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Integration Techniques, L’Hopital’s Rule, and Improper Integrals

8.6 Integration by Tables and Other Integration Techniques

Focus on choose an integration strategy, reduction formulas, verify a table result.

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Ideas in this lesson

01Choose an integration strategyWorked steps · five questions02Reduction formulasWorked steps · five questions03Verify a table resultWorked steps · five questions
Idea 1 of 3

Choose an integration strategy

Inspect structure before computing. Substitution, parts, and partial fractions solve different patterns.

Worked example 1

Choose the most direct technique for xexxe^{x}.
  1. u=x,dv=exdx  ⟹  ∫xexdx=xex−ex+Cu = x, dv = e^{x} dx \implies \int x e^{x} dx = x e^{x} - e^{x} + C
  2. ∫xexdx  ⟹  \int xe^{x} dx \implies Integration by parts

Answer: Integration by parts

Worked example 2

Choose the most direct technique for 1x(x+3)\dfrac{1}{x(x + 3)}.
  1. 1x(x+3)=13[1x−1x+3]\dfrac{1}{x(x + 3)} = \dfrac{1}{3}[\dfrac{1}{x} - \dfrac{1}{x + 3}]
  2. ∫1x(x+3)dx  ⟹  \int \dfrac{1}{x(x + 3)} dx \implies Partial fractions

Answer: Partial fractions

Idea 2 of 3

Reduction formulas

A recurrence lowers a power step by step. Supply the base integral and track every multiplier.

Worked example 1

Evaluate ∫0π22cos⁡2x dx\int_{0}^{\dfrac{\pi }{2}} 2 \cos ^{2}x\,dx.
  1. I0=π2I_{0} = \dfrac{\pi }{2}; In=n−1nInI_{n} = \dfrac{n - 1}{n}I_{n}₋2_{2}
  2. I2=12I0I_2 = \dfrac{1}{2}I_0
  3. 2I2=π2≈1.572I_2 = \dfrac{\pi }{2} \approx 1.57

Answer: π2≈1.57\dfrac{\pi }{2} \approx 1.57

Worked example 2

Evaluate ∫0π23cos⁡4x dx\int_{0}^{\dfrac{\pi }{2}} 3 \cos ^{4}x\,dx.
  1. I0=π2I_{0} = \dfrac{\pi }{2}; In=n−1nInI_{n} = \dfrac{n - 1}{n}I_{n}₋2_{2}
  2. I2=12I0I_2 = \dfrac{1}{2}I_0
  3. I4=34I2I_4 = \dfrac{3}{4}I_2
  4. 3I4=9π16≈1.773I_4 = \dfrac{9\pi }{16} \approx 1.77

Answer: 9π16≈1.77\dfrac{9\pi }{16} \approx 1.77

Idea 3 of 3

Verify a table result

Differentiate a proposed antiderivative and check scale factors and variable substitutions before using it.

Worked example 1

Evaluate ∫012x(x2+2) dx\int_{0}^{1} 2x(x^{2} + 2)\,dx.
  1. u=x2+2u = x^{2} + 2; du=2xdxdu = 2x dx; x=0  ⟹  u=2,x=1  ⟹  u=3x = 0 \implies u = 2, x = 1 \implies u = 3
  2. ∫012x(x2+2) dx=∫23u du\int_{0}^{1} 2x(x^{2} + 2)\,dx = \int_{2}^{3} u\,du
  3. 32−222=52≈2.5\dfrac{3^{2} - 2^{2}}{2} = \dfrac{5}{2} \approx 2.5

Answer: 52≈2.5\dfrac{5}{2} \approx 2.5

Worked example 2

Evaluate ∫012x(x2+3)2 dx\int_{0}^{1} 2x(x^{2} + 3)^{2}\,dx.
  1. u=x2+3u = x^{2} + 3; du=2xdxdu = 2x dx; x=0  ⟹  u=3,x=1  ⟹  u=4x = 0 \implies u = 3, x = 1 \implies u = 4
  2. ∫012x(x2+3)2 dx=∫34u2 du\int_{0}^{1} 2x(x^{2} + 3)^{2}\,dx = \int_{3}^{4} u^{2}\,du
  3. 43−333=373≈12.33\dfrac{4^{3} - 3^{3}}{3} = \dfrac{37}{3} \approx 12.33

Answer: 373≈12.33\dfrac{37}{3} \approx 12.33

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §8.6,p8.6, p. 551551. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.