Advanced Substitution Slides Pattern Recognition
Lesson 01: Advanced Substitution
A Tale of Two Integrals
Integral A
\[ \int x \cos(x^2) \, dx \]
Integral B
\[ \int \cos^2(x) \, dx \]
Which one requires "Advanced" techniques? Why?
The Architecture of \(u\)
Mastering the Composite Structure
REF: CALC_SUB_01
1. Identify Inner
Look for \(g(x)\) inside another function \(f(g(x))\).
2. Verify Differential
Does \(g'(x)\) (or a constant multiple) appear elsewhere in the integrand?
3. Transformation
Map the entire integral into the \(u\)-domain. No \(x\) terms may remain.
"Substitution is the chain rule in reverse. If you can see the chain, you can see the \(u\)."
Shifting the Boundaries
When evaluating \(\int_a^b f(g(x))g'(x) dx\):
Define \(u = g(x)\)
New Lower: \(u_{low} = g(a)\)
New Upper: \(u_{up} = g(b)\)
Never "go back" to \(x\) for definite integrals.
EXAMPLE
Evaluate: \[ \int_0^1 x e^{x^2} dx \]
\(u = x^2 \implies du = 2x dx\)
\(x=0 \to u=0\)
\(x=1 \to u=1\)
\[ \int_0^1 \frac{1}{2} e^u du \]
3
Advanced Pattern: Inverse Trig
Type A: ArcSine
\[ \int \frac{du}{\sqrt{1-u^2}} = \arcsin(u) + C \]
Type B: ArcTangent
\[ \int \frac{du}{1+u^2} = \arctan(u) + C \]
Diagnosis Check:
Often requires "forcing" the structure:
\[ \int \frac{e^x}{1 + e^{2x}} dx \]
Diagnosis Drills
Identify \(u\) and \(du\). Do NOT solve.
1. \(\int \frac{\ln(x)}{x} dx\)
\(u = \text{?}\)
2. \(\int \sin^4(x)\cos(x) dx\)
\(u = \text{?}\)
3. \(\int \frac{x}{1+x^4} dx\)
\(u = \text{?}\)
Substitution Drill Sheet Substitution Drill Sheet
Calculus II // Advanced Pattern Recognition
Name:
Date:
Part I: The Diagnosis
Identify the optimal substitution \(u\) and its differential \(du\). Do not solve the integral in this section.
\[ \int x^2 \sqrt{x^3 + 5} \, dx \]
Target \(u\)
Differential \(du\)
\[ \int \frac{\sin(\sqrt{x})}{\sqrt{x}} \, dx \]
Target \(u\)
Differential \(du\)
\[ \int \frac{e^x}{1 + e^{2x}} \, dx \]
Target \(u\)
Differential \(du\)
Part II: Limit Transformation
Evaluate each definite integral. Note: You must explicitly show the transformation of the limits of integration.
