What is completing the square?

When the quadratic is not yet in a^2 (linear)^2 form, complete the square first. For example 1x^2 + 4x + 13 = 1(x + 2)^2 + 9, which then matches the arctan form with a = 3 and a shift x → x + 2.

What are integrals leading to logarithms?

A fraction whose numerator is the derivative of the denominator integrates to a logarithm: f'(x)f(x)\,dx = ln|f(x)| + C. Spotting this pattern (or engineering it by adjusting a constant) avoids unnecessary substitution.

What is sign of the surd?

sqrt(a^2 - a^2sin^2θ) = acosθ only when cosθ ≥ 0; keep θ in the principal range.

Write (1)/(sqrt(9 - x^2))\,dx. [1 mark]

Which substitution suits sqrt(1 + x^2)\,dx? [1 mark]

Complete the square for the denominator of (1)/(x^2 - 2x + 5)\,dx. [1 mark]

SEAB Original-style practice: Find (1)/(sqrt(4 - x^2))\,dx and (1)/(9 + x^2)\,dx.

The first matches the standard form (1)/(sqrt(a^2 - x^2))\,dx = (x)/(a) + C with a = 2: $int (1)/(sqrt(4 - x^2))\,dx = (x)/(2) + C.$ The second matches (1)/(a^2 + x^2)\,dx = (1)/(a)(x)/(a) + C with a = 3: $int (1)/(9 + x^2)\,dx = (1)/(3)(x)/(3) + C.$ Markers reward recognising each standard form, identifying a = 2 and a = 3, and the correct inverse-trig results with the constant.

SEAB Original-style practice: Using the substitution x = 2sinθ, evaluate _0^1 sqrt(4 - x^2)\,dx.

With x = 2sinθ, dx = 2cosθ\,dθ and sqrt(4 - x^2) = sqrt(4 - 4sin^2θ) = 2cosθ (taking the positive root for -π2 ≤ θ ≤ π2). Change the limits: x = 0 θ = 0; x = 1 sinθ = 12 θ = π6. $_0^1sqrt(4 - x^2)\,dx = _0^π/6 2cosθ· 2cosθ\,dθ = 4_0^π/6cos^2θ\,dθ.$ Use cos^2θ = 12(1 + cos 2θ): $= 4·12[θ + 12sin 2θ]_0^π/6 = 2(π6 + 12sinπ3) = 2(π6 + sqrt(3)4) = π3 + sqrt(3)2.$ Markers reward the substitution and sqrt(4 - x^2) = 2cosθ, changing the limits, the double-angle reduction, and the exact value π3 + sqrt(3)2.

SingaporeFurther MathsSyllabus dot point

How do trigonometric and hyperbolic substitutions and standard inverse-function integrals extend our integration toolkit?

Integrate using trigonometric and hyperbolic substitutions and recognise standard integrals giving inverse trigonometric and logarithmic forms

A focused answer to the H2 Further Mathematics outcome on further integration. Standard integrals giving inverse sine and inverse tangent, completing the square, trigonometric and hyperbolic substitutions, and integrals leading to logarithms.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-06

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

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What this dot point is asking

SEAB wants you to extend integration to forms that produce inverse trigonometric functions and logarithms, to recognise the standard integrals, to complete the square when the quadratic is not in standard form, and to use trigonometric and hyperbolic substitutions to handle expressions involving $\sqrt{a^2 - x^2}$ , $\sqrt{a^2 + x^2}$ and $\sqrt{x^2 - a^2}$ .

The answer

Standard integrals giving inverse trig

Two results to recognise on sight:

\int \frac{1}{\sqrt{a^2 - x^2}}\,dx = \arcsin\frac{x}{a} + C, \qquad \int \frac{1}{a^2 + x^2}\,dx = \frac{1}{a}\arctan\frac{x}{a} + C.

Completing the square

When the quadratic is not yet in $a^2 \pm (\text{linear})^2$ form, complete the square first. For example $\dfrac{1}{x^2 + 4x + 13} = \dfrac{1}{(x + 2)^2 + 9}$ , which then matches the arctan form with $a = 3$ and a shift $x \to x + 2$ .

Trigonometric substitutions

The substitution is chosen to make the surd a single trig function via a Pythagorean identity:

$\sqrt{a^2 - x^2}$ : substitute $x = a\sin\theta$ , so $\sqrt{a^2 - x^2} = a\cos\theta$ ;
$\sqrt{a^2 + x^2}$ : substitute $x = a\tan\theta$ , so $\sqrt{a^2 + x^2} = a\sec\theta$ .

After substituting, change the limits to $\theta$ -values for a definite integral and reduce any $\cos^2\theta$ or $\sec^2\theta$ with a double-angle identity.

Hyperbolic substitutions

For $\sqrt{a^2 + x^2}$ and $\sqrt{x^2 - a^2}$ , hyperbolic substitutions $x = a\sinh u$ and $x = a\cosh u$ exploit $\cosh^2 u - \sinh^2 u = 1$ , often leading to a logarithmic form for the result. These give the inverse hyperbolic functions, which can be written as logarithms.

