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Cambridge International AS & A Level · Topic 2.4
A-Level Further Maths 9231 / Topic 2.4 / AS & A Level

Four square roots, four standard answers.

Integrate \(\sinh\), \(\cosh\), \(\operatorname{sech}^{2}\) and \(\tanh\), then learn to recognise the four standard forms whose answers are \(\sin^{-1}\), \(\tan^{-1}\), \(\sinh^{-1}\) and \(\cosh^{-1}\). Completing the square and the substitutions \(x = a\sin u\), \(x = a\sinh u\) and \(x = a\cosh u\) bring many more integrals into those forms, with exact answers in logs and \(\pi\).

Topic 2.4All further integration lessons →
The key idea

\(\displaystyle\int \frac{dx}{\sqrt{a^{2} - x^{2}}} = \sin^{-1}\frac{x}{a}\), \(\displaystyle\int \frac{dx}{x^{2} + a^{2}} = \frac{1}{a}\tan^{-1}\frac{x}{a}\), \(\displaystyle\int \frac{dx}{\sqrt{x^{2} + a^{2}}} = \sinh^{-1}\frac{x}{a}\), \(\displaystyle\int \frac{dx}{\sqrt{x^{2} - a^{2}}} = \cosh^{-1}\frac{x}{a}\) (each \(+\,c\)). Only the \(\tan^{-1}\) form has a factor \(\frac{1}{a}\). Make the \(x^{2}\) coefficient 1 and complete the square to reach these forms; for \(\sqrt{a^{2} - x^{2}}\), \(\sqrt{a^{2} + x^{2}}\) or \(\sqrt{x^{2} - a^{2}}\) on top, substitute \(x = a\sin u\), \(a\sinh u\) or \(a\cosh u\) and use a double-angle identity.

x y O 1 2 3 4 5 6 ⅓ y = 1 / √(x² + 9) area = sinh⁻¹ ⁴⁄₃ = ln 3 ≈ 1.099
Fig. 1: the area under y = 1/√(x² + 9) from x = 0 to x = 4 is sinh⁻¹(4/3) = ln(4/3 + 5/3) = ln 3, an exact answer from a standard integral
Section 01

Hyperbolic integrals and the four standard forms.

Every derivative from section 2.3 read backwards is an integral. \(\dfrac{d}{dx}\cosh x = \sinh x\) and \(\dfrac{d}{dx}\sinh x = \cosh x\) (no minus sign, unlike \(\cos\)), \(\dfrac{d}{dx}\tanh x = \operatorname{sech}^{2} x\), and \(\tanh x = \dfrac{\sinh x}{\cosh x}\) is of the form \(\dfrac{f'(x)}{f(x)}\). The derivatives of \(\sin^{-1}\), \(\tan^{-1}\), \(\sinh^{-1}\) and \(\cosh^{-1}\) give the four standard forms below, all in the formula list.

IntegrandIntegral (\(+\,c\))IntegrandIntegral (\(+\,c\))
\(\sinh x\)\(\cosh x\)\(\dfrac{1}{\sqrt{a^{2} - x^{2}}}\)\(\sin^{-1}\dfrac{x}{a}\)   \((|x| \lt a)\)
\(\cosh x\)\(\sinh x\)\(\dfrac{1}{x^{2} + a^{2}}\)\(\dfrac{1}{a}\tan^{-1}\dfrac{x}{a}\)
\(\operatorname{sech}^{2} x\)\(\tanh x\)\(\dfrac{1}{\sqrt{x^{2} + a^{2}}}\)\(\sinh^{-1}\dfrac{x}{a}\) or \(\ln\!\big(x + \sqrt{x^{2} + a^{2}}\big)\)
\(\tanh x\)\(\ln \cosh x\)\(\dfrac{1}{\sqrt{x^{2} - a^{2}}}\)\(\cosh^{-1}\dfrac{x}{a}\) or \(\ln\!\big(x + \sqrt{x^{2} - a^{2}}\big)\)   \((x \gt a)\)

The log versions come from section 2.1: \(\sinh^{-1}\dfrac{x}{a} = \ln\!\left(\dfrac{x + \sqrt{x^{2} + a^{2}}}{a}\right)\), and the \(-\ln a\) is absorbed into the constant. For definite integrals, either version gives the same exact number. No modulus is needed in \(\ln \cosh x\), because \(\cosh x \gt 0\).