4. \[ \int_0^1 x^3(x^4 + 1)^2 \, dx \]
UNIT_01_DEF_01
5. \[ \int_0^{\pi/2} \sin(x)e^{\cos(x)} \, dx \]
UNIT_01_DEF_02
Part III: The Inverse Climb
6. Evaluate: \[ \int \frac{1}{\sqrt{9 - x^2}} \, dx \]
HINT: FACTOR OUT 9
Substitution Drill Key Answer Key
Calculus II // Advanced Pattern Recognition // Drill Sheet
Instructor Use Only
Part I: The Diagnosis
\[ \int x^2 \sqrt{x^3 + 5} \, dx \]
Target \(u\)
\(u = x^3 + 5\)
Differential \(du\)
\(du = 3x^2 dx\)
\[ \int \frac{\sin(\sqrt{x})}{\sqrt{x}} \, dx \]
Target \(u\)
\(u = \sqrt{x}\)
Differential \(du\)
\(du = \frac{1}{2\sqrt{x}} dx\)
\[ \int \frac{e^x}{1 + e^{2x}} \, dx \]
Target \(u\)
\(u = e^x\)
Differential \(du\)
\(du = e^x dx\)
Part II: Limit Transformation
4. \[ \int_0^1 x^3(x^4 + 1)^2 \, dx \]
Let \(u = x^4 + 1 \implies du = 4x^3 dx \implies \frac{1}{4}du = x^3 dx\)
5. \[ \int_0^{\pi/2} \sin(x)e^{\cos(x)} \, dx \]
Let \(u = \cos(x) \implies du = -\sin(x) dx \implies -du = \sin(x) dx\)
Part III: The Inverse Climb
6. Evaluate: \[ \int \frac{1}{\sqrt{9 - x^2}} \, dx \]
\[ \int \frac{1}{\sqrt{9(1 - \frac{x^2}{9})}} \, dx = \int \frac{1}{3\sqrt{1 - (\frac{x}{3})^2}} \, dx \]
Integration by Parts Slides Parts & Tables
Lesson 02: Reverse Product Rule
Damped Oscillation
In physics, a swinging pendulum in a viscous fluid experiences damping .
Its position is described by: \[ y(t) = e^{-t} \cos(t) \]
To find the total distance traveled (energy loss), we must integrate the product of two incompatible functions.
\[ \int e^{-x} \cos(x) \, dx \]
Substitution fails. We need a new tool.
The Integration by Parts Formula
Derived from the Product Rule
\[ \int u \, dv = uv - \int v \, du \]
Choosing \(u\)
Something that becomes simpler when differentiated.
Choosing \(dv\)
Something that is easy to integrate .
Hierarchy of Selection: LIATE
L
Logarithmic Functions
\(\ln(x), \log_2(x)\)
I
Inverse Trig Functions
\(\arctan(x), \arcsin(x)\)
A
Algebraic Functions
\(x^2, 3x^5, \sqrt{x}\)
T
Trigonometric Functions
\(\sin(x), \cos(x)\)
E
Exponential Functions
\(e^x, 2^x\)
Pick \(u\) based on what appears highest in this list.
The Tabular Method
Efficiency Hack
Used when \(u\) is a polynomial that eventually differentiates to zero.
1. Create columns for Signs , \(u\) , and \(dv\) .
2. Differentiate \(u\) until 0.
3. Integrate \(dv\) matching times.
4. Multiply diagonally.
\[ \int x^2 \sin(x) \, dx \]
+/- \(D\) \(I\) + \(x^2\) \(\sin(x)\) - \(2x\) \(-\cos(x)\) + \(2\) \(-\sin(x)\) - \(0\) \(\cos(x)\)
Parts and Tables Worksheet Parts and Tables
Calculus II // Integration by Parts
Name:
Date:
Part I: LIATE Selection
For each integral, identify \(u\) and \(dv\) using the LIATE hierarchy.
1. \[ \int x^2 \ln(x) \, dx \]
\(u\):
\(dv\):
2. \[ \int \arctan(x) \, dx \]
\(u\):
\(dv\):
Part II: Standard Method
3. Solve: \[ \int x \cos(x) \, dx \]
Part III: Tabular Method
Use the tabular method to evaluate these repeated-parts integrals.
