Quiz 2

Integration Techniques

1011 words
5 min read
Python Week 1: the first filter for runtime behavior
Visual companion
Python
Type and operator map

Python Week 1: the first filter for runtime behavior

Open galleryRelated surface
View
Revision summary

What this note is really saying

Short form

# Integration Techniques ## 🎯 Learning Objectives By the end of this topic, you will be able to: 1. **Apply** substitution method for composite functions 2.

Integration Techniques

🎯 Learning Objectives

By the end of this topic, you will be able to:
  1. Apply substitution method for composite functions
  2. Use integration by parts for products of functions
  3. Decompose rational functions using partial fractions
  4. Integrate piecewise-defined functions

📋 Prerequisites

📖 Core Content

17.1 Review of Substitution

∫f(g(x))⋅g′(x) dx=∫f(u) duwhere u=g(x)\int f(g(x)) \cdot g'(x)\,dx = \int f(u)\,du \quad \text{where } u = g(x)
Example: ∫2xcos⁡(x2) dx\int 2x \cos(x^2)\,dx Let u=x2u = x^2, du=2x dxdu = 2x\,dx. ∫cos⁡u du=sin⁡u+C=sin⁡(x2)+C\int \cos u\,du = \sin u + C = \sin(x^2) + C

17.2 Integration by Parts

∫u dv=uv−∫v du\int u\,dv = uv - \int v\,du
Intuition: The reverse of the product rule. Choose uu as the part that simplifies when differentiated. LIATE rule (order of choosing uu): Logarithms, Inverse trig, Algebraic, Trigonometric, Exponential.

Worked Examples

Example 2.1: ∫xex dx\int x e^x\,dx Let u=xu = x, dv=ex dxdv = e^x\,dx. Then du=dxdu = dx, v=exv = e^x. ∫xex dx=xex−∫ex dx=xex−ex+C=ex(x−1)+C\int x e^x\,dx = x e^x - \int e^x\,dx = x e^x - e^x + C = e^x(x-1) + C
ex(x−1)+C\boxed{e^x(x-1) + C}
Example 2.2: ∫ln⁡x dx\int \ln x\,dx Let u=ln⁡xu = \ln x, dv=dxdv = dx. Then du=1xdxdu = \frac{1}{x}dx, v=xv = x. ∫ln⁡x dx=xln⁡x−∫x⋅1x dx=xln⁡x−∫1 dx=xln⁡x−x+C\int \ln x\,dx = x\ln x - \int x \cdot \frac{1}{x}\,dx = x\ln x - \int 1\,dx = x\ln x - x + C
xln⁡x−x+C\boxed{x\ln x - x + C}
Example 2.3: ∫x2ex dx\int x^2 e^x\,dx (repeated parts) Let u=x2u = x^2, dv=ex dxdv = e^x\,dx. du=2x dxdu = 2x\,dx, v=exv = e^x. =x2ex−∫2xex dx=x2ex−2(xex−ex)+C=ex(x2−2x+2)+C= x^2 e^x - \int 2x e^x\,dx = x^2 e^x - 2(x e^x - e^x) + C = e^x(x^2 - 2x + 2) + C
ex(x2−2x+2)+C\boxed{e^x(x^2 - 2x + 2) + C}

17.3 Partial Fractions

For ∫P(x)Q(x) dx\int \frac{P(x)}{Q(x)}\,dx where degree of PP < degree of QQ: Step 1: Factor Q(x)Q(x) Step 2: Write partial fraction decomposition Step 3: Integrate each term
Factor in Q(x)Q(x)Term in Decomposition
(ax+b)(ax + b)Aax+b\frac{A}{ax+b}
(ax+b)k(ax+b)^kA1ax+b+A2(ax+b)2+⋯+Ak(ax+b)k\frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2} + \cdots + \frac{A_k}{(ax+b)^k}
(ax2+bx+c)(ax^2+bx+c) irreducibleAx+Bax2+bx+c\frac{Ax+B}{ax^2+bx+c}

