CP 5.6 - Integrals giving inverse trigonometric functions | Edexcel A-Level Further Mathematics

Some algebraic integrands have inverse trigonometric antiderivatives. You will recognise the arcsine and arctangent forms, choose sine or tangent substitutions for related integrals, and distinguish these from a reciprocal difference of squares, which gives a logarithm.

A square root giving arcsine

Why does $1/\sqrt{9-x^2}$ integrate to an angle, while $1/(9-x^2)$ does not give the same answer? The square root changes the structure completely. Recognising exactly what is in the denominator is the first decision.

You already know

The derivative of $\arcsin u$ is $u'/\sqrt{1-u^2}$ , and that of $\arctan u$ is $u'/(1+u^2)$ . Substitution requires changing both the integrand and the differential. A derivative check is a reliable way to recover a missing constant factor.

For $a>0$ , differentiating $\arcsin(x/a)$ gives

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

because $\sqrt{a^2-x^2}=a\sqrt{1-x^2/a^2}$ . The chain factor cancels the scale inside the square root.

Arcsine integral

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

\int\frac1{\sqrt{a^2-x^2}}\,\mathrm{d}x=\arcsin\left(\frac xa\right)+C

In the formula booklet.

The positive parameter $a$ is the square root of the constant term, not the constant itself. For $9-x^2$ , use $a=3$ .

Worked example 1: a standard form

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

Plan. Match $a^2=9$ to the standard arcsine integral.

\begin{aligned} a&=3 && \text{choose the positive root}\\ \int\frac{\mathrm{d}x}{\sqrt{9-x^2}} &=\arcsin\left(\frac x3\right)+C && \text{apply the standard result} \end{aligned}

Why no factor $\frac13$ outside? The chain factor is cancelled by the square-root scale.

Answer. $\arcsin(x/3)+C$ .

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

Worked example 2 changes only the coefficient of $x^2$ . A substitution is now needed to expose the standard form.

Worked example 2: scaling the variable

Question. Find $\int(9-4x^2)^{-1/2}\,\mathrm{d}x$ , for $|x|<3/2$ .

Plan. Put $u=2x$ , including the factor $\mathrm{d}x=\frac12\mathrm{d}u$ .

\begin{aligned} \int\frac{\mathrm{d}x}{\sqrt{9-4x^2}} &=\frac12\int\frac{\mathrm{d}u}{\sqrt{9-u^2}} && u=2x\\ &=\frac12\arcsin\left(\frac u3\right)+C && \text{standard form}\\ &=\frac12\arcsin\left(\frac{2x}{3}\right)+C && \text{return to }x \end{aligned}

Why this step? Replacing the argument by $2x$ doubles its derivative, so the outside factor $\frac12$ is necessary.

Answer. $\frac12\arcsin(2x/3)+C$ .

Check. Differentiation gives $\frac12\times\frac{2/3}{\sqrt{1-4x^2/9}}=1/\sqrt{9-4x^2}$ .

Matching the appearance is only half the task; differentiating the proposed answer checks its scale.

Watch out

Wrong: $\int(9-4x^2)^{-1/2}\,\mathrm{d}x=\arcsin(2x/3)+C$ . Right: include $\frac12$ outside. An argument change brings a chain factor with it.

Change both scales in the matched check.

A sum of squares giving arctangent

A reciprocal sum of squares has a different antiderivative. Without a square root, the scale from the denominator does not cancel the chain factor in the same way.

Arctangent integral

For $a>0$ ,

\int\frac1{a^2+x^2}\,\mathrm{d}x =\frac1a\arctan\left(\frac xa\right)+C

This holds for every real $x$ .

In the formula booklet.

Differentiating the answer gives $(1/a)(1/a)/(1+x^2/a^2)=1/(a^2+x^2)$ . Both factors of $1/a$ are needed before the denominator is simplified.

Worked example 3: a reciprocal sum of squares

Question. Find $\int\frac1{9+x^2}\,\mathrm{d}x$ .

Plan. Match $a=3$ and use both the outside factor and the scaled argument.

\int\frac{\mathrm{d}x}{9+x^2} =\frac13\arctan\left(\frac x3\right)+C

Why this factor? $\arctan(x/3)$ alone differentiates to $3/(9+x^2)$ , three times the required integrand.

Answer. $\frac13\arctan(x/3)+C$ .

Check. At $x=0$ , the answer's derivative is $\frac13\times\frac13=\frac19$ , matching the integrand.