4. Solve: \[ \int x^3 e^x \, dx \]
+/- Diff (\(u\)) Int (\(dv\)) + \(x^3\) \(e^x\) - + - +
Final Antiderivative
Parts and Tables Key Answer Key
Calculus II // Integration by Parts // Drill Sheet
Instructor Use Only
Part I: LIATE Selection
1. \[ \int x^2 \ln(x) \, dx \]
\(u\):
\(\ln(x)\)
\(dv\):
\(x^2 dx\)
2. \[ \int \arctan(x) \, dx \]
\(u\):
\(\arctan(x)\)
\(dv\):
\(dx\)
Part II: Standard Method
3. Solve: \[ \int x \cos(x) \, dx \]
\(u = x \implies du = dx\)
Part III: Tabular Method
4. Solve: \[ \int x^3 e^x \, dx \]
+/- Diff (\(u\)) Int (\(dv\)) + \(x^3\) \(e^x\) - \(3x^2\) \(e^x\) + \(6x\) \(e^x\) - \(6\) \(e^x\) + \(0\) \(e^x\)
Combine diagonal products:
\[ (x^3)(e^x) - (3x^2)(e^x) + (6x)(e^x) - (6)(e^x) + C \]
Factor out \(e^x\):
Trig Integration Slides Trig Transformations
Lesson 03: Radical Geometry
The Problem of the Circle
A circle with radius \(r\) has the equation: \[ x^2 + y^2 = r^2 \]
To find the area using calculus, we integrate: \[ A = 4 \int_0^r \sqrt{r^2 - x^2} \, dx \]
This radical is "locked" in the algebraic domain. We need to unlock it using angles.
\(r\)
Identity Tactics
The "Odd Power" Rule
If \(\cos(x)\) has an odd power, save one \(\cos(x)\) and convert the rest to \(\sin(x)\).
\[ \cos^2(x) = 1 - \sin^2(x) \]
The "Even Power" Rule
If both powers are even, use half-angle identities to reduce power.
\[ \cos^2(x) = \frac{1 + \cos(2x)}{2} \] \[ \sin^2(x) = \frac{1 - \cos(2x)}{2} \]
The Substitution Map
Radical Form Substitution Identity Used \[ \sqrt{a^2 - x^2} \] \[ x = a \sin(\theta) \] \[ 1 - \sin^2 \theta = \cos^2 \theta \] \[ \sqrt{a^2 + x^2} \] \[ x = a \tan(\theta) \] \[ 1 + \tan^2 \theta = \sec^2 \theta \] \[ \sqrt{x^2 - a^2} \] \[ x = a \sec(\theta) \] \[ \sec^2 \theta - 1 = \tan^2 \theta \]
The Final Reversion
STEP: θ \(\to\) x
\(\theta\)
Adjacent
Opposite
Hypotenuse
Once you find the antiderivative in terms of \(\theta\):
Draw a right triangle based on your original substitution.
If \(x = a \sin(\theta)\), then \(\sin(\theta) = \frac{x}{a}\).
Label sides as \(Opp = x\) and \(Hyp = a\).
Find the third side using Pythagorean theorem.
Trig Substitution Worksheet Trig Transformations
Calculus II // Lesson 03 Worksheet
Name:
Part I: Powers and Identities
Rewrite the integrand using trig identities to prepare for u-substitution.
1. \[ \int \sin^3(x) \cos^4(x) \, dx \]
Identity Step & Substitution Choice:
2. \[ \int \cos^2(x) \, dx \]
Reduction Formula Step:
Part II: Strategic Selection
Choose the best substitution \(x = f(\theta)\) and find \(dx\).
3. \[ \int \frac{\sqrt{25 - x^2}}{x} \, dx \]
\(x = \)
\(dx = \)
4. \[ \int \frac{1}{\sqrt{x^2 + 4}} \, dx \]
\(x = \)
\(dx = \)
Part III: The Full Climb
5. Solve: \[ \int \frac{1}{x^2 \sqrt{x^2 - 9}} \, dx \]
Show your substitution, triangle, and final algebraic answer:
Trig Substitution Key Answer Key
Calculus II // Trig Transformations // Drill Sheet
Instructor Use Only
Part I: Powers and Identities
1. \[ \int \sin^3(x) \cos^4(x) \, dx \]
Save one sine: \(\int \sin^2(x) \cos^4(x) \sin(x) dx\)
2. \[ \int \cos^2(x) \, dx \]
Use Identity: \[ \int \frac{1 + \cos(2x)}{2} \, dx \]
Part II: Strategic Selection
3. \[ \int \frac{\sqrt{25 - x^2}}{x} \, dx \]
\(x = 5 \sin(\theta)\)
\(dx = 5 \cos(\theta) d\theta\)
4. \[ \int \frac{1}{\sqrt{x^2 + 4}} \, dx \]
\(x = 2 \tan(\theta)\)
\(dx = 2 \sec^2(\theta) d\theta\)
Part III: The Full Climb
5. Solve: \[ \int \frac{1}{x^2 \sqrt{x^2 - 9}} \, dx \]
Sub: \(x = 3 \sec\theta \implies dx = 3 \sec\theta \tan\theta d\theta\)
Triangle: \(x = 3 \sec\theta \implies \cos\theta = 3/x \implies Adj=3, Hyp=x, Opp=\sqrt{x^2-9}\)
Partial Fractions Slides Rational Decomposition
Lesson 04: Partial Fractions
Logistic Systems
Populations don't grow forever. They are limited by Carrying Capacity (K) .
