Integrating Hyperbolic Functions

Integration is where all the earlier hyperbolic work pays off. Reversing the derivatives of the previous lesson gives the direct integrals $\int\cosh x\,\mathrm dx=\sinh x+c$∫coshxdx=sinhx+c and friends; reversing the inverse-derivatives produces three of the most important standard integrals in the course — $\int\tfrac{1}{\sqrt{x^2+a^2}}\,\mathrm dx=\operatorname{arsinh}\tfrac xa+c$∫x2+a2​1​dx=arsinhax​+c and its relatives — which crack a whole class of integrals that have no elementary antiderivative in terms of polynomials and $\arctan$arctan alone. This lesson assembles the standard results, shows how the power-reduction identities handle $\sinh^2$sinh2 and $\cosh^2$cosh2, uses completing-the-square to extend the inverse-hyperbolic forms, and works the hyperbolic substitution $x=a\sinh t$x=asinht (the hyperbolic analogue of $x=a\sin\theta$x=asinθ).

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1 Spec mapping

Hyperbolic integration is compulsory pure content for Papers 1 and 2 of AQA Further Mathematics (7367). Carrying out a standard integral or a stated substitution is AO1; choosing the right method (recognise a standard form, complete the square, or substitute) and interpreting the result is AO2/AO3. It is the direct continuation of the differentiation lesson — every standard integral here is a derivative from there reversed — and it builds on integration by substitution, definite integrals, and completing the square from A-Level Maths. Examiners love the inverse-hyperbolic integrals because they reward students who can recognise the structure $\sqrt{x^2\pm a^2}$x2±a2​ and reach for the right standard form rather than flailing with a generic substitution. Beyond the pure papers, these integrals are the bread and butter of the further-calculus strand — arc lengths, surface areas of revolution and certain volumes all reduce to exactly the forms practised here — so the recognition skill earns marks well outside the immediate topic.

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2 Core theory — the direct standard integrals

Reversing each standard derivative immediately gives:

$\int\cosh x\,\mathrm dx=\sinh x+c,\qquad \int\sinh x\,\mathrm dx=\cosh x+c,\qquad \int\operatorname{sech}^2 x\,\mathrm dx=\tanh x+c,$∫coshxdx=sinhx+c,∫sinhxdx=coshx+c,∫sech2xdx=tanhx+c,

$\int\operatorname{cosech}^2 x\,\mathrm dx=-\coth x+c,\quad \int\operatorname{sech} x\tanh x\,\mathrm dx=-\operatorname{sech} x+c,\quad \int\operatorname{cosech} x\coth x\,\mathrm dx=-\operatorname{cosech} x+c.$∫cosech2xdx=−cothx+c,∫sechxtanhxdx=−sechx+c,∫cosechxcothxdx=−cosechx+c.

The first two are pleasingly free of the minus sign that trips students in trig: $\int\sinh=+\cosh$∫sinh=+cosh (whereas $\int\sin=-\cos$∫sin=−cos). The reciprocal-squared and product integrals all do carry a minus, matching their negative derivatives. A quick way to reconstruct any of these under pressure is to think "what differentiates to give the integrand?" — the integral table is just the derivative table read right-to-left, with every entry verifiable on the spot by differentiating your proposed answer.

This "reverse the derivative" mindset is worth making explicit, because it means there is nothing new to memorise in this section beyond the previous lesson. If you can differentiate $\tanh x$tanhx to get $\operatorname{sech}^2 x$sech2x, then you automatically know $\int\operatorname{sech}^2 x\,\mathrm dx=\tanh x+c$∫sech2xdx=tanhx+c; if you know $\tfrac{\mathrm d}{\mathrm dx}\operatorname{sech} x=-\operatorname{sech} x\tanh x$dxd​sechx=−sechxtanhx, then the slightly awkward integral $\int\operatorname{sech} x\tanh x\,\mathrm dx$∫sechxtanhxdx is just $-\operatorname{sech} x+c$−sechx+c. Whenever you meet an unfamiliar hyperbolic integrand, ask which standard function has that as its derivative; almost always the answer is staring back from the derivative table. The genuinely new skills in this lesson are the three things you cannot do by pure reversal: integrating even powers (power reduction, §3), recognising the inverse-hyperbolic forms (§4), and the substitution method (§6).

Worked Example 1 — a direct definite integral

Evaluate $\displaystyle\int_0^1\cosh x\,\mathrm dx$∫01​coshxdx.

$\int_0^1\cosh x\,\mathrm dx=\big[\sinh x\big]_0^1=\sinh 1-\sinh 0=\frac{e-e^{-1}}{2}\approx1.175.$∫01​coshxdx=[sinhx]01​=sinh1−sinh0=2e−e−1​≈1.175.

