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🎯⭐ INTERACTIVE LESSON

Advanced Integration

Learn step-by-step with interactive practice!

Advanced Integration - Complete Interactive Lesson

Part 1: Core Concepts

Advanced Integration Techniques — BC Level

Part 1 of 7 — Choosing the Right Method

Integration Strategy Flowchart

When facing an integral on the BC exam, use this decision process:

What you seeMethod to use
∫f(g(x)) g′(x) dx\int f(g(x))\,g'(x)\,dxuu-substitution
Product of unlike functionsIntegration by parts
P(x)Q(x)\frac{P(x)}{Q(x)} with deg⁡P≥deg⁡Q\deg P \ge \deg QLong division first
P(x)Q(x)\frac{P(x)}{Q(x)} with factorable QQPartial fractions
a2−x2\sqrt{a^2 - x^2}, a2+x2\sqrt{a^2 + x^2}, x2−a2\sqrt{x^2 - a^2}Trig substitution
Powers of sin⁡x\sin x and cos⁡x\cos xReduction formulas / identities
Infinite limits or discontinuitiesImproper integral

Step 1: Simplify→Step 2: Identify pattern→Step 3: Apply method\boxed{\text{Step 1: Simplify} \to \text{Step 2: Identify pattern} \to \text{Step 3: Apply method}}

Advanced uu-Substitution

Beyond basic substitution, BC-level problems may require:

Completing the square first: ∫dxx2+4x+8=∫dx(x+2)2+4\int \frac{dx}{x^2 + 4x + 8} = \int \frac{dx}{(x+2)^2 + 4}

Let u=x+2u = x + 2: ∫duu2+4=12arctan⁡u2+C=12arctan⁡x+22+C\int \frac{du}{u^2 + 4} = \frac{1}{2}\arctan\frac{u}{2} + C = \frac{1}{2}\arctan\frac{x+2}{2} + C

Back-substitution in definite integrals: ∫01x1−x2 dx→u=1−x2∫10u⋅−du2=12∫01u1/2 du=13\int_0^1 x\sqrt{1-x^2}\,dx \xrightarrow{u=1-x^2} \int_1^0 \sqrt{u}\cdot\frac{-du}{2} = \frac{1}{2}\int_0^1 u^{1/2}\,du = \frac{1}{3}

AP Tip: When using uu-sub in a definite integral, you can either change the limits OR back-substitute. Changing limits is usually faster.

Integrals Producing Inverse Trig Functions

∫dua2−u2=arcsin⁡ua+C\boxed{\int \frac{du}{\sqrt{a^2 - u^2}} = \arcsin\frac{u}{a} + C}

∫dua2+u2=1aarctan⁡ua+C\boxed{\int \frac{du}{a^2 + u^2} = \frac{1}{a}\arctan\frac{u}{a} + C}

IntegralResult
∫dx9−x2\int \frac{dx}{\sqrt{9 - x^2}}arcsin⁡(x/3)+C\arcsin(x/3) + C
∫dx4+x2\int \frac{dx}{4 + x^2}12arctan⁡(x/2)+C\frac{1}{2}\arctan(x/2) + C
∫dx1−4x2\int \frac{dx}{\sqrt{1 - 4x^2}}12arcsin⁡(2x)+C\frac{1}{2}\arcsin(2x) + C

Key Fact: Recognize the pattern: denominator has a2±u2a^2 \pm u^2 (with or without square root). This is a high-frequency BC topic.

