4.08h - Inverse trig and hyperbolic integrals | OCR A-Level Further Mathematics A

Recognise the four standard integrals that give inverse trigonometric or hyperbolic functions, including after completing the square. You will explain them using substitutions and choose trigonometric or hyperbolic substitutions for related definite and indefinite integrals that contain square roots.

Integrals giving inverse sine

Can you integrate $\sqrt{9-x^2}$ exactly without expanding it into a series? The semicircle suggests that angles may help. First learn its simpler reciprocal, then use the same substitution to tackle the original square root.

You already know

$(\sin^{-1}x)'=1/\sqrt{1-x^2}$ , $(\tan^{-1}x)'=1/(1+x^2)$ , $(\sinh^{-1}x)'=1/\sqrt{1+x^2}$ and $(\cosh^{-1}x)'=1/\sqrt{x^2-1}$ on their derivative domains. Reverse the chain rule or substitute, changing both the differential and any definite limits.

Take $x=3\sin\theta$ , with $-\pi/2<\theta<\pi/2$ . Then $dx=3\cos\theta\,d\theta$ and $\sqrt{9-x^2}=3\cos\theta$ , because cosine is positive on that interval. Thus

\int\frac{dx}{\sqrt{9-x^2}} =\int\frac{3\cos\theta}{3\cos\theta}\,d\theta =\theta+c=\sin^{-1}\left(\frac{x}{3}\right)+c

The cancellation explains why this integral has no factor $1/3$ . Replace $3$ by any positive constant $a$ to obtain the standard form.

The inverse sine integral

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

\int\frac{dx}{\sqrt{a^2-x^2}} =\sin^{-1}\left(\frac{x}{a}\right)+c

Here and throughout the lesson, $c$ is an arbitrary constant.
In the formula booklet.

An inner coefficient needs its own compensation. A completed square identifies the centre and the value of $a$ ; neither can be read reliably from an uncompleted quadratic.

Worked example 1: extracting an inner scale

Question. Find $\displaystyle\int\frac{dx}{\sqrt{25-4x^2}}$ .

Plan. Put $u=2x$ , so the denominator becomes the standard form with $a=5$ .

\begin{aligned} u&=2x,\quad dx=\tfrac12\,du && \text{substitute} \\ I&=\frac12\int\frac{du}{\sqrt{25-u^2}} && \text{transform the differential} \\ &=\frac12\sin^{-1}\left(\frac{u}{5}\right)+c && \text{standard integral} \end{aligned}

Why this step? The standard formula's absence of an outer $1/a$ does not remove the factor caused by $u=2x$ .

Answer. $\frac12\sin^{-1}(2x/5)+c$ , on $-5/2<x<5/2$ .

Check. Differentiating gives $\frac12(2/5)/\sqrt{1-4x^2/25}=1/\sqrt{25-4x^2}$ .

Change only the inner argument from $2x$ to $2x-1$ . Completing the square reveals that shift before you use the formula.

Worked example 2: a shifted square

Question. Find $\displaystyle\int\frac{dx}{\sqrt{24+4x-4x^2}}$ .

Plan. Complete the square and set $u=2x-1$ .

\begin{aligned} 24+4x-4x^2&=25-(2x-1)^2 && \text{complete the square} \\ u&=2x-1,\quad dx=\tfrac12\,du && \text{inner derivative }2 \\ I&=\frac12\int\frac{du}{\sqrt{25-u^2}} && \text{standard form} \\ &=\frac12\sin^{-1}\left(\frac{2x-1}{5}\right)+c && \text{return to }x \end{aligned}

Why this step? $25$ , not $24$ , is the square of the radius after completing the square.

Answer. $\frac12\sin^{-1}((2x-1)/5)+c$ , for $-2<x<3$ .

Check. $|2x-1|<5$ gives exactly that interval, and differentiating recovers the original denominator.

The derivative of the answer is a quick safeguard against a stray scale factor.

Watch out

Wrong: $\int dx/\sqrt{a^2-x^2}=(1/a)\sin^{-1}(x/a)+c$ . Differentiation supplies a factor $1/a$ and the denominator supplies another factor that cancels it. The correct coefficient is $1$ .

Keep the inner scale and shift, but change the radius.

Integrals giving inverse tangent

When the quadratic is not under a square root, a sum of squares leads to inverse tangent, with a different scale factor.

For $a>0$ , set $x=a\tan\theta$ , choosing $-\pi/2<\theta<\pi/2$ . Then $dx=a\sec^2\theta\,d\theta$ , and $a^2+x^2=a^2\sec^2\theta$ , so

\int\frac{dx}{a^2+x^2} =\frac1a\int d\theta =\frac1a\tan^{-1}\left(\frac{x}{a}\right)+c

Here the differential contributes one factor of $a$ , while the denominator contributes two. One factor remains below the line.

The inverse tangent integral

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

\int\frac{dx}{a^2+x^2} =\frac1a\tan^{-1}\left(\frac{x}{a}\right)+c

In the formula booklet.

Complete the square before choosing the inverse. A positive quadratic with an $x$ -term may still be a sum of squares after a shift.

Worked example 3: a scaled sum of squares

Question. Find $\displaystyle\int\frac{dx}{9+4x^2}$ .

Plan. Set $u=2x$ and use the inverse tangent integral with $a=3$ .

\begin{aligned} u&=2x,\quad dx=\tfrac12\,du && \text{substitute} \\ I&=\frac12\int\frac{du}{9+u^2} && \text{include }dx \\ &=\frac16\tan^{-1}\left(\frac{2x}{3}\right)+c && \text{standard integral} \end{aligned}

Why this step? The factors $1/2$ and $1/3$ have different sources: the inner scale and the standard inverse tangent integral.

