CP 8.5 - Integrals giving inverse hyperbolic functions | Edexcel A-Level Further Mathematics

Recognise reciprocal square-root integrals giving $\operatorname{arsinh}$ or $\operatorname{arcosh}$ , and choose hyperbolic substitutions for associated integrals (CP 8.5). Completing the square reveals the structure; the real domain and the sign of a square root determine the correct branch.

A sum of squares gives arsinh

Why does

\int\frac{\mathrm dx}{\sqrt{x^2+4}}

give an inverse hyperbolic sine, while changing the plus sign to a minus sign gives a different inverse and a restricted domain? The right substitution turns the square root into a single hyperbolic factor, which cancels against the differential.

For a sum of squares, $\cosh^2t-\sinh^2t=1$ rearranges to $1+\sinh^2t=\cosh^2t$ . This suggests $x=a\sinh t$ when $a>0$ .

You already know

Substitution changes the differential as well as the integrand; inverse trigonometric integrals have different denominator structures, as in $\int(a^2-x^2)^{-1/2}\,\mathrm dx=\arcsin(x/a)+C$ . From hyperbolic functions, $\cosh^2t-\sinh^2t=1$ , $\cosh t>0$ , and the inverse logarithmic forms have specific real domains.

With $x=a\sinh t$ ,

\begin{aligned} \mathrm dx&=a\cosh t\,\mathrm dt && \text{differentiate the substitution}\\ \sqrt{x^2+a^2}&=a\sqrt{1+\sinh^2t}=a\cosh t && \cosh t>0\\ \int\frac{\mathrm dx}{\sqrt{x^2+a^2}} &=\int1\,\mathrm dt=t+C && \text{cancel identical factors} \end{aligned}

Since $t=\operatorname{arsinh}(x/a)$ , this yields the standard result for every real $x$ .

The reciprocal sum-of-squares square root

For $a>0$ and real $x$ ,

\int\frac{\mathrm dx}{\sqrt{x^2+a^2}} =\operatorname{arsinh}\left(\frac xa\right)+C

An equivalent logarithmic answer is $\ln(x+\sqrt{x^2+a^2})+C$ . The inverse form contains an additional constant $-\ln a$ , which is absorbed into the arbitrary integration constant.

In the formula booklet.

The cancellation explains why no factor $1/a$ appears outside the inverse. A separate scale multiplying $x$ , however, still creates its own differential factor.

Worked example 1: a standard sum of squares

Question. Find $\int(x^2+9)^{-1/2}\,\mathrm dx$ .

Plan. Use $x=3\sinh t$ , which makes the radical $3\cosh t$ .

\begin{aligned} x&=3\sinh t,\qquad \mathrm dx=3\cosh t\,\mathrm dt && \text{chosen substitution}\\ \sqrt{x^2+9}&=3\cosh t && \text{fundamental identity}\\ \int\frac{\mathrm dx}{\sqrt{x^2+9}} &=\int1\,\mathrm dt=\operatorname{arsinh}\frac x3+C && \text{cancel and return} \end{aligned}

Why this step? Both the square root and differential contain the same factor $3\cosh t$ ; there is no leftover $1/3$ .

Answer. $\operatorname{arsinh}(x/3)+C$ , for all real $x$ .

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

Change only the coefficient of $x^2$ . The radical now contains $2x$ , so scaling the variable contributes an outside factor.

Worked example 2: a scale inside the radical

Question. Find $\int(4x^2+9)^{-1/2}\,\mathrm dx$ .

Plan. Put $u=2x$ , then apply the standard sum-of-squares result.

\begin{aligned} u&=2x,\qquad \mathrm dx=\frac12\,\mathrm du && \text{scale the variable}\\ \int\frac{\mathrm dx}{\sqrt{4x^2+9}} &=\frac12\int\frac{\mathrm du}{\sqrt{u^2+9}} && \text{include the differential factor}\\ &=\frac12\operatorname{arsinh}\left(\frac{2x}{3}\right)+C && \text{standard result} \end{aligned}

Why this step? The factor $1/2$ comes from replacing $x$ by $u/2$ , not from the radius parameter $3$ .

Answer. $\frac12\operatorname{arsinh}(2x/3)+C$ , for all real $x$ .

Check. Differentiation gives

\frac12\frac{2/3}{\sqrt{1+4x^2/9}}=\frac1{\sqrt{4x^2+9}}

Compare the whole denominator before choosing an inverse. A reciprocal sum of squares without a square root gives an arctangent form, not this $\operatorname{arsinh}$ form.

Watch out

Wrong: $\int(x^2+9)^{-1/2}\,\mathrm dx=\frac13\operatorname{arsinh}(x/3)+C$ . Differentiation leaves an unwanted $1/3$ . Right: $\operatorname{arsinh}(x/3)+C$ .

Match a new pair of scales.

A difference of squares and its branches

The reciprocal square root of $x^2-a^2$ is real only on the two exterior intervals $x>a$ and $x<-a$ , and the substitution must respect which interval you are using.