Worked example 4 changes only the coefficient of $x^2$ , just as in the arcsine pair. Now the standard integral already has an outside scale factor too.

Worked example 4: two scale factors

Question. Find $\int\frac1{9+4x^2}\,\mathrm{d}x$ .

Plan. Put $u=2x$ , then apply the arctangent formula with $a=3$ .

\begin{aligned} \int\frac{\mathrm{d}x}{9+4x^2} &=\frac12\int\frac{\mathrm{d}u}{9+u^2} && \text{change the differential}\\ &=\frac12\times\frac13\arctan\left(\frac u3\right)+C && \text{standard integral}\\ &=\frac16\arctan\left(\frac{2x}{3}\right)+C && \text{combine the factors} \end{aligned}

Why this step? One factor comes from scaling $x$ , the other from the standard arctangent integral. They have separate jobs.

Answer. $\frac16\arctan(2x/3)+C$ .

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

The denominator $a^2+x^2$ has no real zero, so there is no interval restriction from a denominator singularity here.

Watch out

Wrong: use $1/9$ outside arctangent because the denominator begins with $9$ . Right: the standard form uses $a=3$ , giving $1/3$ before any additional variable scaling.

Practise keeping both scales visible.

Choosing a sine substitution for a difference of squares

When a root contains $a^2-x^2$ , choose $x=a\sin\theta$ . The identity $1-\sin^2\theta=\cos^2\theta$ turns the root into something manageable.

Choose $-\pi/2\le\theta\le\pi/2$ . Then $\cos\theta\ge0$ , so $\sqrt{a^2-x^2}=a\cos\theta$ , rather than an unresolved absolute value. Also $\mathrm{d}x=a\cos\theta\,\mathrm{d}\theta$ . This substitution strategy is not listed in the formula booklet: learn how the identity motivates it.

[DIAGRAM: asset_name: Triangles behind sine and tangent substitutions; asset_slug: 9fm0-cp56-substitution-triangles; caption: Sine matches a difference under a root; tangent matches a sum of squares; description: Two right triangles for positive x and a are shown. The sine triangle has opposite side x, hypotenuse a and adjacent side square root of a squared minus x squared, with x=a sin theta. The tangent triangle has opposite side x, adjacent side a and hypotenuse square root of a squared plus x squared, with x=a tan theta. Each theta is placed at the lower-left vertex.]
Sine matches a difference under a root

The familiar identities $\cos^2\theta=(1+\cos2\theta)/2$ and $\sin^2\theta=(1-\cos2\theta)/2$ then reduce squared trigonometric functions to integrable terms. These identities are assumed A-level Mathematics knowledge.

Worked example 5: integrating a semicircle height

Question. Use a trigonometric substitution to find $\int\sqrt{9-x^2}\,\mathrm{d}x$ , for $|x|<3$ .

Plan. Put $x=3\sin\theta$ , integrate $9\cos^2\theta$ , then return to $x$ .

\begin{aligned} \int\sqrt{9-x^2}\,\mathrm{d}x &=9\int\cos^2\theta\,\mathrm{d}\theta && \text{root and differential each contribute }3\cos\theta\\ &=\frac92\theta+\frac94\sin2\theta+C && \text{double-angle identity} \end{aligned}

Here $\theta=\arcsin(x/3)$ and $\sin2\theta=2(x/3)(\sqrt{9-x^2}/3)$ .

Why this substitution? It replaces the difference of squares by a squared cosine, whose sign is controlled by the chosen branch.

Answer.

\frac x2\sqrt{9-x^2}+\frac92\arcsin\left(\frac x3\right)+C

Check. At $x=0$ , differentiating the first term gives $3/2$ and the second gives $3/2$ , totalling the integrand's value $3$ .

Worked example 6 uses the same square root in the denominator and changes the algebraic numerator to $x^2$ . The same substitution now gives a squared sine.

Worked example 6: a squared numerator

Question. Find $\int\frac{x^2}{\sqrt{9-x^2}}\,\mathrm{d}x$ , for $|x|<3$ , using $x=3\sin\theta$ .

Plan. Cancel the cosine factors from the root and differential, then integrate $9\sin^2\theta$ .

\begin{aligned} \int\frac{x^2}{\sqrt{9-x^2}}\,\mathrm{d}x &=9\int\sin^2\theta\,\mathrm{d}\theta && \text{cancel }3\cos\theta\\ &=\frac92\theta-\frac94\sin2\theta+C && \text{use the sine-square identity} \end{aligned}

Why this step? The root and differential cancel, but the numerator becomes $9\sin^2\theta$ ; it must not disappear too.