Mark scheme: M1 antiderivative $\sinh x$sinhx; A1 apply limits; A1 exact form $\tfrac{e-e^{-1}}{2}$2e−e−1​ (a decimal alone, if an exact value is asked, would lose the final mark).

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3 Integrating $\sinh^2 x$sinh2x and $\cosh^2 x$cosh2x — power reduction

Just as $\int\sin^2 x$∫sin2x needs the double-angle identity, the even powers of $\sinh$sinh and $\cosh$cosh are integrated by first applying the power-reduction identities from the identities lesson:

$\sinh^2 x=\frac{\cosh 2x-1}{2},\qquad \cosh^2 x=\frac{\cosh 2x+1}{2}.$sinh2x=2cosh2x−1​,cosh2x=2cosh2x+1​.

Integrating term by term (and using $\int\cosh 2x=\tfrac12\sinh 2x$∫cosh2x=21​sinh2x):

$\boxed{\ \int\sinh^2 x\,\mathrm dx=\frac{\sinh 2x}{4}-\frac{x}{2}+c,\qquad \int\cosh^2 x\,\mathrm dx=\frac{\sinh 2x}{4}+\frac{x}{2}+c\ }$ ∫sinh2xdx=4sinh2x​−2x​+c,∫cosh2xdx=4sinh2x​+2x​+c ​

The only difference between the two is the sign of the $\tfrac x2$2x​ term — minus for $\sinh^2$sinh2, plus for $\cosh^2$cosh2 — which you can anchor by noting that $\cosh^2 x\ge\sinh^2 x$cosh2x≥sinh2x, so its integral should be the larger. The reason a naive attempt fails is instructive: you cannot integrate $\cosh^2 x$cosh2x as though it were a power of $x$x (there is no "$\tfrac13\cosh^3 x$31​cosh3x" rule, because the chain rule would demand an extra $\sinh$sinh factor that is not present). The power-reduction identity sidesteps this by trading the square for a linear combination of $\cosh 2x$cosh2x and a constant, both of which integrate trivially. The same trick — convert a power into a multiple angle — is exactly how $\int\sin^2,\int\cos^2$∫sin2,∫cos2 are handled in A-Level Maths, so the method should feel familiar; only the identity changes. For higher even powers such as $\cosh^4 x$cosh4x you simply apply power reduction twice, and for odd powers such as $\sinh^3 x$sinh3x you peel off one factor and use the fundamental identity (e.g. $\sinh^3 x=\sinh x(\cosh^2 x-1)$sinh3x=sinhx(cosh2x−1), then substitute $u=\cosh x$u=coshx).

Worked Example 2 — power reduction with limits

Evaluate $\displaystyle\int_0^{\ln 2}\sinh^2 x\,\mathrm dx$∫0ln2​sinh2xdx.

$\int_0^{\ln 2}\sinh^2 x\,\mathrm dx=\left[\frac{\sinh 2x}{4}-\frac{x}{2}\right]_0^{\ln 2}.$∫0ln2​sinh2xdx=[4sinh2x​−2x​]0ln2​.

Evaluate $\sinh(2\ln 2)=\sinh(\ln 4)=\dfrac{e^{\ln 4}-e^{-\ln 4}}{2}=\dfrac{4-\tfrac14}{2}=\dfrac{15}{8}$sinh(2ln2)=sinh(ln4)=2eln4−e−ln4​=24−41​​=815​. So

$=\frac{1}{4}\cdot\frac{15}{8}-\frac{\ln 2}{2}-0=\frac{15}{32}-\frac{\ln 2}{2}.$=41​⋅815​−2ln2​−0=3215​−2ln2​.

Mark scheme: M1 use $\sinh^2 x=\tfrac{\cosh 2x-1}{2}$sinh2x=2cosh2x−1​; A1 antiderivative $\tfrac{\sinh 2x}{4}-\tfrac x2$4sinh2x​−2x​; M1 evaluate $\sinh(\ln 4)=\tfrac{15}{8}$sinh(ln4)=815​ using the exponential definition; A1 $\tfrac{15}{32}-\tfrac{\ln 2}{2}$3215​−2ln2​.