Worked Example

Example 3.1: ∫2x+3x2+x−2 dx\int \frac{2x+3}{x^2 + x - 2}\,dx Step 1 — Factor: x2+x−2=(x−1)(x+2)x^2+x-2 = (x-1)(x+2) Step 2 — Decompose: 2x+3(x−1)(x+2)=Ax−1+Bx+2\frac{2x+3}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2} 2x+3=A(x+2)+B(x−1)=(A+B)x+(2A−B)2x+3 = A(x+2) + B(x-1) = (A+B)x + (2A - B) A+B=2A+B = 2, 2A−B=3  ⟹  A=5/32A-B = 3 \implies A = 5/3, B=1/3B = 1/3 Step 3 — Integrate: ∫5/3x−1 dx+∫1/3x+2 dx=53ln⁡∣x−1∣+13ln⁡∣x+2∣+C\int \frac{5/3}{x-1}\,dx + \int \frac{1/3}{x+2}\,dx = \frac{5}{3}\ln|x-1| + \frac{1}{3}\ln|x+2| + C
13(5ln⁡∣x−1∣+ln⁡∣x+2∣)+C\boxed{\frac{1}{3}\left(5\ln|x-1| + \ln|x+2|\right) + C}

17.4 Trigonometric Integrals

TypeStrategy
∫sin⁡mxcos⁡nx dx\int \sin^m x \cos^n x\,dxIf mm odd, save sin⁡x\sin x , convert rest to cos⁡\cos
If nn odd, save cos⁡x\cos x , convert rest to sin⁡\sin
If both even, use half-angle formulas
∫tan⁡mxsec⁡nx dx\int \tan^m x \sec^n x\,dxIf nn even, save sec⁡2x\sec^2 x
If mm odd, save sec⁡xtan⁡x\sec x \tan x
Example: ∫sin⁡2x dx=∫1−cos⁡2x2 dx=x2−sin⁡2x4+C\int \sin^2 x\,dx = \int \frac{1-\cos 2x}{2}\,dx = \frac{x}{2} - \frac{\sin 2x}{4} + C

17.5 Integration of Piecewise Functions

Split the integral at the piece boundaries and integrate each piece separately. Example: ∫−12f(x) dx\int_{-1}^2 f(x)\,dx where
f(x)={xx<0x2x≥0f(x) = \begin{cases} x & x < 0 \\ x^2 & x \geq 0 \end{cases}
∫−10x dx+∫02x2 dx=[x22]−10+[x33]02=(0−12)+(83−0)=−12+83=136\int_{-1}^0 x\,dx + \int_0^2 x^2\,dx = \left[\frac{x^2}{2}\right]_{-1}^0 + \left[\frac{x^3}{3}\right]_0^2 = (0 - \frac{1}{2}) + (\frac{8}{3} - 0) = -\frac{1}{2} + \frac{8}{3} = \frac{13}{6}

📐 Key Formulas — Summary Table

TechniqueFormulaWhen to Use
Substitution∫f(g(x))g′(x) dx=∫f(u) du\int f(g(x))g'(x)\,dx = \int f(u)\,duComposite functions
Parts∫u dv=uv−∫v du\int u\,dv = uv - \int v\,duProduct of functions
Partial fractionsP(x)Q(x)=∑Aifactor\frac{P(x)}{Q(x)} = \sum \frac{A_i}{\text{factor}}Rational functions
Trig identitiessin⁡2x=1−cos⁡2x2\sin^2 x = \frac{1-\cos 2x}{2}Powers of trig

⚠️ Common Pitfalls

Pitfall 1: Wrong Choice of uu in Parts

Choose uu that simplifies when differentiated. Use LIATE: ln⁡x\ln x before algebraic, exponential last.

Pitfall 2: Forgetting Absolute Values in Log Integrals

∫1x dx=ln⁡∣x∣+C\int \frac{1}{x}\,dx = \ln|x| + C, not ln⁡x+C\ln x + C (which is only valid for x>0x > 0).

