Tutorial on Integration Techniques

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Tutorial on Integration Techniques

Trigonometric Substitution

  • Used to simplify integrals containing forms like $\sqrt{a^2 - x^2}$, $\sqrt{a^2 + x^2}$, or $\sqrt{x^2 - a^2}$.
  • The substitution replaces the variable with a trigonometric function to eliminate the square root.

$$ \text{Key Substitutions:} \begin{cases} \sqrt{a^2 - x^2} \rightarrow x = a \sin \theta \ \sqrt{a^2 + x^2} \rightarrow x = a \tan \theta \ \sqrt{x^2 - a^2} \rightarrow x = a \sec \theta \end{cases} $$

  • 1(a): $\int \frac{dx}{x \sqrt{x^2 - 25}}$

    • Substitution: $x = 5 \sec \theta$, $dx = 5 \sec \theta \tan \theta , d\theta$
    • Result: $\frac{1}{5} \sec^{-1} \frac{x}{5} + C$

  • 1(b): $\int \frac{dx}{x^2 \sqrt{4 - x^2}}$

    • Substitution: $x = 2 \sin \theta$, $dx = 2 \cos \theta , d\theta$
    • Result: $-\frac{\sqrt{4 - x^2}}{4x} + C$

  • 1(c): $\int \frac{dx}{x \sqrt{x^2 + 4}}$

    • Substitution: $x = 2 \tan \theta$, $dx = 2 \sec^2 \theta , d\theta$
    • Result: $\frac{1}{2} \ln \left| \frac{x}{2 + \sqrt{x^2 + 4}} \right| + C$
    • Hint used: $\int \csc \alpha , d\alpha = -\ln |\csc \alpha + \cot \alpha| + c$

  • 1(d): $\int \frac{x^3}{\sqrt{16 - x^2}} dx$

    • Substitution: $x = 4 \sin \theta$, $dx = 4 \cos \theta , d\theta$
    • Result: $-\frac{1}{3} (16 - x^2)^{3/2} + 16 \sqrt{16 - x^2} + C$

  • 1(e): $\int \frac{dx}{(4x^2 - 9)^{3/2}}$

    • Substitution: $2x = 3 \sec \theta$, $2dx = 3 \sec \theta \tan \theta , d\theta$
    • Result: $\frac{x}{9 \sqrt{4x^2 - 9}} + C$

  • 1(f): $\int \frac{dx}{x^2 \sqrt{x^2 + 9}}$

    • Substitution: $x = 3 \tan \theta$, $dx = 3 \sec^2 \theta , d\theta$
    • Result: $-\frac{\sqrt{x^2 + 9}}{9x} + C$

  • 2: Using $\int \sec \alpha , d\alpha = \ln |\sec \alpha + \tan \alpha| + c$, find $\int \frac{1}{\sqrt{2x^2 - 4}} dx$

    • Substitution: $\sqrt{2}x = 2 \sec \theta$
    • Result: $\frac{1}{\sqrt{2}} \ln \left| \sqrt{2}x + \sqrt{2x^2 - 4} \right| + C$

Definite Integrals with Absolute Values

  • For $\int_a^b |f(x)| dx$, split the integral at points where $f(x) = 0$.
  • 3: $\int_{0}^{4} |x^2 - 2| , dx$
    • Breakpoint: $x^2 - 2 = 0 \implies x = \sqrt{2}$ (positive root in $[0,4]$)
    • Split: $$\int_{0}^{\sqrt{2}} (2 - x^2) dx + \int_{\sqrt{2}}^{4} (x^2 - 2) dx$$
    • Result: $= \frac{4\sqrt{2}}{3} + \left( \frac{64}{3} - 8 - \frac{2\sqrt{2}}{3} + 2\sqrt{2} \right) = \frac{76}{3} - \frac{8\sqrt{2}}{3}$

Integration by Partial Fractions

  • Decompose a rational function $\frac{P(x)}{Q(x)}$ into simpler fractions.

  • Steps:

    1. Ensure degree of $P(x)$ < degree of $Q(x)$ (if not, perform polynomial division).
    2. Factor $Q(x)$.
    3. Write the partial fraction decomposition.

  • 4(a): $\int \frac{dx}{(x+4)(x-1)}$

    • Decomposition: $\frac{1}{(x+4)(x-1)} = \frac{A}{x+4} + \frac{B}{x-1}$
    • Solve: $1 = A(x-1) + B(x+4)$.
      • Set $x=1$: $1 = 5B \implies B = 1/5$.
      • Set $x=-4$: $1 = -5A \implies A = -1/5$.
    • Integral: $\int \left( -\frac{1}{5(x+4)} + \frac{1}{5(x-1)} \right) dx$
    • Result: $\frac{1}{5} \ln \left| \frac{x-1}{x+4} \right| + C$

  • 4(b): $\int \frac{x+5}{x^2 - 1} dx$

    • Factor denominator: $x^2 - 1 = (x-1)(x+1)$.
    • Decomposition: $\frac{x+5}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$
    • Solve: $x+5 = A(x+1) + B(x-1)$.
      • Set $x=1$: $6 = 2A \implies A = 3$.
      • Set $x=-1$: $4 = -2B \implies B = -2$.
    • Integral: $\int \left( \frac{3}{x-1} - \frac{2}{x+1} \right) dx$
    • Result: $3 \ln |x-1| - 2 \ln |x+1| + C$

  • 5(a): $\int \frac{x^3 - 2x^2 + 3}{x^4 + x^2} dx$

    • Factor denominator: $x^4 + x^2 = x^2(x^2 + 1)$.
    • Partial fractions need forms: $\frac{A}{x} + \frac{B}{x^2} + \frac{Cx + D}{x^2 + 1}$.
    • Result: $\ln|x| + \frac{3}{x} - 2 \tan^{-1} x + C$

  • 5(b): $\int \frac{5x^3 - 4x^2 + 2x - 3}{x^4 + x^2} dx$

    • Partial fractions: $\frac{5x^3 - 4x^2 + 2x - 3}{x^2(x^2+1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{Cx + D}{x^2+1}$
    • Result: $2 \ln|x| + \frac{3}{x} + 3 \tan^{-1} x + C$

More Definite Integrals

  • 6(a): $\int_{1}^{2} \left( x - \frac{1}{x} \right) \left( x + \frac{1}{x} \right) dx$

    • Simplify integrand: $(x - 1/x)(x + 1/x) = x^2 - 1/x^2$.
    • Integral: $\int_{1}^{2} (x^2 - x^{-2}) dx$
    • Result: $\left[ \frac{x^3}{3} + x^{-1} \right]_{1}^{2} = \left( \frac{8}{3} + \frac{1}{2} \right) - \left( \frac{1}{3} + 1 \right) = \frac{11}{6}$

  • 6(b): $\int_{0}^{1} (1-x) \sqrt{x} , dx$

    • Expand: $\int_0^1 (x^{1/2} - x^{3/2}) dx$
    • Result: $\left[ \frac{2}{3} x^{3/2} - \frac{2}{5} x^{5/2} \right]_{

Source: Soalan Tut 4 FAD1014(25.26)..pdf