The rate of change is: \[ \frac{dP}{dt} = k P \left(1 - \frac{P}{K}\right) \]
To solve for \(P(t)\), we must integrate: \[ \int \frac{1}{P(1 - P/K)} \, dP \]
\[ \frac{1}{P(1 - P/K)} = \frac{A}{P} + \frac{B}{1 - P/K} \]
We "decomposed" a complex fraction into two simpler ones.
The Decomposition Catalog
1. Distinct Linear
Factors like \((x-1)(x+2)\)
\[ \frac{A}{x-1} + \frac{B}{x+2} \]
2. Repeated Linear
Factors like \((x-1)^2\)
\[ \frac{A}{x-1} + \frac{B}{(x-1)^2} \]
3. Irreducible Quad
Factors like \((x^2+1)\)
\[ \frac{Ax + B}{x^2+1} \]
Solving for the Unknowns
Method A: Zero Selection
Pick values of \(x\) that make terms vanish (e.g., \(x=1, x=-2\)).
Method B: Systems
Expand the numerator and equate coefficients of like powers of \(x\).
\[ 1 = A(x+2) + B(x-1) \]
Plug in x = 1 \(\to\) 1 = 3A
The Output Terminals
Linear Factors
\[ \int \frac{A}{x-c} dx \]
\[ A \ln|x-c| + C \]
Quadratic Factors
\[ \int \frac{1}{x^2+a^2} dx \]
\[ \frac{1}{a} \arctan(\frac{x}{a}) + C \]
Partial Fractions Worksheet Rational Decomposition
Calculus II // Lesson 04 Worksheet
Name:
Part I: The Blueprint
Write the form of the partial fraction decomposition. Do not solve for the coefficients.
\[ \frac{x+1}{(x-3)(x+4)} \]
\[ \frac{5}{x(x+1)^2} \]
\[ \frac{x^2 + 1}{x(x^2 + 4)} \]
Part II: Coefficient Solver
4. Integrate: \[ \int \frac{4}{x^2 - 4} \, dx \]
Tip: Factor the denominator first.
Part III: Quadratic Stability
5. Integrate: \[ \int \frac{x+5}{x(x^2 + 1)} \, dx \]
Show full algebraic decomposition and integration:
Partial Fractions Key Answer Key
Calculus II // Rational Decomposition // Drill Sheet
Instructor Use Only
Part I: The Blueprint
\[ \frac{x+1}{(x-3)(x+4)} \]
\[ \frac{A}{x-3} + \frac{B}{x+4} \]
\[ \frac{5}{x(x+1)^2} \]
\[ \frac{A}{x} + \frac{B}{x+1} + \frac{C}{(x+1)^2} \]
\[ \frac{x^2 + 1}{x(x^2 + 4)} \]
\[ \frac{A}{x} + \frac{Bx + C}{x^2 + 4} \]
Part II: Coefficient Solver
4. Integrate: \[ \int \frac{4}{x^2 - 4} \, dx \]
Factor: \((x-2)(x+2)\). Setup: \(\frac{4}{(x-2)(x+2)} = \frac{A}{x-2} + \frac{B}{x+2}\)
Part III: Quadratic Stability
5. Integrate: \[ \int \frac{x+5}{x(x^2 + 1)} \, dx \]
Setup: \(\frac{x+5}{x(x^2+1)} = \frac{A}{x} + \frac{Bx+C}{x^2+1}\)
Equating coeffs:
\[ \int \left( \frac{5}{x} + \frac{-5x+1}{x^2+1} \right) dx = \int \frac{5}{x} dx - 5 \int \frac{x}{x^2+1} dx + \int \frac{1}{x^2+1} dx \]
Integration Strategy Slides Integration Synthesis
Lesson 05: Strategic Diagnosis
Integration Bee
Identify the FIRST STEP for this integral in 10 seconds:
\[ \int \frac{\sin(\ln x)}{x} \, dx \]
By Parts
Substitution
Trig Sub
Mission Control: Strategy Map
1
Substitution
Is there a composite function \(f(g(x))\) whose derivative \(g'(x)\) is present?