On $x>a>0$ , put $x=a\cosh t$ with $t>0$ . Then

\begin{aligned} \mathrm dx&=a\sinh t\,\mathrm dt && \text{differentiate}\\ \sqrt{x^2-a^2}&=a\sqrt{\sinh^2t}=a\sinh t && t>0\\ \int\frac{\mathrm dx}{\sqrt{x^2-a^2}} &=\int1\,\mathrm dt=\operatorname{arcosh}\frac xa+C && \text{cancel and return} \end{aligned}

The strict domain excludes $x=a$ , where the reciprocal integrand is undefined.

The positive exterior branch

For $a>0$ and $x>a$ ,

\int\frac{\mathrm dx}{\sqrt{x^2-a^2}} =\operatorname{arcosh}\left(\frac xa\right)+C

An equivalent logarithmic primitive is $\ln(x+\sqrt{x^2-a^2})+C$ , with the constant shift absorbed.

In the formula booklet.

For $x<-a$ , the real integrand still exists, but $x/a<-1$ is outside the domain of $\operatorname{arcosh}$ . Instead put $x=-a\cosh t$ , $t>0$ . The square root is still $a\sinh t>0$ , while $\mathrm dx=-a\sinh t\,\mathrm dt$ . Their quotient is now $-\mathrm dt$ .

The negative exterior branch

For $a>0$ and $x<-a$ ,

\int\frac{\mathrm dx}{\sqrt{x^2-a^2}} =-\operatorname{arcosh}\left(\frac{-x}{a}\right)+C

On either exterior interval, a logarithmic primitive is

\ln\left|x+\sqrt{x^2-a^2}\right|+C

The two intervals may have different arbitrary constants. This extends the booklet's stated $x>a$ form.

Not in the formula booklet: learn it.

The minus sign in the negative branch makes its derivative positive, as the original integrand is positive on both intervals. Simply writing $\operatorname{arcosh}(-x/a)$ without that minus sign would produce a negative derivative.

[DIAGRAM: asset_name: Domains of reciprocal square-root integrands; asset_slug: 9fm0-hyperbolic-integral-domains; caption: A sum of squares permits every real input; a difference permits only the two exterior intervals; description: Two Cartesian panels show y equals one over root of x squared plus 4 as a smooth positive curve for all real x, and y equals one over root of x squared minus 4 as two positive exterior branches with vertical asymptotes x equals minus 2 and x equals 2. The interval between those asymptotes is pale grey and labelled no real integrand; neither endpoint belongs to the domain.]
A sum of squares permits every real input

Worked example 3: the positive branch

Question. Find $\int(x^2-9)^{-1/2}\,\mathrm dx$ , for $x>3$ .

Plan. Put $x=3\cosh t$ , with $t>0$ , so the radical and differential share a positive factor.

\begin{aligned} \mathrm dx&=3\sinh t\,\mathrm dt,\qquad \sqrt{x^2-9}=3\sinh t && t>0\\ \int\frac{\mathrm dx}{\sqrt{x^2-9}} &=\int1\,\mathrm dt=t+C && \text{cancel}\\ &=\operatorname{arcosh}\frac x3+C && \text{positive inverse branch} \end{aligned}

Why this step? The restriction $t>0$ justifies replacing $\sqrt{\sinh^2t}$ by $\sinh t$ , instead of an absolute value with unknown sign.

Answer. $\operatorname{arcosh}(x/3)+C$ , for $x>3$ .

Check. Its derivative is $(1/3)/\sqrt{x^2/9-1}=1/\sqrt{x^2-9}$ .

Change only the interval of $x$ . The radicand is unchanged, but the substitution and the inverse answer acquire the appropriate signs.

Worked example 4: the negative branch

Question. Find $\int(x^2-9)^{-1/2}\,\mathrm dx$ , for $x<-3$ .

Plan. Use $x=-3\cosh t$ , $t>0$ , so that the inverse's input is positive.

\begin{aligned} \mathrm dx&=-3\sinh t\,\mathrm dt,\qquad \sqrt{x^2-9}=3\sinh t && t>0\\ \int\frac{\mathrm dx}{\sqrt{x^2-9}} &=-\int1\,\mathrm dt=-t+C && \text{retain the differential sign}\\ &=-\operatorname{arcosh}\left(\frac{-x}{3}\right)+C && \text{return to }x \end{aligned}

Why this step? The square root is a non-negative length; its sign does not follow the negative sign of $x$ . That sign belongs in the differential.

Answer. $-\operatorname{arcosh}(-x/3)+C$ , for $x<-3$ .

Check. Differentiating gives

-\frac{-1/3}{\sqrt{x^2/9-1}}=\frac1{\sqrt{x^2-9}}

At $x=-5$ , this is $1/4>0$ .

The same sign issue underlies $\sqrt{u^2}=|u|$ . Choosing a range for the substitution before simplifying the radical prevents a later branch correction from being forgotten.

Watch out

Wrong: $\operatorname{arcosh}(x/a)$ on $x<-a$ . That inverse has no real value there. Right: use the negative-branch answer, or the logarithmic primitive with an absolute value, and state the interval.

Use a new parameter on the negative interval.

Completing the square and choosing a substitution

A quadratic expression may hide its sum or difference of squares, so complete the square before selecting an inverse or a substitution.

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CP 8.5 - Integrals giving inverse hyperbolic functions

A sum of squares gives arsinh

A difference of squares and its branches

Completing the square and choosing a substitution

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