What is wrong substitution choice?

Use for sqrt(x^2 + a^2) and for sqrt(x^2 - a^2); swapping them leaves a square root you cannot simplify.

State the substitution to integrate 1sqrt(x^2 - 16). [1 mark]

OCR OCR 2018-style practice: Find 1sqrt(x^2 + 9)\,dx using a hyperbolic substitution.

Substitute x = 3 u, so dx = 3 u\,du (M1). Then x^2 + 9 = 9^2 u + 9 = 9(^2 u + 1) = 9^2 u, so sqrt(x^2 + 9) = 3 u (M1, A1). The integral becomes 3 u3 u\,du = 1\,du = u + C (A1). Reverse the substitution (u = arsinhx3) (A1): = arsinhx3 + C, equivalently ln(x + sqrt(x^2 + 9)) + C'. Markers reward the substitution, using ^2 + 1 = ^2, simplifying the integrand to 1, and reversing the substitution.

OCR OCR 2022-style practice: Find ^2 x\,dx.

Use the double-angle identity 2x = 2^2 x - 1, so ^2 x = 12( 2x + 1) (M1, A1). Integrate term by term (M1): ^2 x\,dx = 12( 2x + 1)\,dx = 12(12 2x + x) + C (A1, A1). Simplify (A1): = 14 2x + 12x + C. Markers reward the double-angle identity, integrating 2x to 12 2x, integrating the constant, and the tidy final form.

EnglandFurther MathsSyllabus dot point

How do you differentiate and integrate hyperbolic functions, and how do they help integrate certain algebraic functions?

Differentiation and integration of hyperbolic and inverse hyperbolic functions, and using hyperbolic substitutions to integrate functions involving the square root of x squared plus or minus a squared.

A focused answer to the OCR A-Level Further Mathematics A content on calculus with hyperbolic functions, covering the derivatives and integrals of sinh, cosh and tanh and their inverses, the standard integrals giving inverse hyperbolic functions, and using hyperbolic substitutions to integrate functions involving the square root of x squared plus or minus a constant.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

The derivatives are $\dfrac{d}{dx}\sinh x = \cosh x$ , $\dfrac{d}{dx}\cosh x = \sinh x$ (no sign change, unlike trig), and $\dfrac{d}{dx}\tanh x = \text{sech}^2 x$ . Integrating reverses these: $\int\sinh x\,dx = \cosh x + C$ , $\int\cosh x\,dx = \sinh x + C$ . The inverse hyperbolic functions arise as standard integrals: $\displaystyle\int\dfrac{1}{\sqrt{x^2 + a^2}}\,dx = \text{arsinh}\dfrac{x}{a} + C$ and $\displaystyle\int\dfrac{1}{\sqrt{x^2 - a^2}}\,dx = \text{arcosh}\dfrac{x}{a} + C$ . For integrals containing $\sqrt{x^2 + a^2}$ use $x = a\sinh u$ (since $\sinh^2 u + 1 = \cosh^2 u$ ); for $\sqrt{x^2 - a^2}$ use $x = a\cosh u$ . Powers of $\cosh$ or $\sinh$ are integrated after a double-angle reduction such as $\cosh^2 x = \tfrac{1}{2}(\cosh 2x + 1)$ .

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What this dot point is asking
Derivatives of hyperbolic functions
Integrals and the standard inverse-hyperbolic integrals
Hyperbolic substitution
Putting it together
Try this

What this dot point is asking

OCR wants you to differentiate and integrate the hyperbolic functions and their inverses, to recognise the standard integrals that produce inverse hyperbolic functions, and to use hyperbolic substitutions ( $x = a\sinh u$ or $x = a\cosh u$ ) to integrate functions involving $\sqrt{x^2 + a^2}$ or $\sqrt{x^2 - a^2}$ .

Derivatives of hyperbolic functions

The derivatives mirror the trigonometric ones but without the sign change between sine and cosine, which is one of the conveniences of the hyperbolic functions.

Integrals and the standard inverse-hyperbolic integrals

Reversing the derivatives gives the basic integrals, and the inverse-function derivatives give two standard integrals that turn awkward square-root denominators into inverse hyperbolic functions.

Hyperbolic substitution

The big application is integrating expressions with $\sqrt{x^2 \pm a^2}$ . Choosing the right hyperbolic substitution turns the square root into a single hyperbolic function via the identity $\cosh^2 u - \sinh^2 u = 1$ .

Putting it together

A typical integral, say $\int\sqrt{x^2 + 4}\,dx$ , combines the substitution $x = 2\sinh u$ with a double-angle reduction of $\cosh^2 u$ , then reverses the substitution. Recognising which standard integral or substitution fits is the key exam skill. The decision tree is short: if the integrand is a single hyperbolic function or a low power, integrate directly or with a double-angle identity; if it is $\dfrac{1}{\sqrt{x^2 \pm a^2}}$ , quote the standard inverse-hyperbolic integral; and if it is $\sqrt{x^2 \pm a^2}$ itself (or that square root in a denominator with other factors), use the matching hyperbolic substitution. Always state the substitution explicitly, transform $dx$ correctly, and finish in terms of $x$ .

Hyperbolic calculus completes the polar and hyperbolic strand and supplies standard integrals used across further calculus, including in volumes and improper integrals.

Try this

Q1. Differentiate $\cosh 3x$ . [1 mark]

Cue. By the chain rule, $3\sinh 3x$ .

Q2. State the substitution to integrate $\dfrac{1}{\sqrt{x^2 - 16}}$ . [1 mark]

Cue. $x = 4\cosh u$ , since $\cosh^2 u - 1 = \sinh^2 u$ .

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20185 marksFind

\displaystyle\int \dfrac{1}{\sqrt{x^2 + 9}}\,dx

using a hyperbolic substitution.

Show worked answer →

Substitute $x = 3\sinh u$ , so $dx = 3\cosh u\,du$ (M1).

Then $x^2 + 9 = 9\sinh^2 u + 9 = 9(\sinh^2 u + 1) = 9\cosh^2 u$ , so $\sqrt{x^2 + 9} = 3\cosh u$ (M1, A1).

The integral becomes $\displaystyle\int\dfrac{3\cosh u}{3\cosh u}\,du = \displaystyle\int 1\,du = u + C$ (A1).

Reverse the substitution ( $u = \text{arsinh}\dfrac{x}{3}$ ) (A1): $= \text{arsinh}\dfrac{x}{3} + C$ , equivalently $\ln\left(x + \sqrt{x^2 + 9}\right) + C'$ .

Markers reward the substitution, using $\sinh^2 + 1 = \cosh^2$ , simplifying the integrand to $1$ , and reversing the substitution.

OCR 20226 marksFind

\displaystyle\int \cosh^2 x\,dx

Show worked answer →

Use the double-angle identity $\cosh 2x = 2\cosh^2 x - 1$ , so $\cosh^2 x = \tfrac{1}{2}(\cosh 2x + 1)$ (M1, A1).

Integrate term by term (M1): $\displaystyle\int \cosh^2 x\,dx = \tfrac{1}{2}\displaystyle\int(\cosh 2x + 1)\,dx = \tfrac{1}{2}\left(\tfrac{1}{2}\sinh 2x + x\right) + C$ (A1, A1).

Simplify (A1): $= \tfrac{1}{4}\sinh 2x + \tfrac{1}{2}x + C$ .

Markers reward the double-angle identity, integrating $\cosh 2x$ to $\tfrac{1}{2}\sinh 2x$ , integrating the constant, and the tidy final form.

Related dot points

Sources & how we know this

OCR A Level Further Mathematics A (H245) specification — OCR (2017)