Worked example Hyperbolic functions
Find the exact value of \(\displaystyle\int_{0}^{\ln 2} (\cosh 2x + \tanh x)\,dx\).
  1. \[\int (\cosh 2x + \tanh x)\,dx = \tfrac{1}{2}\sinh 2x + \ln \cosh x + c.\]
  2. \(\sinh(2\ln 2) = \sinh(\ln 4) = \dfrac{4 - \frac{1}{4}}{2} = \dfrac{15}{8}\), and \(\cosh(\ln 2) = \dfrac{2 + \frac{1}{2}}{2} = \dfrac{5}{4}\).
  3. At \(x = 0\): \(\tfrac{1}{2}\sinh 0 + \ln\cosh 0 = 0 + \ln 1 = 0\).
  4. \[\Big[\tfrac{1}{2}\sinh 2x + \ln \cosh x\Big]_{0}^{\ln 2} = \tfrac{1}{2}\cdot\tfrac{15}{8} + \ln\tfrac{5}{4} - 0.\]
Answer: \(\dfrac{15}{16} + \ln\dfrac{5}{4}\)
Worked example Recognise the form
Find the exact value of each integral. (a) \(\displaystyle\int_{0}^{4} \frac{dx}{\sqrt{x^{2} + 9}}\) (b) \(\displaystyle\int_{5/2}^{4} \frac{dx}{\sqrt{x^{2} - 4}}\) (c) \(\displaystyle\int_{0}^{\sqrt{3}} \frac{dx}{x^{2} + 3}\) (d) \(\displaystyle\int_{0}^{1} \frac{dx}{\sqrt{4 - x^{2}}}\)
  1. (a) \(\sqrt{x^{2} + a^{2}}\) with \(a = 3\): \(\Big[\sinh^{-1}\dfrac{x}{3}\Big]_{0}^{4} = \sinh^{-1}\dfrac{4}{3} = \ln\!\left(\dfrac{4}{3} + \sqrt{\dfrac{16}{9} + 1}\right) = \ln\!\left(\dfrac{4}{3} + \dfrac{5}{3}\right) = \ln 3\) (Fig. 1).
  2. (b) \(\sqrt{x^{2} - a^{2}}\) with \(a = 2\): \(\cosh^{-1} 2 - \cosh^{-1}\dfrac{5}{4} = \ln\!\big(2 + \sqrt{3}\big) - \ln\!\left(\dfrac{5}{4} + \dfrac{3}{4}\right)\).
  3. So (b) \(= \ln\!\big(2 + \sqrt{3}\big) - \ln 2 = \ln\!\left(\dfrac{2 + \sqrt{3}}{2}\right)\).
  4. (c) \(x^{2} + a^{2}\) with \(a = \sqrt{3}\): \(\Big[\dfrac{1}{\sqrt{3}}\tan^{-1}\dfrac{x}{\sqrt{3}}\Big]_{0}^{\sqrt{3}} = \dfrac{1}{\sqrt{3}}\tan^{-1} 1 = \dfrac{\pi}{4\sqrt{3}}\).
  5. (d) \(\sqrt{a^{2} - x^{2}}\) with \(a = 2\): \(\Big[\sin^{-1}\dfrac{x}{2}\Big]_{0}^{1} = \sin^{-1}\dfrac{1}{2} = \dfrac{\pi}{6}\).
Answer: (a) \(\ln 3\)   (b) \(\ln\!\left(\frac{2 + \sqrt{3}}{2}\right)\)   (c) \(\frac{\pi}{4\sqrt{3}}\)   (d) \(\frac{\pi}{6}\)
Your turn
Find the exact value of \(\displaystyle\int_{0}^{\ln 3} \operatorname{sech}^{2} x\,dx\).
Examiner trap