Answer.

\frac92\arcsin\left(\frac x3\right)-\frac x2\sqrt{9-x^2}+C

Check. Adding the integrands in Examples 5 and 6 gives $9/\sqrt{9-x^2}$ . Their answers add to $9\arcsin(x/3)+C$ , exactly as the standard integral predicts.

Square roots can hide $\pi$ : using the same method with $a=1$ gives $\int_0^1\sqrt{1-x^2}\,\mathrm{d}x=\pi/4$ . This is also the area of a quarter of the unit circle. The algebraic and geometric answers meet at the same constant.

Watch out

Wrong: $\sqrt{9\cos^2\theta}=3\cos\theta$ without controlling the angle. In general it is $3|\cos\theta|$ . Our principal substitution range makes the cosine non-negative, so the simpler expression is justified.

Change the radius in the next sine substitution.

Choosing a tangent substitution for a sum of squares

For $a^2+x^2$ , choose $x=a\tan\theta$ . Then $a^2+x^2=a^2\sec^2\theta$ and $\mathrm{d}x=a\sec^2\theta\,\mathrm{d}\theta$ . Take $-\pi/2<\theta<\pi/2$ , so $\theta=\arctan(x/a)$ is unambiguous.

This strategy is not listed in the formula booklet. Its value is that the differential cancels part of a denominator containing a power of $a^2+x^2$ .

Worked example 7: a squared denominator

Question. Use a trigonometric substitution to find $\int\frac1{(9+x^2)^2}\,\mathrm{d}x$ .

Plan. Put $x=3\tan\theta$ , cancel secant factors, then integrate a squared cosine.

\begin{aligned} \int\frac{\mathrm{d}x}{(9+x^2)^2} &=\frac1{27}\int\cos^2\theta\,\mathrm{d}\theta && \frac{3\sec^2\theta}{81\sec^4\theta}=\frac1{27}\cos^2\theta\\ &=\frac1{54}\theta+\frac1{108}\sin2\theta+C && \text{integrate} \end{aligned}

Since $\tan\theta=x/3$ , $\sin2\theta=2\tan\theta/(1+\tan^2\theta)=6x/(9+x^2)$ .

Why tangent? Its differential contains $\sec^2\theta$ , the same factor produced by the sum of squares.

Answer.

\frac{x}{18(9+x^2)}+\frac1{54}\arctan\left(\frac x3\right)+C

Check. At $x=0$ , the two derivative contributions are $1/162$ each, totalling $1/81$ , the original integrand's value.

Worked example 8 changes only the numerator to $x^2$ . The transformed integral becomes a squared sine instead.

Worked example 8: the numerator survives substitution

Question. Find $\int\frac{x^2}{(9+x^2)^2}\,\mathrm{d}x$ using $x=3\tan\theta$ .

Plan. Transform the numerator as well as the denominator and differential, then simplify to $\sin^2\theta$ .

\begin{aligned} \int\frac{x^2}{(9+x^2)^2}\,\mathrm{d}x &=\frac13\int\sin^2\theta\,\mathrm{d}\theta && \tan^2\theta/\sec^2\theta=\sin^2\theta\\ &=\frac16\theta-\frac1{12}\sin2\theta+C && \text{integrate} \end{aligned}

Why this step? The numerator contributes $9\tan^2\theta$ , changing the coefficient from $1/27$ to $1/3$ .

Answer.

\frac16\arctan\left(\frac x3\right)-\frac{x}{2(9+x^2)}+C

Check. Differentiation gives

\frac1{2(9+x^2)}-\frac{9-x^2}{2(9+x^2)^2} =\frac{x^2}{(9+x^2)^2}

For a definite integral, either change both limits into $\theta$ -values or return to $x$ before using the original limits. Mixing one variable's limits with another variable's antiderivative changes the problem.

Watch out

Wrong: from $x=3\tan\theta$ , write $\mathrm{d}x=3\sec\theta\,\mathrm{d}\theta$ . The derivative of tangent is secant squared, so this would leave the wrong powers after cancellation. Right: $\mathrm{d}x=3\sec^2\theta\,\mathrm{d}\theta$ .

The differential is part of choosing the substitution, so write it before transforming the fraction.

Problem-solving

Choose a substitution for the expression you want to simplify: $x=a\sin\theta$ for $\sqrt{a^2-x^2}$ , and $x=a\tan\theta$ for $a^2+x^2$ . Write the differential immediately, then cancel factors.

Use the tangent method with a new scale.

