Techniques of Integration Lecture Notes

Techniques of Integration

Lecture Overview

  • Course: Math 203
  • Location: German University in Cairo (GUC)
  • Topic: Techniques of Integration

Recall on Integration

Lecture Outline

  1. Recall on Integration
    • Indefinite Integrals
    • Evaluation of Indefinite Integrals
    • Evaluation of Definite Integrals
  2. Techniques of Integration
    • Substitution Method
    • Integration by Parts

Recommended Reading

  • Thomas’ Calculus (13th ed.): sections 5.5, 8.1 and 8.2
  • Stewart Calculus (7th ed.): sections 4.4, 4.5, and 7.1

Indefinite Integrals

Definition

  • The indefinite integral of a function $f(x)$ on an interval $I$, denoted $ rac{1}{ ext{dx}} igg[ f(x) igg]$, represents the most general antiderivative of $f$ on $I$.

Antiderivatives

  • A function $F$ is an antiderivative of $f$ on an interval $I$ if:
    • F'(x) = f(x) for all $x ext{ in } I$.

Examples of Antiderivatives

  1. rac{d}{dx} igg[ rac{1}{5} x^5 igg] = x^4
    • Therefore, $F(x) = rac{1}{5} x^5$ is an antiderivative of $f(x) = x^4$ on $ ext{R}$.
  2. rac{d}{dx} igg[ rac{1}{2} ext{sin}(2x) igg] = ext{cos}(2x)
    • Therefore, $F(x) = rac{1}{2} ext{sin}(2x)$ is an antiderivative of $f(x) = ext{cos}(2x)$ on $ ext{R}$.
  3. rac{d}{dx} [ ext{ln}(x)] = rac{1}{x}
    • Therefore, $F(x) = ext{ln}(x)$ is an antiderivative of $f(x) = rac{1}{x}$ on $(0, + ext{∞})$.

Properties of Antiderivatives

  1. If $F(x)$ is an antiderivative of $f(x)$ on an interval $I$, then $F(x) + C$ is also an antiderivative for any constant $C$.
  2. If $F$ and $G$ are two antiderivatives of $f$ over $I$ then:
    • G(x) = F(x) + C for some constant $C$.

Proof of the Claim

  • If $F'(x) = G'(x)$, then:
    • G'(x) - F'(x) = 0
      ightarrow ext{asserts: } rac{d}{dx}[G(x) - F(x)] = 0
    • Thus, G(x) - F(x) = C (valid only over intervals).

Remarks

  • The premise of Property (2) indicates that any two different antiderivatives of a function on an interval will differ only by a constant.
  • Property (2) does not hold if $I$ is not an interval.
    • Question: Can you give a counterexample?

Basic Indefinite Integrals

Key Basic Integrals

  • These should be memorized as they are fundamental:
    • rac{ ext{dx}}{x^n} = rac{x^{n+1}}{n + 1} + C ext{ for } n
      eq -1 on $ ext{R}$
    • rac{ ext{dx}}{x^ ext{α}} = rac{x^{ ext{α}+1}}{ ext{α} + 1} + C ext{ for } ext{α}
      eq -1 on $(0, + ext{∞})
    • rac{1}{x} ext{dx} = ext{ln}|x| + C on $(0, + ext{∞})$ or $(- ext{∞}, 0)$
    • e^{kx} ext{dx} = rac{1}{k}e^{kx} + C on $ ext{R}$
    • a^{kx} ext{dx} = rac{1}{k} ext{ln}(a) a^{kx} + C ext{ (where a > 0 and a ≠ 1)} on $ ext{R}$
    • rac{ ext{dx}}{ ext{cos}(k ext{x})} = rac{1}{k} ext{sin}(k ext{x}) + C on $ ext{R}$
    • rac{ ext{dx}}{ ext{sin}(k ext{x})} = - rac{1}{k} ext{cos}(k ext{x}) + C on $ ext{R}$
    • rac{ ext{dx}}{ ext{cosh}(k ext{x})} = rac{1}{k} ext{sinh}(k ext{x}) + C on $ ext{R}$
    • rac{ ext{dx}}{ ext{sinh}(k ext{x})} = rac{1}{k} ext{cosh}(k ext{x}) + C on $ ext{R}$
    • For $a > 0$: rac{ ext{dx}}{ ext{a}^2 - x^2} = ext{sin}^{-1}igg( rac{x}{a}igg) + C ext{ on } |x| < a
    • rac{ ext{dx}}{ ext{a}^2 + x^2} = rac{1}{a} ext{tan}^{-1}igg( rac{x}{a}igg) + C ext{ on } ext{R}
    • rac{ ext{dx}}{x^2 + a^2} = ext{sinh}^{-1}igg( rac{x}{a}igg) + C ext{ on } ext{R}
    • rac{ ext{dx}}{x^2 - a^2} = ext{cosh}^{-1}igg( rac{x}{a}igg) + C ext{ on } x > a

Remarks on Evaluation of Non-Basic Integrals

  • Evaluating integrals like $ rac{1}{ ext{f(x)}}$ where $f(x)$ isn't a function with an established derivative can be complex.
    • It’s noted that no general recipe exists for finding explicit expressions for all antiderivatives.
    • However, some specific types of integrals can be solved using known techniques that will be explored in the semester.

