Worked Pattern: Algebraic times Exponential

For integral of x e^x dx: set u = x, dv = e^x dx. Then du = dx, v = e^x. Result: x e^x - integral of e^x dx = x e^x - e^x + C.

Repeated Application

Some integrals need parts applied twice or more (e.g. integral of t^2 e^(-t) dt). Each round reduces the polynomial degree by one.

The Cycling Trick

For integral of e^x cos x dx, applying parts twice brings back the original integral. Call it I. After two rounds you get I = e^x cos x + e^x sin x - I, so 2I = e^x (sin x + cos x) and I = (e^x / 2)(sin x + cos x) + C.

Integration by Parts for Definite Integrals

The boundary term is evaluated at the limits:

\int_a^b u\,dv = \bigl[uv\bigr]_a^b - \int_a^b v\,du

The "Paradox" Warning

Applying parts to integral of 1/x dx with u = 1/x and dv = dx appears to yield 0 = 1. The error is omitting + C from the indefinite integrals on both sides. The constants on each side differ by 1, which resolves the contradiction. Moral: never drop + C.

Core Content, Part 3: Trigonometric Integrals

Strategy for sin^m x cos^n x dx

The approach depends on whether m and n are odd or even.

• (a) Cosine power is odd (n = 2k+1): Save one cos x factor. Convert the remaining cos^2k x to (1 - sin^2 x)^k. Substitute u = sin x, du = cos x dx.

• (b) Sine power is odd (m = 2k+1): Save one sin x factor. Convert the remaining sin^2k x to (1 - cos^2 x)^k. Substitute u = cos x, du = -sin x dx.

• (c) Both powers are even: Use half-angle identities: cos^2 x = (1 + cos 2x)/2 and sin^2 x = (1 - cos 2x)/2. The identity sin x cos x = (1/2) sin 2x can also simplify things. Repeated application may be needed.

If both powers are odd, either (a) or (b) works; choose whichever is simpler.

Strategy for tan^m x sec^n x dx

• (a) Secant power is even (n = 2k, k >= 1): Save sec^2 x. Convert the remaining sec^(2k-2) x using sec^2 x = 1 + tan^2 x. Substitute u = tan x.

• (b) Tangent power is odd (m = 2k+1): Save sec x tan x. Convert the remaining tan^2k x using tan^2 x = sec^2 x - 1. Substitute u = sec x.

Product-to-Sum Formulas

For integrals of the form sin(Ax) cos(Bx) with different arguments:

• sin A cos B = (1/2)[sin(A+B) + sin(A-B)]

• cos A cos B = (1/2)[cos(A+B) + cos(A-B)]

• sin A sin B = (1/2)[cos(A-B) - cos(A+B)]

These convert a product into a sum you can integrate term by term.

The "Enigma" of Multiple Correct Answers

Integral of sin x cos x dx can be done four ways (sub w = cos x, sub w = sin x, use sin 2x identity, or integration by parts) producing results that look different: -cos^2 x / 2, sin^2 x / 2, -cos(2x)/4. These are all correct; they differ by a constant, which the +C absorbs.

Formulas Reference

Useful Trigonometric Identities

Integration by Parts

\int u\,dv = uv - \int v\,du

The Substitution Rule

\int f(g(x))\,g'(x)\,dx = \int f(u)\,du \quad \text{where } u = g(x)

Common Misconceptions and Exam Flags

• Students often try to split the integral of a product into the product of two integrals. This is false. There is no "product rule" for integration in that sense; integration by parts is the correct tool.

• When using u-sub on a definite integral, a common mistake is changing the integrand to u but leaving the limits in terms of x. Either convert the limits or substitute back to x before evaluating.

• In trig integrals, students sometimes apply the half-angle identities when one power is odd. Check the parity first: if either m or n is odd, the save-one-factor strategy is faster and cleaner.

• The cycling trick (for e^x cos x, e^x sin x) requires you to be consistent with your choice of u and dv in both rounds of parts. Switching roles midway will loop you back to the start without progress.

• Expect u-substitution to appear on every exam, often nested inside another technique.

• Integration by parts with ln x is classic: u = ln x, dv = dx (or x^n dx). Integral of ln x dx = x ln x - x + C shows up constantly.

• The sin^m x cos^n x strategy table (odd cosine, odd sine, both even) is worth memorising cold.

• Symmetric integral shortcuts (even/odd) are frequently tested as quick-check problems.

• The antiderivative of sec x = ln|sec x + tan x| + C comes up often and is worth committing to memory.

Quick Self-Test

1. True or False: To integrate sin^5 x cos^2 x dx, save one sin x factor and substitute u = cos x. (True. Sine power is odd, so save sin x and convert sin^4 x = (1 - cos^2 x)^2.)

1. Fill in the blank: Integration by parts states integral of u dv = ______. (uv - integral of v du.)

1. True or False: Integral from -3 to 3 of x^4 sin x dx = 0. (True. x^4 sin x is odd since x^4 is even and sin x is odd.)

1. Fill in the blank: When both sin and cos have even powers, use the ______ identities to reduce. (Half-angle.)

1. True or False: If you use u-sub on a definite integral and change the limits to u-values, you must substitute back to x at the end. (False. The limits are already in u; evaluate directly.)

Practice Q&A

Q: Evaluate integral of 2x(1 + x^2)^23 dx.

A: Set u = 1 + x^2, du = 2x dx. The integral becomes integral of u^23 du = u^24/24 + C = (1 + x^2)^24 / 24 + C.

Q: Evaluate integral of x e^x dx.

A: Parts with u = x, dv = e^x dx. du = dx, v = e^x. Result: x e^x - e^x + C.

Q: Evaluate integral of ln x dx.

A: Parts with u = ln x, dv = dx. du = (1/x) dx, v = x. Result: x ln x - x + C.

Q: Evaluate integral of sin^3 x dx.

A: Sine power is odd. Save one sin x: sin^3 x = sin^2 x times sin x = (1 - cos^2 x) sin x. Set u = cos x, du = -sin x dx. Integral = -integral of (1 - u^2) du = -u + u^3/3 + C = -cos x + cos^3 x / 3 + C.

Q: Evaluate integral of cos^4 B dB.

A: Both powers even (cos^4 = (cos^2)^2). Use half-angle: cos^2 B = (1 + cos 2B)/2. So cos^4 B = (1 + cos 2B)^2 / 4 = (1 + 2cos 2B + cos^2 2B)/4. Apply half-angle again to cos^2 2B. Result: 3B/8 + sin(2B)/4 + sin(4B)/32 + C.

Q: Evaluate integral of e^x cos x dx.

A: Apply parts twice (u = cos x first, then u = sin x in the second round), getting I = e^x cos x + e^x sin x - I. Solve: I = (e^x / 2)(sin x + cos x) + C.

Connections to Other Topics

U-substitution is a prerequisite for every technique that follows: trig substitution (Section 7.3) is just u-sub with x = a sin theta or x = a tan theta, and partial fractions (Section 7.4) produces terms you integrate via u-sub or arctan formulas. Integration by parts reappears in deriving reduction formulas and in solving certain differential equations. Trig integral strategies feed directly into trig substitution, where you end up with integrals of exactly these sin/cos/tan/sec products.

Related Terms / Search Tags

u-substitution, reverse chain rule, integration by parts, LIATE, tabular integration, repeated integration by parts, cycling trick, trig integrals, sin^m cos^n strategy, tan^m sec^n strategy, half-angle identities, product-to-sum formulas, even and odd functions, symmetric integrals, Calculus II, NYU Calculus for Engineers, Chapter 5.5, Chapter 7.1, Chapter 7.2