The factor \(\dfrac{1}{a}\) belongs ONLY to the \(\tan^{-1}\) form. Writing \(\dfrac{1}{3}\sinh^{-1}\dfrac{x}{3}\) for \(\displaystyle\int \frac{dx}{\sqrt{x^{2} + 9}}\) carries it over by mistake; differentiate your answer to check. Also, \(\displaystyle\int \tanh x\,dx = \ln\cosh x\), not \(\operatorname{sech}^{2} x\) (that is its derivative) and not \(-\ln\cosh x\) (the minus belongs to \(\int \tan x\,dx = -\ln|\cos x|\)).

Section 02

Coefficients and completing the square.

The standard forms need \(x^{2}\) with coefficient 1. If it is not, take the coefficient out of the square root (or the denominator) first: \(\sqrt{9 - 4x^{2}} = 2\sqrt{\tfrac{9}{4} - x^{2}}\), so \(a = \tfrac{3}{2}\). If there is an \(x\) term, complete the square, then the standard form applies with \(x\) replaced by the bracket (a substitution \(t = x + p\) with \(dt = dx\)).

Worked example A coefficient in front of x²
Find the exact value of (a) \(\displaystyle\int_{0}^{3/4} \frac{dx}{\sqrt{9 - 4x^{2}}}\) (b) \(\displaystyle\int_{0}^{2} \frac{dx}{\sqrt{4x^{2} + 9}}\).
  1. (a) \(\sqrt{9 - 4x^{2}} = 2\sqrt{\tfrac{9}{4} - x^{2}}\), so \(a = \tfrac{3}{2}\) and a factor \(\tfrac{1}{2}\) comes out:
  2. \[\int_{0}^{3/4} \frac{dx}{\sqrt{9 - 4x^{2}}} = \frac{1}{2}\Big[\sin^{-1}\frac{2x}{3}\Big]_{0}^{3/4} = \frac{1}{2}\sin^{-1}\frac{1}{2} = \frac{\pi}{12}.\]
  3. (b) \(\sqrt{4x^{2} + 9} = 2\sqrt{x^{2} + \tfrac{9}{4}}\):
  4. \[\int_{0}^{2} \frac{dx}{\sqrt{4x^{2} + 9}} = \frac{1}{2}\Big[\sinh^{-1}\frac{2x}{3}\Big]_{0}^{2} = \frac{1}{2}\sinh^{-1}\frac{4}{3}.\]
  5. From Example 2(a), \(\sinh^{-1}\dfrac{4}{3} = \ln 3\).
Answer: (a) \(\dfrac{\pi}{12}\)   (b) \(\dfrac{1}{2}\ln 3\)
Worked example Complete the square first
(a) Find \(\displaystyle\int \frac{dx}{\sqrt{x^{2} + x}}\) for \(x \gt 0\). (b) Find the exact value of \(\displaystyle\int_{-1}^{0} \frac{dx}{\sqrt{3 - 2x - x^{2}}}\).
  1. (a) \(x^{2} + x = \left(x + \tfrac{1}{2}\right)^{2} - \tfrac{1}{4}\): the \(\cosh^{-1}\) form with \(a = \tfrac{1}{2}\).
  2. \[\int \frac{dx}{\sqrt{\left(x + \frac{1}{2}\right)^{2} - \frac{1}{4}}} = \cosh^{-1}\!\left(\frac{x + \frac{1}{2}}{\frac{1}{2}}\right) + c = \cosh^{-1}(2x + 1) + c.\]
  3. In log form: \(\cosh^{-1}(2x + 1) = \ln\!\Big(2x + 1 + \sqrt{(2x + 1)^{2} - 1}\Big) = \ln\!\Big(2x + 1 + 2\sqrt{x^{2} + x}\Big)\).
  4. (b) \(3 - 2x - x^{2} = 4 - (x^{2} + 2x + 1) = 4 - (x + 1)^{2}\): the \(\sin^{-1}\) form with \(a = 2\).
  5. \[\int_{-1}^{0} \frac{dx}{\sqrt{4 - (x + 1)^{2}}} = \Big[\sin^{-1}\frac{x + 1}{2}\Big]_{-1}^{0} = \sin^{-1}\frac{1}{2} - \sin^{-1} 0 = \frac{\pi}{6}.\]
Answer: (a) \(\cosh^{-1}(2x + 1) + c\), or \(\ln\!\big(2x + 1 + 2\sqrt{x^{2} + x}\big) + c\)   (b) \(\dfrac{\pi}{6}\)
Your turn
Find the exact value of \(\displaystyle\int_{-3}^{-1} \frac{dx}{x^{2} + 6x + 13}\).
Examiner trap