Pitfall 3: Partial Fractions Requires Proper Rational Functions

If degree of numerator ≥ degree of denominator, divide first.

📝 Practice Questions

Q1: Find ∫xsin⁡x dx\int x \sin x\,dx.
Parts: u=xu=x, dv=sin⁡x dxdv=\sin x\,dx, du=dxdu=dx, v=−cos⁡xv=-\cos x. =−xcos⁡x+∫cos⁡x dx=−xcos⁡x+sin⁡x+C= -x\cos x + \int \cos x\,dx = -x\cos x + \sin x + C
−xcos⁡x+sin⁡x+C\boxed{-x\cos x + \sin x + C} Q2: Find ∫dxx2−1\int \frac{dx}{x^2 - 1}.
1(x−1)(x+1)=1/2x−1−1/2x+1\frac{1}{(x-1)(x+1)} = \frac{1/2}{x-1} - \frac{1/2}{x+1} ∫=12ln⁡∣x−1∣−12ln⁡∣x+1∣+C=12ln⁡∣x−1x+1∣+C\int = \frac{1}{2}\ln|x-1| - \frac{1}{2}\ln|x+1| + C = \frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| + C
12ln⁡∣x−1x+1∣+C\boxed{\frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| + C} Q3: Find ∫ln⁡(x2) dx\int \ln(x^2)\,dx.
ln⁡(x2)=2ln⁡x\ln(x^2) = 2\ln x 2∫ln⁡x dx=2(xln⁡x−x)+C2\int \ln x\,dx = 2(x\ln x - x) + C
2xln⁡x−2x+C\boxed{2x\ln x - 2x + C} Q4: Find ∫xe2x dx\int xe^{2x}\,dx.
Parts: u=xu=x, dv=e2xdxdv=e^{2x}dx, du=dxdu=dx, v=e2x/2v=e^{2x}/2. =xe2x2−12∫e2xdx=xe2x2−e2x4+C= \frac{xe^{2x}}{2} - \frac{1}{2}\int e^{2x}dx = \frac{xe^{2x}}{2} - \frac{e^{2x}}{4} + C
e2x4(2x−1)+C\boxed{\frac{e^{2x}}{4}(2x - 1) + C} Q5: Find ∫01xex dx\int_0^1 x e^x\,dx.
[xex−ex]01=(e−e)−(0−1)=1[xe^x - e^x]_0^1 = (e - e) - (0 - 1) = 1
1\boxed{1} Q6: Find ∫dxx2+4\int \frac{dx}{x^2 + 4}.
12arctan⁡(x2)+C\frac{1}{2}\arctan\left(\frac{x}{2}\right) + C
12arctan⁡(x2)+C\boxed{\frac{1}{2}\arctan\left(\frac{x}{2}\right) + C} **Q7: Find ∫−11f(x) dx\int_{-1}^1 f(x)\,dx where
>f(x)={1x<02x≥0>> f(x) = \begin{cases} 1 & x<0 \\ 2 & x \geq 0 \end{cases} >
.**
∫−101 dx+∫012 dx=1+2=3\int_{-1}^0 1\,dx + \int_0^1 2\,dx = 1 + 2 = 3
3\boxed{3} Q8: Find ∫cos⁡2x dx\int \cos^2 x\,dx.
1+cos⁡2x2\frac{1+\cos 2x}{2} → x2+sin⁡2x4+C\frac{x}{2} + \frac{\sin 2x}{4} + C
x2+sin⁡2x4+C\boxed{\frac{x}{2} + \frac{\sin 2x}{4} + C}

🔗 Cross-References

Document outline

Keep your place and jump directly to a heading.

Table of Contents
System Normal // Awaiting Context

Intelligence Hub

Navigate the knowledge graph to generate context. The Hub adapts dynamically to surface backlinks, related notes, and metadata insights.

Open library →Launch Assessment Engine