Check Your Understanding

Method Selection

Practice

Key Takeaways

Recognize the form→Choose the method→Execute carefully\boxed{\text{Recognize the form} \to \text{Choose the method} \to \text{Execute carefully}}

PatternMethod
1/(a2+u2)1/(a^2 + u^2)1aarctan⁡(u/a)\frac{1}{a}\arctan(u/a)
1/a2−u21/\sqrt{a^2 - u^2}arcsin⁡(u/a)\arcsin(u/a)
Degree top ≥ bottomLong division first
Factorable denominatorPartial fractions

Next: Part 2 — Trigonometric Integrals and Substitution

Part 2: Worked Examples

Trigonometric Integrals and Substitution

Part 2 of 7 — Working with Trig Functions

Powers of Sine and Cosine

∫sin⁡mxcos⁡nx dx\int \sin^m x \cos^n x\,dx

CaseStrategy
mm oddSave one sin⁡x\sin x, convert rest to cos⁡\cos via sin⁡2=1−cos⁡2\sin^2 = 1 - \cos^2, u=cos⁡xu = \cos x
nn oddSave one cos⁡x\cos x, convert rest to sin⁡\sin via cos⁡2=1−sin⁡2\cos^2 = 1 - \sin^2, u=sin⁡xu = \sin x
Both evenUse half-angle: sin⁡2x=1−cos⁡2x2\sin^2 x = \frac{1-\cos 2x}{2}, cos⁡2x=1+cos⁡2x2\cos^2 x = \frac{1+\cos 2x}{2}

Key Fact: "Save one" means factor out one copy to pair with dxdx for substitution.

Worked Example

∫sin⁡3xcos⁡2x dx\int \sin^3 x \cos^2 x\,dx (m=3m = 3 is odd)

StepWork
Save one sin⁡x\sin x∫sin⁡2xcos⁡2x⋅sin⁡x dx\int \sin^2 x \cos^2 x \cdot \sin x\,dx
Convert sin⁡2x\sin^2 x∫(1−cos⁡2x)cos⁡2xsin⁡x dx\int (1 - \cos^2 x)\cos^2 x \sin x\,dx
Let u=cos⁡xu = \cos xdu=−sin⁡x dxdu = -\sin x\,dx
Substitute−∫(1−u2)u2 du=−∫(u2−u4) du-\int (1 - u^2)u^2\,du = -\int (u^2 - u^4)\,du
Integrate−u33+u55+C-\frac{u^3}{3} + \frac{u^5}{5} + C
Back-sub−cos⁡3x3+cos⁡5x5+C-\frac{\cos^3 x}{3} + \frac{\cos^5 x}{5} + C

Trigonometric Substitution

ExpressionSubstitutionIdentity used
a2−x2\sqrt{a^2 - x^2}x=asin⁡θx = a\sin\theta1−sin⁡2θ=cos⁡2θ1 - \sin^2\theta = \cos^2\theta
a2+x2\sqrt{a^2 + x^2}x=atan⁡θx = a\tan\theta1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta
x2−a2\sqrt{x^2 - a^2}x=asec⁡θx = a\sec\thetasec⁡2θ−1=tan⁡2θ\sec^2\theta - 1 = \tan^2\theta

Trig Substitution Example

∫dx4−x2\int \frac{dx}{\sqrt{4 - x^2}}

Let x=2sin⁡θx = 2\sin\theta, dx=2cos⁡θ dθdx = 2\cos\theta\,d\theta.

4−x2=4−4sin⁡2θ=2cos⁡θ\sqrt{4 - x^2} = \sqrt{4 - 4\sin^2\theta} = 2\cos\theta

∫2cos⁡θ dθ2cos⁡θ=∫dθ=θ+C=arcsin⁡x2+C\int \frac{2\cos\theta\,d\theta}{2\cos\theta} = \int d\theta = \theta + C = \arcsin\frac{x}{2} + C

AP Tip: Many trig substitution problems on the BC exam reduce to inverse trig results. Recognizing the pattern 1/a2−x21/\sqrt{a^2 - x^2} directly gives arcsin⁡(x/a)\arcsin(x/a) without going through the full substitution.

