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Logarithmic, Exponential, and Other Transcendental Functions

5.7 Inverse Trigonometric Functions: Integration

Focus on arctangent integration forms, arcsine integration forms, choose a substitution.

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Ideas in this lesson

01Arctangent integration formsWorked steps · five questions02Arcsine integration formsWorked steps · five questions03Choose a substitutionWorked steps · five questions
Idea 1 of 3

Arctangent integration forms

Recognize 1a2+x2\frac{1}{a^{2} + x^{2}}. The antiderivative is arctan⁡(xa)a\frac{\arctan (\frac{x}{a})}{a} for a>0a > 0.

Worked example 1

Evaluate ∫0214+x2 dx\int_{0}^{2} \dfrac{1}{4 + x^{2}}\,dx.
  1. F(x)=arctan⁡(x2)2F(x) = \dfrac{\arctan (\dfrac{x}{2})}{2}
  2. F(2)−F(0)=π8≈0.39F(2) - F(0) = \dfrac{\pi }{8} \approx 0.39

Answer: π8≈0.39\dfrac{\pi }{8} \approx 0.39

Worked example 2

Evaluate ∫01.519−x2 dx\int_{0}^{1.5} \dfrac{1}{\sqrt{9 - x^{2}}}\,dx.
  1. F(x)=arcsin⁡(x3)F(x) = \arcsin (\dfrac{x}{3})
  2. F(1.5)−F(0)=π6≈0.52F(1.5) - F(0) = \dfrac{\pi }{6} \approx 0.52

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

Idea 2 of 3

Arcsine integration forms

Recognize 1a2−x2\frac{1}{\sqrt{a^{2} - x^{2}}}. The antiderivative is arcsin⁡(xa)\arcsin (\frac{x}{a}) within its real domain.

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

Choose a substitution

Scale the variable to match a standard inverse-trig form and adjust the differential consistently.

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

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §5.7,p5.7, p. 375375. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.