A reciprocal difference of squares gives a logarithm

Removing the square root from $a^2-x^2$ changes the answer from arcsine to a logarithm. Factorising gives two linear denominators, which you can integrate using A-level Mathematics partial fractions.

For $a>0$ ,

\frac1{a^2-x^2}=\frac1{2a}\left(\frac1{a+x}+\frac1{a-x}\right)

The second fraction integrates with a negative sign because its denominator has derivative $-1$ .

Reciprocal difference of squares

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

\int\frac{\mathrm{d}x}{a^2-x^2} =\frac1{2a}\ln\left(\frac{a+x}{a-x}\right)+C

In the formula booklet.

The fraction inside the logarithm is positive on this interval. On other intervals avoiding $x=\pm a$ , the same derivation uses $\ln\left|\frac{a+x}{a-x}\right|$ . The absolute-value version extends the displayed logarithmic formula; never integrate across a denominator's zero as an ordinary definite integral.

Worked example 9: the root is absent

Question. Find $\int\frac1{9-x^2}\,\mathrm{d}x$ , for $|x|<3$ .

Plan. Split the difference of squares into linear fractions, integrate, and combine the logarithms.

\begin{aligned} \int\frac{\mathrm{d}x}{9-x^2} &=\frac16\int\left(\frac1{3+x}+\frac1{3-x}\right)\mathrm{d}x && \text{partial fractions}\\ &=\frac16\bigl(\ln(3+x)-\ln(3-x)\bigr)+C && \text{inner derivative gives a minus}\\ &=\frac16\ln\left(\frac{3+x}{3-x}\right)+C && \text{combine logarithms} \end{aligned}

Why a logarithm? Each fraction is the reciprocal of a linear expression; no square root remains to produce arcsine.

Answer. $\frac16\ln((3+x)/(3-x))+C$ .

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

Worked example 10 changes only the coefficient of $x^2$ . Keep the variable-scaling factor as well as the logarithmic formula's factor.

Worked example 10: a scaled difference

Question. Find $\int\frac1{9-4x^2}\,\mathrm{d}x$ , for $|x|<3/2$ .

Plan. Put $u=2x$ , then use the preceding logarithmic integral.

\begin{aligned} \int\frac{\mathrm{d}x}{9-4x^2} &=\frac12\int\frac{\mathrm{d}u}{9-u^2} && \text{scale the differential}\\ &=\frac1{12}\ln\left(\frac{3+u}{3-u}\right)+C && \text{integrate}\\ &=\frac1{12}\ln\left(\frac{3+2x}{3-2x}\right)+C && \text{return to }x \end{aligned}

Why this step? The factor $1/2$ comes from the substitution, independently of the factor $1/6$ in the standard logarithmic result.

Answer. $\frac1{12}\ln((3+2x)/(3-2x))+C$ .

Check. At $x=0$ , differentiation gives $\frac1{12}(2/3+2/3)=1/9$ , matching the integrand.

The opening distinction can now be organised into three decisions.

Denominator	Antiderivative type	Reason
$\sqrt{a^2-x^2}$	Arcsine	Inverse-sine derivative
$a^2+x^2$	Arctangent	Inverse-tangent derivative
$a^2-x^2$	Logarithm	Two linear partial fractions

Use the sign and the square root together; recognising only “a squared expression” is not enough.

Watch out

Wrong: $\int\mathrm{d}x/(9-x^2)=\arcsin(x/3)+C$ . Arcsine differentiates to a reciprocal square root, which is absent here. Right: on $|x|<3$ , the integral is $\frac16\ln((3+x)/(3-x))+C$ .

A derivative check distinguishes these two denominator structures immediately.

Quick Recap

$1/\sqrt{a^2-x^2}$ gives arcsine; $1/(a^2+x^2)$ gives arctangent, with an outside factor $1/a$ .
Use $x=a\sin\theta$ for a difference under a root and $x=a\tan\theta$ for a sum of squares.
Change the differential, control the angle branch, and return to the original variable or change both limits.
$1/(a^2-x^2)$ gives a logarithm through partial fractions; retain interval restrictions.
Differentiate your final expression to check every constant factor.

The opening two integrands differ by a square root: that is precisely why one gives an inverse angle and the other a logarithm.

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CP 5.6 - Integrals giving inverse trigonometric functions

A square root giving arcsine

A sum of squares giving arctangent

Choosing a sine substitution for a difference of squares

Choosing a tangent substitution for a sum of squares

A reciprocal difference of squares gives a logarithm

Quick Recap

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