H5 - Hyperbolic substitutions | AQA A-Level Further Mathematics

A carefully chosen hyperbolic substitution turns a square root into a single sinh or cosh factor. You will integrate reciprocal square roots and related functions, choose substitutions from the completed square, and handle the separate positive and negative branches of a difference of squares.

Integrating reciprocal square roots

How can $\int\mathrm dx/\sqrt{x^2+9}$ become $\int\mathrm du$ , with no square root left? The substitution $x=3\sinh u$ will put the same factor in the differential and the denominator. We will explain that cancellation before changing the expression inside the root.

More generally, take $a>0$ . The identity $1+\sinh^2u=\cosh^2u$ suggests $x=a\sinh u$ when a root contains $x^2+a^2$ . Since $\cosh u>0$ ,

\sqrt{x^2+a^2}=a\cosh u,\qquad \mathrm dx=a\cosh u\,\mathrm du

The two factors cancel. Sinh takes every real value, so this substitution works for every real $x$ .

For $\sqrt{x^2-a^2}$ on $x>a$ , take $x=a\cosh u$ with $u>0$ . Then $\sinh u>0$ , so the principal square root is $a\sinh u$ , and again it cancels with $\mathrm dx=a\sinh u\,\mathrm du$ . The sign condition matters: generally $\sqrt{a^2\sinh^2u}=a|\sinh u|$ .

Two standard integrals, with conditions

For $a>0$ ,

\int\frac{\mathrm dx}{\sqrt{x^2+a^2}} =\sinh^{-1}\left(\frac xa\right)+C\qquad(x\in\mathbb R)

\int\frac{\mathrm dx}{\sqrt{x^2-a^2}} =\cosh^{-1}\left(\frac xa\right)+C\qquad(x>a)

The second displayed inverse-cosh form applies to the positive branch only; the negative branch is developed in the final part.

In the formula booklet, with $x>a$ on the difference-of-squares form.

The scale $a$ appears in both the substitution and the final inverse input, as this pair shows.

Worked example 1: a sum of squares

Question. Find $\int\mathrm dx/\sqrt{x^2+9}$ .

Plan. Put $x=3\sinh u$ ; the root and differential will share the same positive cosh factor.

\sqrt{x^2+9}=3\cosh u,\qquad \mathrm dx=3\cosh u\,\mathrm du

\int\frac{\mathrm dx}{\sqrt{x^2+9}} =\int\mathrm du=u+C

Answer. $\sinh^{-1}(x/3)+C$ , for all real $x$ . Equivalently, $\ln(x+\sqrt{x^2+9})+C$ , since the omitted $-\ln3$ is absorbed into the constant.

Check. Differentiating the inverse form gives $(1/3)/\sqrt{1+x^2/9}=1/\sqrt{x^2+9}$ , using $3>0$ .

Change only the sign of the constant inside the root. The substitution changes and the real domain becomes two separate intervals.

Worked example 2: a difference of squares on its positive branch

Question. Find $\int\mathrm dx/\sqrt{x^2-9}$ , for $x>3$ .

Plan. Put $x=3\cosh u$ with $u>0$ , so the square root is $3\sinh u$ .

\sqrt{x^2-9}=3\sinh u,\qquad \mathrm dx=3\sinh u\,\mathrm du

\int\frac{\mathrm dx}{\sqrt{x^2-9}}=\int\mathrm du=u+C

Answer. $\cosh^{-1}(x/3)+C$ , for $x>3$ , or $\ln(x+\sqrt{x^2-9})+C$ on this interval.

Check. Its derivative is $(1/3)/\sqrt{x^2/9-1}=1/\sqrt{x^2-9}$ . The inverse input exceeds 1 as required.

The formula booklet gives these inverse and logarithmic forms, with the positive-branch condition on the difference of squares. The substitution explains both the scale and the condition.

Watch out

Wrong: $\int\mathrm dx/\sqrt{x^2+9}=\tfrac13\sinh^{-1}(x/3)+C$ . The differential and root already cancel their factors of 3, so the answer is $\sinh^{-1}(x/3)+C$ , with no extra multiplier.

Use the scale inside the inverse function for this new root.

Choosing substitutions for associated functions

The same substitutions help when the square root is multiplied rather than inverted. Write the differential explicitly: a square root alone becomes a square of sinh or cosh, which the double-angle identities can integrate.

With $x=a\sinh u$ ,

\sqrt{x^2+a^2}\,\mathrm dx=a^2\cosh^2u\,\mathrm du =\frac{a^2}{2}(\cosh2u+1)\,\mathrm du

With $x=a\cosh u$ , $u>0$ ,

\sqrt{x^2-a^2}\,\mathrm dx=a^2\sinh^2u\,\mathrm du =\frac{a^2}{2}(\cosh2u-1)\,\mathrm du

The sign in the square root becomes the sign of the linear term after integration. Back-substitute using $\sinh2u=2\sinh u\cosh u$ , which expresses the result using the original root.

Worked example 3: integrating a square root of a sum

Question. Find $\int\sqrt{x^2+4}\,\mathrm dx$ .

Plan. Put $x=2\sinh u$ , then replace cosh squared by a double angle.

\begin{aligned} \int\sqrt{x^2+4}\,\mathrm dx &=4\int\cosh^2u\,\mathrm du\\ &=2\int(\cosh2u+1)\,\mathrm du =\sinh2u+2u+C \end{aligned}

Here $\sinh u=x/2$ , $\cosh u=\sqrt{x^2+4}/2$ , and $u=\sinh^{-1}(x/2)$ .

Answer. $\frac{x}{2}\sqrt{x^2+4}+2\sinh^{-1}(x/2)+C$ , for all real $x$ .

Check. Differentiation gives $\sqrt{x^2+4}/2+x^2/(2\sqrt{x^2+4})+2/\sqrt{x^2+4}=\sqrt{x^2+4}$ .

Change the plus to a minus, using the positive branch. The final inverse term changes sign.

Worked example 4: integrating a square root of a difference

Question. Find $\int\sqrt{x^2-4}\,\mathrm dx$ , for $x>2$ .

Plan. Put $x=2\cosh u$ , with $u>0$ , then reduce sinh squared.

\begin{aligned} \int\sqrt{x^2-4}\,\mathrm dx &=4\int\sinh^2u\,\mathrm du\\ &=2\int(\cosh2u-1)\,\mathrm du =\sinh2u-2u+C \end{aligned}

Now $\cosh u=x/2$ , $\sinh u=\sqrt{x^2-4}/2$ , and $u=\cosh^{-1}(x/2)$ .

Answer. $\frac{x}{2}\sqrt{x^2-4}-2\cosh^{-1}(x/2)+C$ , for $x>2$ .

Check. Its derivative is $\sqrt{x^2-4}/2+x^2/(2\sqrt{x^2-4})-2/\sqrt{x^2-4}=\sqrt{x^2-4}$ .

A quadratic may first need completing the square. For example, $x^2+6x+13=(x+3)^2+4$ , so use $x+3=2\sinh u$ , not $x=2\sinh u$ . Then $\mathrm dx=2\cosh u\,\mathrm du$ , and an integral of this square root becomes $4\int\cosh^2u\,\mathrm du$ , exactly the structure of Worked example 3. A completed square $(x+b)^2-a^2$ instead suggests $x+b=a\cosh u$ on $x+b>a$ .

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H5 - Hyperbolic substitutions

Integrating reciprocal square roots

Choosing substitutions for associated functions

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