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Topic 6 of 8 · Form 6

Further calculus: hyperbolic functions, reduction formulae and applications

sinh x = (eˣ − e⁻ˣ)/2; cosh x = (eˣ + e⁻ˣ)/2; tanh x = sinh x/cosh x.

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Hyperbolic functions

sinh x = (eˣ − e⁻ˣ)/2; cosh x = (eˣ + e⁻ˣ)/2; tanh x = sinh x/cosh x.

Identities: cosh²x − sinh²x = 1 (Osborn's rule converts trig identities, changing the sign of products of two sines).

Derivatives: d/dx sinh x = cosh x; d/dx cosh x = sinh x.

Inverses: arsinh x = ln(x + √(x² + 1)); arcosh x = ln(x + √(x² − 1)) (x ≥ 1); artanh x = ½ ln((1 + x)/(1 − x)).

Integrals: ∫1/√(x² + a²) dx = arsinh(x/a) + c; ∫1/√(x² − a²) dx = arcosh(x/a) + c.

Reduction formulae

Worked example: Iₙ = ∫₀^(π/2) sinⁿx dx satisfies Iₙ = ((n − 1)/n)Iₙ₋₂. Find I₄.

I₀ = π/2 → I₂ = ½ × π/2 = π/4 → I₄ = ¾ × π/4 = 3π/16

Applications of integration

  • Arc length: s = ∫√(1 + (dy/dx)²) dx; parametric s = ∫√((dx/dt)² + (dy/dt)²) dt.
  • Surface area of revolution about the x-axis: S = 2π∫y √(1 + (dy/dx)²) dx.
  • Mean value of a function over [a, b]: (1/(b − a))∫f(x) dx.

Worked example: Find the arc length of y = cosh x from x = 0 to x = 1.

1 + sinh²x = cosh²x → s = ∫₀¹ cosh x dx = sinh 1 ≈ 1.175

Exam tip: Hyperbolic identities often simplify arc length integrands – look for 1 + sinh²x = cosh²x.

Check yourself

  1. Show that cosh x + sinh x = eˣ.
  2. Find d/dx(cosh 3x).
  3. Express arsinh 1 in logarithmic form.
Show answers
  1. (eˣ + e⁻ˣ)/2 + (eˣ − e⁻ˣ)/2 = eˣ
  2. 3 sinh 3x
  3. ln(1 + √2)
Next topicComplex numbers and de Moivre's theorem

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