Check Your Understanding

Trig Integral Strategy

Practice

Key Strategies

Odd power→save one, convert, u-subEven powers→half-angle\boxed{\text{Odd power} \to \text{save one, convert, } u\text{-sub} \qquad \text{Even powers} \to \text{half-angle}}

Trig Sub PatternUse when you see
x=asin⁡θx = a\sin\thetaa2−x2\sqrt{a^2 - x^2}
x=atan⁡θx = a\tan\thetaa2+x2\sqrt{a^2 + x^2}
x=asec⁡θx = a\sec\thetax2−a2\sqrt{x^2 - a^2}

Next: Part 3 — Long Division, Completing the Square, and Algebraic Manipulation

Part 3: Problem-Solving Patterns

Algebraic Manipulation Before Integration

Part 3 of 7 — Simplify First, Integrate Second

Long Division for Improper Rational Functions

When deg⁡(numerator)≥deg⁡(denominator)\deg(\text{numerator}) \ge \deg(\text{denominator}), divide first:

∫x2+1x−1 dx\int \frac{x^2 + 1}{x - 1}\,dx

Long division: x2+1x−1=x+1+2x−1\frac{x^2 + 1}{x - 1} = x + 1 + \frac{2}{x-1}

∫(x+1+2x−1)dx=x22+x+2ln⁡∣x−1∣+C\int \left(x + 1 + \frac{2}{x-1}\right)dx = \frac{x^2}{2} + x + 2\ln|x-1| + C

Key Fact: Never attempt partial fractions on an improper fraction. Always divide first.

Completing the Square

Essential for integrals with irreducible quadratics:

∫dxx2+6x+13\int \frac{dx}{x^2 + 6x + 13}

Complete the square: x2+6x+13=(x+3)2+4x^2 + 6x + 13 = (x+3)^2 + 4

∫dx(x+3)2+4=12arctan⁡x+32+C\int \frac{dx}{(x+3)^2 + 4} = \frac{1}{2}\arctan\frac{x+3}{2} + C

When to Complete the Square

Denominator formCompleted formIntegral type
x2+bx+cx^2 + bx + c (no real roots)(x+b/2)2+(c−b2/4)(x + b/2)^2 + (c - b^2/4)arctan⁡\arctan
−(x2+bx+c)-(x^2 + bx + c) under \sqrt{}a2−(x+b/2)2a^2 - (x + b/2)^2arcsin⁡\arcsin

AP Tip: If the quadratic in the denominator doesn't factor over the reals, complete the square.

Splitting Numerators

Sometimes split the numerator to match derivative + constant:

∫2x+5x2+4 dx\int \frac{2x + 5}{x^2 + 4}\,dx

Split: 2xx2+4+5x2+4\frac{2x}{x^2+4} + \frac{5}{x^2+4}

=∫2xx2+4 dx+∫5x2+4 dx= \int \frac{2x}{x^2+4}\,dx + \int \frac{5}{x^2+4}\,dx

=ln⁡(x2+4)+52arctan⁡x2+C= \ln(x^2+4) + \frac{5}{2}\arctan\frac{x}{2} + C

The first integral uses u=x2+4u = x^2 + 4 (numerator is derivative of denominator). The second is an arctan⁡\arctan form.

Adding/Subtracting in the Numerator

∫xx+1 dx=∫(x+1)−1x+1 dx=∫(1−1x+1)dx=x−ln⁡∣x+1∣+C\int \frac{x}{x+1}\,dx = \int \frac{(x+1) - 1}{x+1}\,dx = \int \left(1 - \frac{1}{x+1}\right)dx = x - \ln|x+1| + C

Check Your Understanding

Step-by-Step

Evaluate ∫dxx2+2x+5\int \frac{dx}{x^2 + 2x + 5}.