Taking the coefficient out of a square root gives its SQUARE ROOT outside: \(\sqrt{9 - 4x^{2}} = 2\sqrt{\tfrac{9}{4} - x^{2}}\), not \(4\sqrt{\ldots}\), and \(a = \tfrac{3}{2}\), not \(\tfrac{9}{4}\). For the \(\tan^{-1}\) form there is no root: \(4x^{2} + 9 = 4\left(x^{2} + \tfrac{9}{4}\right)\), so the factor outside is \(\tfrac{1}{4}\), and then \(\tfrac{1}{a} = \tfrac{2}{3}\) as well.

Stage 1 · Learn

Check your understanding

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Find \(\displaystyle\int \tanh x\,dx\).

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Find \(\displaystyle\int \frac{dx}{x^{2} + 16}\).

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Find \(\displaystyle\int \frac{dx}{\sqrt{x^{2} - 25}}\) for \(x \gt 5\).

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Which substitution turns \(\displaystyle\int \sqrt{x^{2} + 4}\,dx\) into an integral of \(\cosh^{2} u\)?

Section 03

Trig and hyperbolic substitutions.

When the square root is on TOP, or a power of it is not a standard form, substitute so that the root becomes a single function. The identities \(1 - \sin^{2} u = \cos^{2} u\), \(1 + \sinh^{2} u = \cosh^{2} u\) and \(\cosh^{2} u - 1 = \sinh^{2} u\) do the work. Remember \(dx\), and change the limits to \(u\)-values.

ExpressionSubstituteThen
\(\sqrt{a^{2} - x^{2}}\)\(x = a\sin u\), \(dx = a\cos u\,du\)\(\sqrt{a^{2} - x^{2}} = a\cos u\)
\(\sqrt{a^{2} + x^{2}}\)\(x = a\sinh u\), \(dx = a\cosh u\,du\)\(\sqrt{a^{2} + x^{2}} = a\cosh u\)
\(\sqrt{x^{2} - a^{2}}\)\(x = a\cosh u\), \(dx = a\sinh u\,du\)\(\sqrt{x^{2} - a^{2}} = a\sinh u\)   \((u \ge 0)\)

Squares are then integrated with double-angle identities: \(\cos^{2} u = \tfrac{1}{2}(1 + \cos 2u)\), \(\cosh^{2} u = \tfrac{1}{2}(\cosh 2u + 1)\), \(\sinh^{2} u = \tfrac{1}{2}(\cosh 2u - 1)\), and \(\sinh 2u = 2\sinh u\cosh u\) to get back to \(\sinh u\) and \(\cosh u\) at the limits.

