Calculus 1AA3 – Integration Techniques & Improper Integrals

Instructor & Course Logistics

  • Instructor: Chris McLean
    • Email: mcleac3@mcmaster.ca
    • Offices: HH 103 / H
    • Office Hours: Wednesday 5:30–6:30 pm (online only)
    • Extra Help: Math Help Centre (Math Café) open 4:30–5:30 pm before office hour.

  • Assessment Scheme
    6\approx 6 Assignments – 20%20\%
    • Test #1 (multiple choice) – 20%20\%
    • Test #2 (multiple choice) – 20%20\%
    • Final Exam (multiple choice) – 40%40\%
    Total100%100\%

  • Textbook: Calculus – Early Transcendentals, 9 E (Stewart, Clegg, Watson, Cengage).
    • Ensure access through Chapter 15.

Quick Reference – Indefinite Integrals

  • Fundamental antiderivatives constantly used in lecture:
    x3/2dx=116x2+C\int x^{\,3/2}\,dx = \frac{1}{\,\,16}x^{2}+C (generic power rule demo)
    11x2dx=artanh(x)+C\int \frac{1}{1-x^{2}}\,dx = \text{artanh}(x)+C
    secxdx=lnsecx+tanx+C\int \sec x\,dx = \ln|\sec x+\tan x|+C
    exdx=ex+C\int e^{x}\,dx = e^{x}+C
    cosxdx=sinx+C\int \cos x\,dx = \sin x + C
    sinxdx=cosx+C\int \sin x\,dx = -\cos x + C
    • Hyperbolic forms:
    sinhxdx=coshx+C\int \sinh x\,dx = \cosh x + C
    coshxdx=sinhx+C\int \cosh x\,dx = \sinh x + C
    tanhxdx=lncoshx+C\int \tanh x\,dx = \ln|\cosh x| + C (implied).

Technique 1 – Substitution (a.k.a. u-Sub)

  • Core idea: pick u=g(x)u=g(x) so integral rewrites as f(u)du\int f(u)\,du.

  • Examples worked:
    x3cos(x2)dxu=x2,  du=2xdx\int \sqrt{x^{3}}\cos(x^{2})\,dx\,\Rightarrow\,u=x^{2},\;du=2x\,dx (adjust constant factors).
    • Piecewise limits substitution example: definite integral with u=4=x2u = \sqrt{4}=x-2 interpreted over x=36x=3\to6.

Technique 2 – Integration by Parts

  • Formula: udv=uvvdu\int u\,dv = u\,v-\int v\,du.

  • “LIATE” / “Thumb rule” used to rank choices:

    1. Log (best) 2. Inverse trig 3. Algebraic 4. Trig 5. Exponential (worst).
      • Best uu: lnx,arctanx\ln x, \arctan x.
      • Mid uu: x,x2,x3x,x^{2},x^{3}\dots
      • Worst uu: ex,sinx,cosx,sinhx,coshxe^{x},\sin x,\cos x,\sinh x,\cosh x.

  • Lecture demo: xexdx=(x1)ex+C\int x e^{x}\,dx = (x-1)e^{x}+C.

Technique 3 – Partial Fractions

  • If degP(x)degQ(x)\deg P(x) \ge \deg Q(x), perform polynomial (synthetic) division first.

  • Decomposition templates:
    • Distinct linear: Axa+Bxb+\dfrac{A}{x-a}+\dfrac{B}{x-b}+\dots
    • Repeated linear: A<em>1xa+A</em>2(xa)2+\dfrac{A<em>{1}}{x-a}+\dfrac{A</em>{2}}{(x-a)^{2}}+\dots
    • Irreducible quadratic: Ax+Bax2+bx+c\dfrac{Ax+B}{ax^{2}+bx+c}
    • Repeated quadratic: extend powers similarly.

  • Class example: 1(x3)(x2)dx\int \dfrac{1}{(x-3)(x-2)}\,dx yielded constants A=1,  B=1A=-1,\;B=1, producing lnx2lnx3+C=lnx2x3+C\ln|x-2|-\ln|x-3|+C = \ln\Big|\dfrac{x-2}{x-3}\Big|+C.

Technique 4 – Trigonometric Integrals (Powers of sin,cos,tan,sec\sin,\cos,\tan,\sec)

  • Decision table (memorise):
    sinmxcosnxdx\int \sin^{m}x\cos^{n}x\,dx
    – If nn odd: let u=sinxu=\sin x.
    – If mm odd: let u=cosxu=\cos x.
    – If neither odd: halve–angle sin2t=1cos2t2,  cos2t=1+cos2t2\sin^{2}t=\tfrac{1-\cos 2t}{2},\;\cos^{2}t=\tfrac{1+\cos 2t}{2} until one power becomes odd.
    tanmxsecnxdx\int \tan^{m}x\sec^{n}x\,dx
    – If nn even: u=tanxu=\tan x (use sec2x=1+tan2x\sec^{2}x=1+\tan^{2}x).
    – If mm odd: u=secxu=\sec x (use tan2x=sec2x1\tan^{2}x=\sec^{2}x-1).
    – If neither trigger: often must integrate secx\sec x directly: secxdx=lnsecx+tanx+C\int \sec x\,dx = \ln|\sec x + \tan x| + C.

  • Worked sample: tan4xsec4xdx\int \tan^{4}x\sec^{4}x\,dx solved via u=tanx,  du=sec2xdxu=\tan x,\;du=\sec^{2}x\,dx.