Practice

Pre-Integration Techniques

TechniqueWhen to use
Long divisiondeg⁡(num)≥deg⁡(den)\deg(\text{num}) \ge \deg(\text{den})
Complete the squareIrreducible quadratic in denominator
Split numeratorNumerator has ax+bax + b over quadratic
Add/subtract trickMake numerator match denominator

Simplify the integrand BEFORE choosing an integration method\boxed{\text{Simplify the integrand BEFORE choosing an integration method}}

Next: Part 4 — Definite Integrals and Accumulation Problems

Part 4: Graphs and Interpretation

Definite Integrals and Accumulation

Part 4 of 7 — FTC Applications at BC Level

The Fundamental Theorem — Both Parts

FTC Part 1 (Evaluation): ∫abf(x) dx=F(b)−F(a)where F′=f\boxed{\int_a^b f(x)\,dx = F(b) - F(a) \quad \text{where } F' = f}

FTC Part 2 (Derivative of accumulation): ddx∫axf(t) dt=f(x)\boxed{\frac{d}{dx}\int_a^x f(t)\,dt = f(x)}

Chain Rule Variation (BC Favorite)

ddx∫ag(x)f(t) dt=f(g(x))⋅g′(x)\boxed{\frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f(g(x)) \cdot g'(x)}

This appears frequently on BC exams and requires careful application of the chain rule.

Key Fact: If BOTH bounds are functions of xx: split using additivity of integrals. ddx∫h(x)g(x)f(t) dt=f(g(x))g′(x)−f(h(x))h′(x)\frac{d}{dx}\int_{h(x)}^{g(x)} f(t)\,dt = f(g(x))g'(x) - f(h(x))h'(x)

Worked Example

Let F(x)=∫1x2sin⁡tt dtF(x) = \int_1^{x^2} \frac{\sin t}{t}\,dt. Find F′(x)F'(x).

Using the chain rule version with g(x)=x2g(x) = x^2:

F′(x)=sin⁡(x2)x2⋅2x=2sin⁡(x2)xF'(x) = \frac{\sin(x^2)}{x^2} \cdot 2x = \frac{2\sin(x^2)}{x}

Accumulation in Context

If R(t)R(t) is a rate (gallons/hour), then:

∫abR(t) dt=total quantity accumulated from t=a to t=b\int_a^b R(t)\,dt = \text{total quantity accumulated from } t = a \text{ to } t = b

Common BC applications:

R(t)R(t) represents∫R dt\int R\,dt gives
VelocityDisplacement
SpeedDistance
Population growth rateChange in population
Flow rateTotal volume

AP Tip: "Rate" in the integrand → the integral gives total change. This is tested in nearly every AP FRQ.

Check Your Understanding

FTC with Variable Bounds

Practice

Key Formulas

ddx∫ag(x)f(t) dt=f(g(x))⋅g′(x)\boxed{\frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f(g(x)) \cdot g'(x)}

ScenarioFormula
Fixed lower boundF′(x)=f(x)F'(x) = f(x)
Upper bound g(x)g(x)F′(x)=f(g(x))⋅g′(x)F'(x) = f(g(x)) \cdot g'(x)
Both bounds variablef(g)g′−f(h)h′f(g)g' - f(h)h'

Next: Part 5 — AP Exam Strategies

Part 5: Applications

AP Exam Strategies — Advanced Integration

Part 5 of 7 — Scoring Maximum Points

Integration on the BC Exam

SectionWhat to expect
MC (no calc)Antiderivative identification, method selection
MC (calc)Definite integrals with complex integrands
FRQ (no calc)Separation + integration, FTC applications
FRQ (calc)Accumulation problems, numerical integrals

Quick Decision Guide

Can I use a known formula?→Is it a u-sub?→By parts?→Partial fractions?\boxed{\text{Can I use a known formula?} \to \text{Is it a } u\text{-sub?} \to \text{By parts?} \to \text{Partial fractions?}}

AP Tip: On calculator sections, you don't need to find antiderivatives. Use fnInt\texttt{fnInt} for definite integrals. On non-calculator sections, you MUST know the techniques.