Fundamental Theorem of Calculus

Statement

  1. If $f$ is continuous over $[a, b]$, then:
    • F(x) = rac{ ext{dx}}{ ext{dx}} igg[ igg( rac{1}{t}(t) ext{dt} igg) is an antiderivative of $f$ on $(a, b)$.
    • This implies $F$ is differentiable and:
    • F'(x) = rac{d}{dx} igg[ rac{ ext{dx}}{dt} (t) igg] = f(x)
  2. If $f$ is continuous over $[a, b]$ and $F$ is any antiderivative of $f$ on that interval:
    • rac{b}{a} f(x) ext{dx} = F(b) - F(a)

Remark

  • Part (2) of the Fundamental Theorem justifies the interest in evaluating indefinite integrals as they facilitate the computation of definite integrals, which are often applied in various contexts.
  • Example:
    • rac{1}{4} ext{dx} igg( rac{1}{x^2 + 1}igg) = ext{tan}^{-1}(x) igg|_0^1 = ext{tan}^{-1}(1) - ext{tan}^{-1}(0) = rac{ ext{π}}{4} - 0 = rac{ ext{π}}{4}

Techniques of Integration

Substitution Method

Formula
  • The substitution (or change of variable) method states:
    • rac{ ext{dx}}{f(u(x))u'(x)} = rac{ ext{du}}{f(u)}
Proof
  • To prove, verify that the derivative of the right-hand side equates to the integrand on the left-hand side:
    • Let $u = u(x)$ and $F(u) = rac{ ext{du}}{f(u)}$
    • Then:
    • rac{d}{dx}F(u) = rac{dF}{du} rac{du}{dx} = f(u) u'
Indications from Substitution method
  • The substitution method is applicable to integrals such as:
    • rac{u^n}{u'}dx = rac{u^{n+1}}{n+1} + C ext{ (for } n ext{ } ext{being natural numbers)}
    • rac{u^{ ext{α}}}{u'}dx = rac{u^{ ext{α}+1}}{ ext{α}+1} + C ext{ (for α ≠ -1)}
    • rac{u'}{u}dx = ext{ln}|u| + C
    • Various integration forms follow this structure as shown in Page 8 (numbered with properties of functions involving $cos$ and $sin$).

Integration by Parts

Formula
  • The integration by parts formula is derived from the product rule for derivatives:
    • rac{d}{dx}[uv] = u'v + uv'
    • Thus, the integration by parts formula can be expressed as:
    • rac{dx}{(uv)'} = uv - rac{dx}{u'}v
Proof
  • Deriving from the product rule yields:
    • rac{d}{dx}(uv) = u'v + uv'
      ightarrow uv' = rac{d}{dx}(uv) - u'v
    • Therefore, integrating leads to:
    • rac{dx}{uv} = u v - rac{dx}{u} v

Examples of Integration by Parts

  1. Example 1: Calculate rac{dx}{xe^x}

    • Let: $u = x$ (then $du = dx$), $v' = e^x
      ightarrow v = e^x$
    • Thus: rac{dx}{xe^x} = x e^x - rac{d}{du} (e^{x})dx
    • Solving results in rac{dx}{x e^x} = e^x (x - 1)

  2. Example 2: Calculate rac{dx}{xe^x} where $I= rac{dx}{xe^x}$

    • The calculation becomes: I = e^x (x bex + (1-I))

Integration by Parts for Definite Integrals

  • By combining the integration by parts formula with Part (2) of the Fundamental Theorem of Calculus:
    • rac{b}{a} u v' dx = uv|_a^b - rac{b}{a} u' v dx

Application of Integration by Parts

Reduction Formula

  1. For $F_n(x)= rac{ ext{dx}}{ ext{cos}^n x}$, where $n ≥ 0$ is an integer:

    • Prove the reduction formula:
    • Fn(x) = rac{1}{n} ext{cos}^{n-1}(x) ext{sin}(x) + rac{n-1}{n} F{n-2}(x)

  2. Using the reduction formula, derive $F_4(x)$ as follows:

    • F4(x) = rac{1}{4} ext{cos}^3 x ext{sin} x + rac{3}{4} F2(x) and repeat evaluation of $F_2$.

  3. To show the integral evaluates as:

    • For $n ≥ 2$, rac{b}{a} ext{cos}^n x dx = rac{n-1}{n} rac{b}{a} ext{cos}^{n-2} x dx

  4. If we set I_n = rac{ ext{dx}}{ ext{cos}^n x} for $b = rac{ ext{π}}{2}$, then recursively apply the pattern shown in step (3) to derive results for odd powers.

    • Periodic application of this gets results such as = rac{2 ext{..} ext{n}}{3.5.7… (2k+1)} yielding results like rac{2.4…2k}{(2k+1)}$$

Conclusion

  • Execute the tasks outlined to apply integration techniques effectively whether by substitution or integration by parts over defined intervals or indefinite forms.
  • Key takeaway: These methods simplify complex integrals.

Closing Remarks

  • The lecture introduced fundamental techniques for integration in calculus and prepared students for more complex applications in the later sessions of the course.