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Integration

4.5 Integration by Substitution

Focus on choose the inner expression, change bounds with the variable, verify by differentiation.

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Ideas in this lesson

01Choose the inner expressionWorked steps · five questions02Change bounds with the variableWorked steps · five questions03Verify by differentiationWorked steps · five questions
Idea 1 of 3

Choose the inner expression

A useful substitution absorbs both an expression and its derivative. Do not leave a mixture of xx and uu in the transformed integral.

Worked example 1

Choose a useful uu for integrating 2xcos⁡(x2+2)2x \cos (x^{2} + 2).
  1. u=x2+2u = x^{2} + 2; du=2xdxdu = 2x dx
  2. Transformed integral: ∫cos⁡udu\int \cos u du

Answer: x2+2x^{2} + 2

Worked example 2

Choose a useful uu for integrating 2xx2+3\dfrac{2x}{x^{2} + 3}.
  1. u=x2+3u = x^{2} + 3; du=2xdxdu = 2x dx
  2. Transformed integral: ∫1udu\int \dfrac{1}{u} du

Answer: x2+3x^{2} + 3

Idea 2 of 3

Change bounds with the variable

For a definite integral, transform both bounds to uu-values or substitute back into an xx-antiderivative.

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

Idea 3 of 3

Verify by differentiation

Differentiate the proposed antiderivative to check that the inner derivative and any scale factor are correct.

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, §4.5,p4.5, p. 292292. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.