Most Common BC Integration Results

IntegralAnswer
∫eax dx\int e^{ax}\,dx1aeax+C\frac{1}{a}e^{ax} + C
∫dxx\int \frac{dx}{x}$\ln
∫dx1−x2\int \frac{dx}{\sqrt{1-x^2}}arcsin⁡x+C\arcsin x + C
∫dx1+x2\int \frac{dx}{1+x^2}arctan⁡x+C\arctan x + C
∫ln⁡x dx\int \ln x\,dxxln⁡x−x+Cx\ln x - x + C
∫xnex dx\int x^n e^x\,dxIBP (tabular method)
∫sec⁡2x dx\int \sec^2 x\,dxtan⁡x+C\tan x + C
∫sec⁡xtan⁡x dx\int \sec x \tan x\,dxsec⁡x+C\sec x + C

Common Traps

  1. ∫dxx=ln⁡∣x∣\int \frac{dx}{x} = \ln|x|, NOT ln⁡x\ln x (absolute value matters)
  2. ∫e3x dx=13e3x\int e^{3x}\,dx = \frac{1}{3}e^{3x}, NOT e3xe^{3x} (don't forget the chain rule factor)
  3. ∫dxx2+4=12arctan⁡(x/2)\int \frac{dx}{x^2+4} = \frac{1}{2}\arctan(x/2), NOT arctan⁡(x/2)\arctan(x/2) (don't forget the 1/a1/a factor)

AP-Style Questions

Speed Round

Practice

Exam Day Integration Checklist

  1. ✓ Can I evaluate directly (known antiderivative)?
  2. ✓ Is it a uu-substitution (chain rule in reverse)?
  3. ✓ Is the numerator the derivative of the denominator? (→ln⁡\to \ln)
  4. ✓ Should I complete the square? (→arctan⁡\to \arctan or arcsin⁡\arcsin)
  5. ✓ Partial fractions? (factor denominator)
  6. ✓ By parts? (product of unlike types)
  7. ✓ Calculator allowed? (just compute numerically)

Next: Part 6 — Problem-Solving Workshop

Part 6: Exam Strategy

Problem-Solving Workshop — Advanced Integration

Part 6 of 7 — Timed Practice

Apply the full toolkit: uu-sub, by parts, partial fractions, completing the square, inverse trig, and FTC.

Workshop Problems

Multi-Step Problem

Evaluate ∫4xx2−1 dx\int \frac{4x}{x^2 - 1}\,dx.

Quick Compute

Workshop Takeaways

  • Check if numerator is derivative of denominator first (saves time)
  • For absolute values, split the integral at the zero
  • LIATE order for integration by parts: Logs, Inverse trig, Algebraic, Trig, Exponential
  • Partial fractions: factor first, then decompose

Next: Part 7 — Comprehensive Review

Part 7: Mixed Review

Comprehensive Review — Advanced Integration

Part 7 of 7 — Final Assessment

Integration Methods Summary

MethodRecognizable by
uu-substitutionComposite function with inner derivative present
By partsProduct of unlike function types
Partial fractionsRational function, factorable denominator
Complete the squareQuadratic denominator, no real roots
Inverse trig formula1/(a2+u2)1/(a^2+u^2) or 1/a2−u21/\sqrt{a^2-u^2}
Trig powerssin⁡mxcos⁡nx\sin^m x \cos^n x
Long divisionDegree of numerator ≥\ge denominator

Simplify→Identify→Execute→Verify by differentiation\boxed{\text{Simplify} \to \text{Identify} \to \text{Execute} \to \text{Verify by differentiation}}

Review Set A

Review Set B

Method Identification

Final Problem

Advanced Integration — Complete

You've mastered:

  • Choosing the right integration technique
  • Inverse trig integrals (arcsin⁡\arcsin, arctan⁡\arctan)
  • Trig integrals (powers of sine/cosine, trig sub)
  • Algebraic preparation (long division, completing the square, splitting numerators)
  • FTC with variable bounds and chain rule
  • AP exam strategies

∫=pattern recognition + algebraic skill + careful execution\boxed{\int = \text{pattern recognition + algebraic skill + careful execution}}