Worked example x = sinh u for √(1 + x²)
Use the substitution \(x = \sinh u\) to find the exact value of \(\displaystyle\int_{0}^{3/4} \sqrt{1 + x^{2}}\,dx\).
  1. \(dx = \cosh u\,du\) and \(\sqrt{1 + \sinh^{2} u} = \cosh u\). Limits: \(x = 0 \Rightarrow u = 0\); \(x = \tfrac{3}{4} \Rightarrow u = \sinh^{-1}\tfrac{3}{4} = \ln 2\).
  2. \[\int_{0}^{3/4} \sqrt{1 + x^{2}}\,dx = \int_{0}^{\ln 2} \cosh^{2} u\,du = \frac{1}{2}\int_{0}^{\ln 2} (\cosh 2u + 1)\,du.\]
  3. \[= \frac{1}{2}\Big[\tfrac{1}{2}\sinh 2u + u\Big]_{0}^{\ln 2} = \frac{1}{2}\Big[\sinh u\cosh u + u\Big]_{0}^{\ln 2}.\]
  4. At \(u = \ln 2\): \(\sinh u = \tfrac{3}{4}\) and \(\cosh u = \sqrt{1 + \tfrac{9}{16}} = \tfrac{5}{4}\), so \(\sinh u\cosh u = \tfrac{15}{16}\). At \(u = 0\) the bracket is 0.
  5. \[\frac{1}{2}\left(\frac{15}{16} + \ln 2\right) = \frac{15}{32} + \frac{1}{2}\ln 2.\]
Answer: \(\dfrac{15}{32} + \dfrac{1}{2}\ln 2\)
Worked example x = a sin u for √(a² - x²)
Use the substitution \(x = 2\sin u\) to find the exact value of \(\displaystyle\int_{0}^{1} \sqrt{4 - x^{2}}\,dx\).
  1. \(dx = 2\cos u\,du\) and \(\sqrt{4 - 4\sin^{2} u} = 2\cos u\) (\(\cos u \ge 0\) here). Limits: \(x = 0 \Rightarrow u = 0\); \(x = 1 \Rightarrow \sin u = \tfrac{1}{2}\), \(u = \tfrac{\pi}{6}\).
  2. \[\int_{0}^{\pi/6} 4\cos^{2} u\,du = 2\int_{0}^{\pi/6} (1 + \cos 2u)\,du = 2\Big[u + \tfrac{1}{2}\sin 2u\Big]_{0}^{\pi/6}.\]
  3. \[= 2\left(\frac{\pi}{6} + \frac{1}{2}\cdot\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3} + \frac{\sqrt{3}}{2}.\]
Answer: \(\dfrac{\pi}{3} + \dfrac{\sqrt{3}}{2}\)
Your turn
Use the substitution \(x = \sinh u\) to show that \(\displaystyle\int_{0}^{1} \frac{dx}{(1 + x^{2})^{3/2}} = \frac{1}{\sqrt{2}}\).
Examiner trap

Three slips cost marks here: forgetting the \(dx = \cosh u\,du\) factor (giving \(\int \cosh u\,du\) instead of \(\int \cosh^{2} u\,du\)); keeping the \(x\)-limits after changing to \(u\); and the sign in the hyperbolic double-angle identity. \(\cosh^{2} u = \tfrac{1}{2}(\cosh 2u + 1)\) and \(\sinh^{2} u = \tfrac{1}{2}(\cosh 2u - 1)\): Osborn's rule flips the sign of \(\sin^{2} u = \tfrac{1}{2}(1 - \cos 2u)\).

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Find the exact value of \(\displaystyle\int_{0}^{2} \frac{dx}{\sqrt{x^{2} + 4}}\).

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Find the exact value of \(\displaystyle\int_{0}^{1} \frac{dx}{\sqrt{4 - 3x^{2}}}\).

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Find \(\displaystyle\int \frac{dx}{x^{2} + 4x + 8}\).

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Use the substitution \(x = \sinh u\) to find the exact value of \(\displaystyle\int_{0}^{4/3} \sqrt{1 + x^{2}}\,dx\).

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Six original Cambridge-style questions
Hyperbolic integrals, the four standard forms with exact answers, coefficients of \(x^{2}\), completing the square, the substitution \(x = 2\sinh u\), and the area inside a hyperbola by \(x = 3\cosh u\).