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L13: Inverse Trig Antiderivatives Overview

Nov 21, 2025

Overview

Lecture covers integrals that yield or use inverse trigonometric functions, primarily via substitution. Includes core antiderivative rules and worked examples.

Elementary Antiderivatives (Inverse Trig)

  • Core idea: Use known derivatives of inverse trig to identify antiderivatives.
  • No separate need for arccos rule since it differs from arcsin by a sign and constant.

Antiderivative–Derivative Pairs

FunctionAntiderivativeJustification (Derivative Rule)
1 / sqrt(1 − x^2)arcsin(x) + Cd/dx[arcsin x] = 1 / sqrt(1 − x^2)
1 / (1 + x^2)arctan(x) + Cd/dx[arctan x] = 1 / (1 + x^2)
1 / (xsqrt(x^2 − 1))
−1 / sqrt(1 − x^2)arccos(x) + Cd/dx[arccos x] = −1 / sqrt(1 − x^2)
  • arcsin(x) and −arccos(x) differ by a constant; same integrand yields either up to C.
  • Similar relationship holds for arccsc and arcsec.

Example A: ∫ 1/(9 + x^2) dx

  • Factor 9: ∫ 1/[9(1 + (x^2/9))] dx = (1/9) ∫ 1/(1 + (x/3)^2) dx.
  • Substitute u = x/3, du = (1/3) dx; adjust with factor 3.
  • Result: (1/3) arctan(x/3) + C.

Example B: ∫ 1/sqrt(25 − 4x^2) dx

  • Factor 25 inside root: sqrt(25(1 − (4x^2/25))) = 5 sqrt(1 − (2x/5)^2).
  • Pull 1/5 outside: (1/5) ∫ 1/sqrt(1 − (2x/5)^2) dx.
  • Substitute u = (2/5)x, du = (2/5) dx; adjust with 5/2.
  • Result: (1/2) arcsin(2x/5) + C.

Example C: ∫ e^x / [e^x sqrt(e^{2x} − 1)] dx

  • Note e^{2x} = (e^x)^2; let u = e^x, du = e^x dx.
  • Integrand becomes ∫ 1/[u sqrt(u^2 − 1)] du.
  • Since u = e^x > 0, |u| = u; apply arcsec rule.
  • Result: arcsec(e^x) + C.
  • Remark: Cancelling e^x yields ∫ 1/sqrt(e^{2x} − 1) dx = arcsec(e^x) + C.

Example D: ∫ 1/[x + x(ln x)^2] dx

  • Factor x: ∫ [1/x] · [1/(1 + (ln x)^2)] dx.
  • Let u = ln x, du = (1/x) dx; integrand becomes ∫ 1/(1 + u^2) du.
  • Result: arctan(ln x) + C.

Example E: ∫ x/(1 + x^4) dx

  • Rewrite x^4 as (x^2)^2: ∫ [1/(1 + (x^2)^2)] · x dx.
  • Let u = x^2, du = 2x dx; insert 2 and 1/2.
  • Result: (1/2) arctan(x^2) + C.

Example F: ∫ [arcsin x]/sqrt(1 − x^2) dx

  • Write as ∫ arcsin(x) · [1/sqrt(1 − x^2)] dx.
  • Let u = arcsin(x), du = [1/sqrt(1 − x^2)] dx.
  • Integrand becomes ∫ u du = u^2/2 + C.
  • Result: (arcsin x)^2 / 2 + C.

Key Terms & Definitions

  • arcsin x: Inverse sine; derivative 1/sqrt(1 − x^2) on (−1, 1).
  • arctan x: Inverse tangent; derivative 1/(1 + x^2) for all real x.
  • arcsec x: Inverse secant; derivative 1/(|x| sqrt(x^2 − 1)) where defined.
  • Substitution: Let u = g(x), rewrite integral in u using du = g′(x) dx.

Action Items / Next Steps

  • Practice identifying patterns matching arcsin, arctan, arcsec forms.
  • Rehearse factoring constants to create 1 ± u^2 structures.
  • Verify substitutions by differentiating final answers to the original integrands.