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4 Integrals that produce inverse hyperbolic functions

Reversing the inverse-hyperbolic derivatives gives the three high-value standard integrals:

$\boxed{\ \int\frac{1}{\sqrt{x^2+a^2}}\,\mathrm dx=\operatorname{arsinh}\frac xa+c,\qquad \int\frac{1}{\sqrt{x^2-a^2}}\,\mathrm dx=\operatorname{arcosh}\frac xa+c\ (x>a)\ }$ ∫x2+a2​1​dx=arsinhax​+c,∫x2−a2​1​dx=arcoshax​+c (x>a) ​

$\boxed{\ \int\frac{1}{a^2-x^2}\,\mathrm dx=\frac1a\operatorname{artanh}\frac xa+c\ (|x|<a)\ }$ ∫a2−x21​dx=a1​artanhax​+c (∣x∣<a) ​

The decisive clue is the sign inside the surd: $x^2+a^2$x2+a2 gives $\operatorname{arsinh}$arsinh; $x^2-a^2$x2−a2 gives $\operatorname{arcosh}$arcosh; and $a^2-x^2$a2−x2 (no surd) gives $\tfrac1a\operatorname{artanh}$a1​artanh. Contrast the inverse-trig forms $\int\tfrac{1}{\sqrt{a^2-x^2}}=\arcsin\tfrac xa$∫a2−x2​1​=arcsinax​ and $\int\tfrac{1}{a^2+x^2}=\tfrac1a\arctan\tfrac xa$∫a2+x21​=a1​arctanax​: reading the $\pm$± pattern correctly is the whole skill.

It is worth laying out the complete decision table side by side, because all five quadratic-surd reciprocals appear interchangeably in exams and the only thing distinguishing them is the position and sign of the squares. With a surd in the denominator: $\sqrt{a^2-x^2}$a2−x2​ (constant minus square) routes to $\arcsin\tfrac xa$arcsinax​; $\sqrt{x^2+a^2}$x2+a2​ (square plus constant) routes to $\operatorname{arsinh}\tfrac xa$arsinhax​; $\sqrt{x^2-a^2}$x2−a2​ (square minus constant) routes to $\operatorname{arcosh}\tfrac xa$arcoshax​. Without a surd, with the squares in the denominator directly: $a^2+x^2$a2+x2 routes to $\tfrac1a\arctan\tfrac xa$a1​arctanax​; $a^2-x^2$a2−x2 routes to $\tfrac1a\operatorname{artanh}\tfrac xa$a1​artanhax​. Memorising this five-row map — and the mnemonic that "a plus under a surd is hyperbolic-sine, a minus under a surd is hyperbolic-cosine, a plus without a surd is arctan, a minus without a surd is artanh" — converts what feels like five separate results into one organised picture, and it is precisely the recognition examiners are probing. Most marks lost in this topic come from grabbing $\arcsin$arcsin when an $\operatorname{arsinh}$arsinh was wanted, simply because a sign was misread under time pressure.

Why are there two hyperbolic forms ($\operatorname{arsinh}$arsinh and $\operatorname{arcosh}$arcosh) but only one circular form ($\arcsin$arcsin) for the surd cases? Because the circular world only has $\sqrt{a^2-x^2}$a2−x2​ as a "nice" surd — the expression $\sqrt{a^2+x^2}$a2+x2​ cannot be tamed by $x=a\sin\theta$x=asinθ, since $a^2+a^2\sin^2\theta$a2+a2sin2θ does not collapse. The hyperbolic identities, with their minus sign, are exactly what is needed to handle the $+$+ and the larger $-$− cases, which is the whole reason hyperbolic functions earn a place in the integration toolkit.

Worked Example 3 — indefinite $\operatorname{arsinh}$arsinh form

Find $\displaystyle\int\frac{1}{\sqrt{x^2+16}}\,\mathrm dx$∫x2+16​1​dx, giving the answer as a logarithm.

Here $a=4$a=4:

$\int\frac{\mathrm dx}{\sqrt{x^2+16}}=\operatorname{arsinh}\frac x4+c=\ln\!\big(x+\sqrt{x^2+16}\big)+c.$∫x2+16​dx​=arsinh4x​+c=ln(x+x2+16​)+c.

Mark scheme: M1 recognise $\operatorname{arsinh}(x/a)$arsinh(x/a) with $a=4$a=4; A1 $\operatorname{arsinh}(x/4)+c$arsinh(x/4)+c; A1 log form (note the additive constant absorbs the $-\ln a$−lna that appears when converting).

Worked Example 4 — definite $\operatorname{arcosh}$arcosh form

Evaluate $\displaystyle\int_5^{13}\frac{1}{\sqrt{x^2-25}}\,\mathrm dx$∫513​x2−25​1​dx.

