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
| Function | Antiderivative | Justification (Derivative Rule) |
|---|
| 1 / sqrt(1 − x^2) | arcsin(x) + C | d/dx[arcsin x] = 1 / sqrt(1 − x^2) |
| 1 / (1 + x^2) | arctan(x) + C | d/dx[arctan x] = 1 / (1 + x^2) |
| 1 / ( | x | sqrt(x^2 − 1)) |
| −1 / sqrt(1 − x^2) | arccos(x) + C | d/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.