2
By Parts
Is the integrand a product of two unrelated types (Algebraic, Trig, Log)?
3
Trig Sub
Does it contain \(\sqrt{a^2 \pm x^2}\) or \(\sqrt{x^2 - a^2}\) without \(du\)?
4
Partial Frac
Is it a rational function with a factorable denominator?
The Path of Least Resistance
Over-Engineering
Using Trig Sub for:
\[ \int \frac{x}{\sqrt{x^2+1}} dx \]
"I'll let \(x = \tan\theta\), then find \(\sec\theta\)..."
COST: 5-8 Minutes
Strategic Thinking
Using \(u\)-Substitution for:
\[ \int \frac{x}{\sqrt{x^2+1}} dx \]
"\(u = x^2+1\), \(du = 2xdx\). Done."
COST: 45 Seconds
Mission: Synthesis
Team up. Solve. Survive.
\[ \int \frac{\arctan(x)}{x^2} \, dx \]
Hint: Combine Integration by Parts and Partial Fractions.
Synthesis Challenge Sheet Synthesis Challenge
Calculus II // Final Mastery Workshop
Name:
Phase I: Strategic Diagnosis
Check the box for the most efficient first technique. Do not solve in this section.
\[ \int \frac{\ln x}{x} dx \]
Sub
Parts
Trig Sub
Partial
\[ \int x e^{2x} dx \]
Sub
Parts
Trig Sub
Partial
\[ \int \frac{1}{x^2 - 1} dx \]
Sub
Parts
Trig Sub
Partial
Phase II: Execution
4. \[ \int e^x \sin(x) \, dx \]
Observation: This will likely loop back.
5. \[ \int \frac{1}{x \sqrt{x^2+1}} \, dx \]
Observation: Consider Trig Substitution.
Phase III: The Synthesis
6. Evaluate: \[ \int \frac{\arctan(x)}{x^2} \, dx \]
ULTRA_GOAL_01
Synthesis Challenge Key Answer Key
Calculus II // Final Mastery Workshop // Key
Instructor Use Only
Phase I: Strategic Diagnosis
\[ \int \frac{\ln x}{x} dx \]
SUBSTITUTE
\[ \int x e^{2x} dx \]
BY PARTS
\[ \int \frac{1}{x^2 - 1} dx \]
PARTIAL FRAC
Phase II: Execution
4. \[ \int e^x \sin(x) \, dx \]
Use Parts twice: \(u = \sin x, dv = e^x dx\).
5. \[ \int \frac{1}{x \sqrt{x^2+1}} \, dx \]
Let \(x = \tan\theta \implies dx = \sec^2\theta d\theta\). Radical becomes \(\sec\theta\).
Phase III: The Synthesis
6. Evaluate: \[ \int \frac{\arctan(x)}{x^2} \, dx \]
Parts: \(u = \arctan x, dv = x^{-2} dx \implies du = \frac{1}{1+x^2} dx, v = -1/x\).