Answer. $\frac16\tan^{-1}(2x/3)+c$ , for all real $x$ .

Check. Its derivative is $\frac16(2/3)/(1+4x^2/9)=1/(9+4x^2)$ .

Change only the inner argument to $2x-1$ , just as in the inverse sine examples.

Worked example 4: completing a positive quadratic

Question. Find $\displaystyle\int\frac{dx}{4x^2-4x+10}$ .

Plan. Complete the square, then account for both scales.

\begin{aligned} 4x^2-4x+10&=(2x-1)^2+9 && \text{complete the square} \\ u&=2x-1,\quad dx=\tfrac12\,du && \text{substitute} \\ I&=\frac12\int\frac{du}{u^2+9} && \text{standard form} \\ &=\frac16\tan^{-1}\left(\frac{2x-1}{3}\right)+c && \text{return to }x \end{aligned}

Why this step? The constant after completing the square is $9$ , so the formula uses its positive square root $3$ , not $9$ .

Answer. $\frac16\tan^{-1}((2x-1)/3)+c$ , for all real $x$ .

Check. At $x=1/2$ , its derivative is $1/9$ , agreeing with the smallest value of the denominator.

The two trigonometric standard forms look similar but have different outer coefficients.

Watch out

Wrong: $\int dx/(x^2+9)=\frac19\tan^{-1}(x/9)+c$ . The standard parameter is $a=3$ , since $a^2=9$ . Right: $\frac13\tan^{-1}(x/3)+c$ .

Change only the positive constant after completing the square.

Integrals giving inverse hyperbolic functions

A sum of squares under a square root needs inverse sine hyperbolic; a difference with the variable square first needs inverse cosine hyperbolic on its positive branch.

The substitution $x=a\sinh t$ gives $dx=a\cosh t\,dt$ and $\sqrt{x^2+a^2}=a\cosh t$ , since $\cosh t>0$ . The factors cancel, leaving $t=\sinh^{-1}(x/a)$ . For $x>a$ , use $x=a\cosh t$ with $t>0$ : now $dx=a\sinh t\,dt$ and $\sqrt{x^2-a^2}=a\sinh t$ , again leaving $t$ .

The inverse hyperbolic standard integrals

For $a>0$ ,

\begin{aligned} \int\frac{dx}{\sqrt{x^2+a^2}}&=\sinh^{-1}\left(\frac{x}{a}\right)+c && x\in\mathbb R \\ \int\frac{dx}{\sqrt{x^2-a^2}}&=\cosh^{-1}\left(\frac{x}{a}\right)+c && x>a \end{aligned}

The booklet also gives the logarithmic alternatives $\ln(x+\sqrt{x^2+a^2})+c$ and $\ln(x+\sqrt{x^2-a^2})+c$ on these respective domains. They differ from the inverse forms by the constant $\ln a$ .
In the formula booklet.

The second integrand also exists for $x<-a$ , but $\cosh^{-1}(x/a)$ does not. On that interval use $-\cosh^{-1}(-x/a)+c$ . A logarithmic expression valid on either separate interval is $\ln|x+\sqrt{x^2-a^2}|+c$ ; the negative-branch adjustment is not printed in the booklet and follows by differentiation. There is no interval crossing either singularity.

Worked example 5: completing a square for sine hyperbolic

Question. Find $\displaystyle\int\frac{dx}{\sqrt{x^2+4x+13}}$ .

Plan. Complete the square, then shift the variable without changing its scale.

\begin{aligned} x^2+4x+13&=(x+2)^2+9 && \text{complete the square} \\ u&=x+2,\quad du=dx && \text{shift only} \\ I&=\int\frac{du}{\sqrt{u^2+9}} && \text{sum of squares} \\ &=\sinh^{-1}\left(\frac{x+2}{3}\right)+c && \text{standard integral} \end{aligned}

Why this step? A translation has inner derivative $1$ , so it introduces no extra coefficient outside the inverse.

Answer. $\sinh^{-1}((x+2)/3)+c$ , for all real $x$ . An equivalent answer is $\ln(x+2+\sqrt{x^2+4x+13})+c$ .

Check. Differentiating gives $(1/3)/\sqrt{1+(x+2)^2/9}=1/\sqrt{x^2+4x+13}$ .

Change only the quadratic's constant from $13$ to $-5$ . The completed-square remainder changes from $+9$ to $-9$ , so the inverse and domain must change.

Worked example 6: a difference of squares

Question. Find $\displaystyle\int\frac{dx}{\sqrt{x^2+4x-5}}$ , for $x>1$ .

Plan. Complete the square and use the positive inverse cosine hyperbolic branch.

\begin{aligned} x^2+4x-5&=(x+2)^2-9 && \text{complete the square} \\ x>1&\Rightarrow x+2>3 && \text{check the branch} \\ I&=\int\frac{du}{\sqrt{u^2-9}}\quad(u=x+2) && \text{substitute} \\ &=\cosh^{-1}\left(\frac{x+2}{3}\right)+c && \text{standard integral} \end{aligned}

Why this step? The stated interval makes the inverse's argument exceed $1$ , so the printed positive-branch formula applies directly.

Answer. $\cosh^{-1}((x+2)/3)+c$ , for $x>1$ . Equivalently, $\ln(x+2+\sqrt{x^2+4x-5})+c$ .

Check. At $x=3$ , the antiderivative's derivative is $(1/3)/\sqrt{25/9-1}=1/4$ , matching the integrand.

These substitutions turn a square root into a single positive hyperbolic function. The same idea works when the square root is in the numerator, as the last part will show.

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4.08h - Inverse trig and hyperbolic integrals

Integrals giving inverse sine

Integrals giving inverse tangent

Integrals giving inverse hyperbolic functions

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