Technique 5 – Trigonometric Substitution

  • Standard substitutions:
    • For a2x2\sqrt{a^{2}-x^{2}} use x=asint,  dx=acostdtx=a\sin t\,,\;dx=a\cos t\,dt.
    • For a2+x2\sqrt{a^{2}+x^{2}} use x=atant,  dx=asec2tdtx=a\tan t\,,\;dx=a\sec^{2} t\,dt.
    • For x2a2\sqrt{x^{2}-a^{2}} use x=asect,  dx=asecttantdtx=a\sec t\,,\;dx=a\sec t\tan t\,dt.

  • Classroom derivations included algebraic re-expressions using sec2t1=tan2t\sec^{2}t-1=\tan^{2}t and 1+tan2t=sec2t1+\tan^{2}t=\sec^{2}t.

Improper Integrals – Overview

  • Occur when interval is infinite or integrand blows up at an endpoint. Two categories:

    1. Type I: infinite limits (to \infty or -\infty).

    2. Type II: integrand has vertical asymptote inside interval.

  • Formal definitions use limits to replace the offending bound.

Type I Definition & Extension

  • If ff continuous on [a,)[a,\infty):
    <em>af(x)dx=lim</em>babf(x)dx\int<em>{a}^{\infty}f(x)\,dx = \lim</em>{b\to\infty}\int_{a}^{b}f(x)\,dx.

  • For (,)(-\infty,\infty) split at any finite point cc:
    <em>f(x)dx=</em>cf(x)dx+cf(x)dx\int<em>{-\infty}^{\infty}f(x)\,dx = \int</em>{-\infty}^{c}f(x)\,dx + \int_{c}^{\infty}f(x)\,dx
    (needs both limits to converge).

Classic Type I Examples
  • 11xdx\int_{1}^{\infty}\dfrac{1}{x}\,dx — diverges (harmonic tail) \Rightarrow \infty.

  • <em>011+x2dx=[arctanx]</em>0=π2\int<em>{0}^{\infty}\dfrac{1}{1+x^{2}}\,dx = \big[\arctan x\big]</em>{0}^{\infty}=\dfrac{\pi}{2} — convergent.

  • Oscillatory counter-example: 2cosxdx\int_{2}^{\infty}\cos x\,dx diverges (limit fails).

Type I “p–Integrals”

  • 1xpdx\displaystyle \int_{1}^{\infty} x^{-p}\,dx converges \Leftrightarrow p>1, diverges for p1p\le1.

  • Heuristic: compare against harmonic 1/x1/x.

Type II Definition

  • If ff continuous on (a,b](a,b] but unbounded as xa+x\to a^{+}:
    <em>abf(x)dx=lim</em>ca+cbf(x)dx\int<em>{a}^{b}f(x)\,dx = \lim</em>{c\to a^{+}}\int_{c}^{b}f(x)\,dx.

  • Likewise when singularity at the right endpoint, or inside interval (split integral).

Classic Type II Examples
  • 01x1/2dx=2\int_{0}^{1} x^{-1/2}\,dx = 2 — convergent.

  • 01x1dx\int_{0}^{1} x^{-1}\,dx diverges (log blow-up).

Type II “p–Integrals”

  • 01xpdx\displaystyle \int_{0}^{1} x^{-p}\,dx converges \Leftrightarrow p<1, diverges for p1p\ge1 (opposite condition to Type I).

Comparison Tests for Improper Integrals

  • Direct Comparison
    • If 0g(x)f(x)0\le g(x) \le f(x) and f\int f converges, then g\int g converges.
    • If 0f(x)g(x)0\le f(x) \le g(x) and g\int g diverges, then f\int f diverges.

  • Mnemonic (lecture’s cat metaphor):
    • “Greater than divergent ⇒ divergent.”
    • “Smaller than convergent ⇒ convergent.”
    • Intermediate comparisons give no info.

  • Example: 02+cosxxdx\displaystyle \int_{0}^{\infty} \dfrac{2+\cos x}{x}\,dx compared to 1/x1/x; since 2+cosx12+\cos x\ge1 and 1/x\int 1/x diverges, the target integral diverges.

  • Another sample: 0arctanxx3dx\int_{0}^{\infty}\dfrac{\arctan x}{x^{3}}dx
    • Note 0<\arctan x<\tfrac{\pi}{2} so integrand <\dfrac{\pi/2}{x^{3}}; since p=3>1 in tail region, the comparison integral converges ⇒ original converges.

Mixed-Type Example (Finite + Infinite)

  • 0e2xx5/2dx\int_{0}^{\infty} \dfrac{e^{-2x}}{x^{5/2}}\,dx required splitting:
    • Near x=0x=0: Type II; compare to 1/x5/21/x^{5/2} (convergent since p>1).
    • Tail xx\to\infty: Type I; exponential decay ensures convergence.
    ⇒ Integral convergent; explicit evaluation via u=xu=\sqrt{x} leads to 22.

Summary Cheat-Sheet

  • Always replace infinite limits or vertical asymptotes by limits before integrating.

  • p–integrals give quick tests:
    • Tail xpx^{-p}: converge if p>1.
    • Origin xpx^{-p}: converge if p<1.

  • Comparison is your Swiss-army knife once integrand is bounded above/below by a known p-integral or exponential.

  • Integration toolbox:

    1. Substitution → simple chain rule inverses.

    2. By parts → product of algebraic & exponential/trig.

    3. Trig power strategies → parity decisions.

    4. Trig substitution → radicals a2±x2\sqrt{a^{2}\pm x^{2}}.

    5. Partial fractions → rational functions.

  • Ethics / Practicalities: treat divergent integrals with caution— they do not produce real finite areas, and Fundamental Theorem of Calculus can fail (e.g., integrand not continuous on whole interval).


End of Lecture 01 (Sections 01.1 – 01.3) study notes.