University of Melbourne · S1 2026 · FACULTY OF SCIENCE

MAST10006 · Calculus 2

Q: Why does differentiating cosh not give a minus sign?

Because cosh is built purely from ex and e−x, and differentiating real exponentials can never introduce a minus. So d/dx cosh x = +sinh x, unlike d/dx cos x = −sin x. Carrying the trig minus across to the hyperbolic case is the most-marked error in this chapter — it silently corrupts every hyperbolic integral.

Q: When should I use a hyperbolic substitution instead of a trig one?

For roots of the form √(x² + a²) (use x = a sinh u) and √(x² − a²) (use x = a cosh u). The master identity removes the root with no leftover radical and no secant integrals, and the answer drops out directly in arcsinh / arccosh (i.e. log) form — usually cleaner than the tanθ route.

Q: What is the point of the complex-exponential integration method?

It collapses ∫ eax cos bx dx into a single antiderivative. The textbook way is integration by parts twice, then solving for the recurring integral; the complex-exponential method writes cos bx = Re(eibx), integrates e(a+ib)x in one step, and takes the real part at the end. It is the technique the examiners reward.

Q: In the complex-exponential method, when do I take the real or imaginary part?

Decide which part you need at the START — Re for a cosine integrand, Im for a sine — but EXTRACT it only at the END, after rationalising 1/(a+ib) and expanding e(a+ib)x = eax(cos bx + i sin bx) and multiplying out. Pulling the part too early drops the cross term and gives the wrong answer.

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Chapter 4 of 8 · MAST10006

Hyperbolic Functions

Split e^x into its even and odd parts and you get the two hyperbolic functions: cosh x = (e^x + e^−x)/2 and sinh x = (e^x − e^−x)/2. They look like cosine and sine, obey almost the same identities — the master identity is cosh² x − sinh² x = 1, with a MINUS sign — and they appear whenever a √(x² + a²) shows up in an integral. The single sign that differs from trig drives the whole topic: differentiating cosh gives +sinh, with no minus. This chapter covers the definitions and identity engine, the derivatives and integrals, the inverse hyperbolic functions (every one is secretly a logarithm), and then Euler's formula and the signature MAST10006 complex-exponential method for integrals like ∫ e^ax cos bx dx — one antiderivative of e^(a+ib)x that beats two rounds of integration by parts.

In this chapter

What this chapter covers

015.1 Definitions and the master identity cosh² − sinh² = 1
02Derivatives and integrals — like trig, minus the sign flips
035.4 Inverse hyperbolic functions and their log forms
045.5 Hyperbolic substitution for √(x² ± a²)
056.1 Euler's formula and the Argand diagram
066.2 The complex-exponential integration method

Worked example · free

Worked example: derive the log form of arcsinh x

Q [4 marks]. Show that arcsinh x = log(x + √(x² + 1)) from the definition.

+1Set y = arcsinh x, i.e. x = sinh y = (e^y − e^−y)/2.
+1Clear the negative exponent: multiply by 2e^y to get 2x e^y = e^2y − 1, a quadratic in e^y: (e^y)² − 2x(e^y) − 1 = 0.
+1Quadratic formula: e^y = x ± √(x² + 1). Since e^y > 0 and √(x²+1) > |x|, take the + root.
+1Take logs: y = log(x + √(x² + 1)), valid for all x.

arcsinh x = log(x + √(x² + 1)). Solving x = sinh y for e^y gives a quadratic; the positive root and a logarithm produce the log form.

Sia tip — If a question says ‘express in logarithmic form’, the inverse-hyperbolic symbol alone is NOT a complete answer — convert to the log expression. And quote the domain: arccosh x needs x ≥ 1 and arctanh x needs |x| < 1, or you lose a mark.

Glossary

Key terms

Hyperbolic functions: sinh x = (e^x − e^−x)/2 and cosh x = (e^x + e^−x)/2, the odd and even halves of e^x, with tanh x = sinh x / cosh x. cosh x ≥ 1 always and −1 < tanh x < 1.
Master identity: cosh² x − sinh² x = 1 — the hyperbolic analogue of cos² + sin² = 1, but with a MINUS sign, because the points (cosh, sinh) trace a hyperbola X² − Y² = 1, not a circle. Dividing it by cosh² or sinh² generates the sech and cosech identities.
Inverse hyperbolic function: The inverse of sinh, cosh or tanh, each expressible as a logarithm: arcsinh x = log(x + √(x²+1)), arccosh x = log(x + √(x²−1)) for x ≥ 1, and arctanh x = ½ log((1+x)/(1−x)) for |x| < 1.
Euler's formula: e^iθ = cos θ + i sin θ, so cos and sin are the real and imaginary parts of the complex exponential. It is the same fact as the Taylor catalogue: putting x = iθ into the series for e^x and splitting even and odd powers reproduces the cos and sin series.
Complex-exponential method: The technique for ∫ e^ax cos bx dx and ∫ e^ax sin bx dx: replace the trig factor by Re or Im of e^ibx, integrate the single exponential e^(a+ib)x, rationalise 1/(a+ib), expand, and take the real or imaginary part at the end — one line instead of integration by parts twice.

FAQ

Hyperbolic Functions FAQ

Why does differentiating cosh not give a minus sign?

Because cosh is built purely from e^x and e^−x, and differentiating real exponentials can never introduce a minus. So d/dx cosh x = +sinh x, unlike d/dx cos x = −sin x. Carrying the trig minus across to the hyperbolic case is the most-marked error in this chapter — it silently corrupts every hyperbolic integral.

When should I use a hyperbolic substitution instead of a trig one?

For roots of the form √(x² + a²) (use x = a sinh u) and √(x² − a²) (use x = a cosh u). The master identity removes the root with no leftover radical and no secant integrals, and the answer drops out directly in arcsinh / arccosh (i.e. log) form — usually cleaner than the tanθ route.

What is the point of the complex-exponential integration method?

It collapses ∫ e^ax cos bx dx into a single antiderivative. The textbook way is integration by parts twice, then solving for the recurring integral; the complex-exponential method writes cos bx = Re(e^ibx), integrates e^(a+ib)x in one step, and takes the real part at the end. It is the technique the examiners reward.

In the complex-exponential method, when do I take the real or imaginary part?

Decide which part you need at the START — Re for a cosine integrand, Im for a sine — but EXTRACT it only at the END, after rationalising 1/(a+ib) and expanding e^(a+ib)x = e^ax(cos bx + i sin bx) and multiplying out. Pulling the part too early drops the cross term and gives the wrong answer.

Study strategy

Exam move

Anchor everything on the master identity cosh² − sinh² = 1 and the one sign that differs from trig: d/dx cosh = +sinh, no minus. Put a sign-flip warning on your A4. Memorise the three inverse-hyperbolic log forms and their domains, because the exam often asks for the answer in log form and the domain restriction carries a mark. For integrals, learn the substitution triggers (√(x²+a²) → sinh, √(x²−a²) → cosh). The marquee skill is the complex-exponential method: drill the five steps until they are automatic — replace the trig factor by Re/Im of an exponential, integrate the single e^(a+ib)x, rationalise the conjugate, expand, and take the part at the very end. That five-step discipline turns the by-parts-twice question into a one-liner.

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