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

8.4 Trigonometric Substitution

Focus on match a radical pattern, transform the differential, transform definite bounds.

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Ideas in this lesson

01Match a radical patternWorked steps · five questions02Transform the differentialWorked steps · five questions03Transform definite boundsWorked steps · five questions
Idea 1 of 3

Match a radical pattern

Choose sine for a2−x2a^{2} - x^{2}, tangent for a2+x2a^{2} + x^{2}, and secant for x2−a2x^{2} - a^{2}, on an appropriate angle interval.

Worked example 1

Choose a trigonometric substitution for 4−x2\sqrt{4 - x^{2}}.
  1. x=2sin⁡θx = 2 \sin \theta ; dx=2cos⁡θdθdx = 2 \cos \theta d\theta
  2. 4−x2=2cos⁡θ\sqrt{4 - x^{2}} = 2 \cos \theta on a suitable θ\theta interval

Answer: x=2sin⁡θx = 2 \sin \theta

Worked example 2

Choose a trigonometric substitution for 9+x2\sqrt{9 + x^{2}}.
  1. x=3tan⁡θx = 3 \tan \theta ; dx=3sec⁡2θdθdx = 3 \sec ^{2}\theta d\theta
  2. 9+x2=3sec⁡θ\sqrt{9 + x^{2}} = 3 \sec \theta on a suitable θ\theta interval

Answer: x=3tan⁡θx = 3 \tan \theta

Idea 2 of 3

Transform the differential

Substitute dxdx as well as xx. Simplify the radical using an identity and the chosen interval’ss sign.

Worked example 1

Evaluate ∫0114−x2 dx\int_{0}^{1} \dfrac{1}{\sqrt{4 - x^{2}}}\,dx.
  1. x=2sin⁡θx = 2 \sin \theta ; dx=2cos⁡θdθdx = 2 \cos \theta d\theta
  2. 4−x2=2cos⁡θ\sqrt{4 - x^{2}} = 2 \cos \theta ; x=1  ⟹  θ=π6x = 1 \implies \theta = \dfrac{\pi }{6}
  3. ∫0π61 dθ=π6≈0.52\int_{0}^{\dfrac{\pi }{6}} 1\,d\theta = \dfrac{\pi }{6} \approx 0.52

Answer: π6≈0.52\dfrac{\pi }{6} \approx 0.52

Worked example 2

Evaluate ∫01.519−x2 dx\int_{0}^{1.5} \dfrac{1}{\sqrt{9 - x^{2}}}\,dx.
  1. x=3sin⁡θx = 3 \sin \theta ; dx=3cos⁡θdθdx = 3 \cos \theta d\theta
  2. 9−x2=3cos⁡θ\sqrt{9 - x^{2}} = 3 \cos \theta ; x=1.5  ⟹  θ=π6x = 1.5 \implies \theta = \dfrac{\pi }{6}
  3. ∫0π61 dθ=π6≈0.52\int_{0}^{\dfrac{\pi }{6}} 1\,d\theta = \dfrac{\pi }{6} \approx 0.52

Answer: π6≈0.52\dfrac{\pi }{6} \approx 0.52

Idea 3 of 3

Transform definite bounds

Convert xx-endpoints to angle endpoints. This avoids an unnecessary inverse substitution at the end.

Worked example 1

Evaluate ∫0114−x2 dx\int_{0}^{1} \dfrac{1}{\sqrt{4 - x^{2}}}\,dx.
  1. x=2sin⁡θx = 2 \sin \theta ; dx=2cos⁡θdθdx = 2 \cos \theta d\theta
  2. 4−x2=2cos⁡θ\sqrt{4 - x^{2}} = 2 \cos \theta ; x=1  ⟹  θ=π6x = 1 \implies \theta = \dfrac{\pi }{6}
  3. ∫0π61 dθ=π6≈0.52\int_{0}^{\dfrac{\pi }{6}} 1\,d\theta = \dfrac{\pi }{6} \approx 0.52

Answer: π6≈0.52\dfrac{\pi }{6} \approx 0.52

Worked example 2

Evaluate ∫01.519−x2 dx\int_{0}^{1.5} \dfrac{1}{\sqrt{9 - x^{2}}}\,dx.
  1. x=3sin⁡θx = 3 \sin \theta ; dx=3cos⁡θdθdx = 3 \cos \theta d\theta
  2. 9−x2=3cos⁡θ\sqrt{9 - x^{2}} = 3 \cos \theta ; x=1.5  ⟹  θ=π6x = 1.5 \implies \theta = \dfrac{\pi }{6}
  3. ∫0π61 dθ=π6≈0.52\int_{0}^{\dfrac{\pi }{6}} 1\,d\theta = \dfrac{\pi }{6} \approx 0.52

Answer: π6≈0.52\dfrac{\pi }{6} \approx 0.52

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §8.4,p8.4, p. 533533. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.