Here $a=5$a=5, and the lower limit $x=5=a$x=5=a is the edge of the domain (the integral is improper there but convergent):

$\int_5^{13}\frac{\mathrm dx}{\sqrt{x^2-25}}=\Big[\operatorname{arcosh}\tfrac x5\Big]_5^{13}=\operatorname{arcosh}\tfrac{13}{5}-\operatorname{arcosh}1=\ln\!\big(\tfrac{13}{5}+\tfrac{12}{5}\big)-0=\ln 5.$∫513​x2−25​dx​=[arcosh5x​]513​=arcosh513​−arcosh1=ln(513​+512​)−0=ln5.

Mark scheme: M1 $\operatorname{arcosh}(x/5)$arcosh(x/5) form; M1 apply limits, noting $\operatorname{arcosh}1=0$arcosh1=0; A1 $\operatorname{arcosh}\tfrac{13}{5}=\ln 5$arcosh513​=ln5 (since $\sqrt{(13/5)^2-1}=\tfrac{12}{5}$(13/5)2−1​=512​).

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5 Completing the square first

When the surd is a general quadratic $\sqrt{ax^2+bx+c}$ax2+bx+c​, complete the square to turn it into one of the standard $x'^2\pm a'^2$x′2±a′2 forms, then substitute the shift. This is the single most common "harder" integral in the topic, and the procedure is mechanical once you see it: complete the square to get $(x+p)^2\pm q^2$(x+p)2±q2, let $u=x+p$u=x+p so that $\mathrm du=\mathrm dx$du=dx, and the integral collapses to a bare standard form in $u$u. The sign of the constant left after completing the square tells you which standard form you land in — $+q^2$+q2 means $\operatorname{arsinh}$arsinh, $-q^2$−q2 means $\operatorname{arcosh}$arcosh — and the shift $u=x+p$u=x+p carries through to the final answer unchanged. Because $\mathrm du=\mathrm dx$du=dx (the shift has gradient $1$1), there is no awkward scaling to track; the only thing to be careful of is reading off $q$q correctly and stating the domain when an $\operatorname{arcosh}$arcosh appears.

Worked Example 5 — completing the square

Find $\displaystyle\int\frac{1}{\sqrt{x^2+6x+13}}\,\mathrm dx$∫x2+6x+13​1​dx.

Complete the square: $x^2+6x+13=(x+3)^2+4$x2+6x+13=(x+3)2+4. With $u=x+3$u=x+3 (so $\mathrm du=\mathrm dx$du=dx) the integral becomes $\int\tfrac{1}{\sqrt{u^2+4}}\,\mathrm du$∫u2+4​1​du, an $\operatorname{arsinh}$arsinh form with $a=2$a=2:

$\int\frac{\mathrm dx}{\sqrt{x^2+6x+13}}=\operatorname{arsinh}\frac{u}{2}+c=\operatorname{arsinh}\frac{x+3}{2}+c.$∫x2+6x+13​dx​=arsinh2u​+c=arsinh2x+3​+c.

Mark scheme: M1 complete the square to $(x+3)^2+4$(x+3)2+4; M1 substitute $u=x+3$u=x+3; A1 $\operatorname{arsinh}\tfrac{x+3}{2}+c$arsinh2x+3​+c. The $+a^2=+4$+a2=+4 signals $\operatorname{arsinh}$arsinh; had it been $(x+3)^2-4$(x+3)2−4 the answer would be $\operatorname{arcosh}\tfrac{x+3}{2}$arcosh2x+3​.

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6 Hyperbolic substitution

For surds that multiply the rest of the integrand (so recognition alone is not enough), use a hyperbolic substitution, the analogue of the trig substitution $x=a\sin\theta$x=asinθ:

Surd	Substitution	Result of the surd
$\sqrt{x^2+a^2}$x2+a2​	$x=a\sinh t$x=asinht	$\sqrt{x^2+a^2}=a\cosh t$x2+a2​=acosht
$\sqrt{x^2-a^2}$x2−a2​	$x=a\cosh t$x=acosht	$\sqrt{x^2-a^2}=a\sinh t$x2−a2​=asinht

The magic is that the substitution makes the surd vanish: with $x=a\sinh t$x=asinht, $x^2+a^2=a^2\sinh^2 t+a^2=a^2(\sinh^2 t+1)=a^2\cosh^2 t$x2+a2=a2sinh2t+a2=a2(sinh2t+1)=a2cosh2t, so $\sqrt{x^2+a^2}=a\cosh t$x2+a2​=acosht (positive, as $\cosh\ge1$cosh≥1). And $\mathrm dx=a\cosh t\,\mathrm dt$dx=acoshtdt is also $a\cosh t$acosht, which usually cancels beautifully.