Integrals leading to logarithms

A fraction whose numerator is the derivative of the denominator integrates to a logarithm: $\displaystyle\int \frac{\mathrm{f}'(x)}{\mathrm{f}(x)}\,dx = \ln|\mathrm{f}(x)| + C$ . Spotting this pattern (or engineering it by adjusting a constant) avoids unnecessary substitution.

Worked example

Find $\displaystyle\int \frac{1}{x^2 + 6x + 25}\,dx$ .

Step 1: Complete the square in the denominator

x^2 + 6x + 25 = (x + 3)^2 + 16.

Step 2: Recognise the arctan form

The integral is $\displaystyle\int \frac{1}{(x + 3)^2 + 16}\,dx$ , which matches $\int \dfrac{1}{u^2 + a^2}\,du$ with $u = x + 3$ and $a = 4$ .

Step 3: Apply the standard result

\int \frac{1}{u^2 + 4^2}\,du = \frac{1}{4}\arctan\frac{u}{4} + C.

Step 4: Revert to x

\int \frac{1}{x^2 + 6x + 25}\,dx = \frac{1}{4}\arctan\frac{x + 3}{4} + C.

Step 5: Sanity check by differentiation

Differentiating $\tfrac{1}{4}\arctan\tfrac{x+3}{4}$ returns $\dfrac{1}{4}\cdot\dfrac{1/4}{1 + ((x+3)/4)^2} = \dfrac{1}{16 + (x+3)^2}$ , confirming the result.

Common traps

Not completing the square first: A general quadratic denominator must be reduced to $a^2 \pm (\text{shifted})^2$ before the standard forms apply.
Forgetting the $\tfrac{1}{a}$ in the arctan result: $\int \dfrac{1}{a^2 + x^2}\,dx = \dfrac{1}{a}\arctan\dfrac{x}{a}$ ; the factor $\tfrac{1}{a}$ is easy to drop.
Leaving the limits in $x$ after substituting: For a definite integral, change the limits to the new variable, or revert fully before substituting numbers.
Choosing the wrong substitution: Using $x = a\tan\theta$ for $\sqrt{a^2 - x^2}$ leads nowhere; match the surd to its identity.
Sign of the surd: $\sqrt{a^2 - a^2\sin^2\theta} = a\cos\theta$ only when $\cos\theta \geq 0$ ; keep $\theta$ in the principal range.

Examples in context

Example 1. Arc length and area of a circle. Integrating $\sqrt{a^2 - x^2}$ with $x = a\sin\theta$ recovers the area of a circular segment and ultimately $\pi a^2$ , showing how the trig substitution underlies the geometry of the circle.

Example 2. Time in an oscillator. Inverting the energy equation of a simple oscillator leads to an integral of the form $\int \dfrac{dx}{\sqrt{a^2 - x^2}}$ , whose arcsine result gives the sinusoidal time dependence, connecting integration directly to oscillatory motion.

Try this

Q1. Write $\displaystyle\int \frac{1}{\sqrt{9 - x^2}}\,dx$ . [1 mark]

Cue. $\arcsin\dfrac{x}{3} + C$ .

Q2. Which substitution suits $\displaystyle\int \sqrt{1 + x^2}\,dx$ ? [1 mark]

Cue. $x = \tan\theta$ (so $\sqrt{1 + x^2} = \sec\theta$ ), or the hyperbolic $x = \sinh u$ .

Q3. Complete the square for the denominator of $\displaystyle\int \frac{1}{x^2 - 2x + 5}\,dx$ . [1 mark]

Cue. $(x - 1)^2 + 4$ , giving an arctan form with $a = 2$ .

Exam-style practice questions

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

Original5 marksFind

\displaystyle\int \frac{1}{\sqrt{4 - x^2}}\,dx

and

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

Show worked answer →

The first matches the standard form $\displaystyle\int \frac{1}{\sqrt{a^2 - x^2}}\,dx = \arcsin\frac{x}{a} + C$ with $a = 2$ :

\int \frac{1}{\sqrt{4 - x^2}}\,dx = \arcsin\frac{x}{2} + C.

The second matches

\displaystyle\int \frac{1}{a^2 + x^2}\,dx = \frac{1}{a}\arctan\frac{x}{a} + C

with

a = 3

\int \frac{1}{9 + x^2}\,dx = \frac{1}{3}\arctan\frac{x}{3} + C.

Markers reward recognising each standard form, identifying $a = 2$ and $a = 3$ , and the correct inverse-trig results with the constant.

Original6 marksUsing the substitution

x = 2\sin\theta

, evaluate

\displaystyle\int_0^{1} \sqrt{4 - x^2}\,dx

Show worked answer →

With $x = 2\sin\theta$ , $dx = 2\cos\theta\,d\theta$ and $\sqrt{4 - x^2} = \sqrt{4 - 4\sin^2\theta} = 2\cos\theta$ (taking the positive root for $-\tfrac{\pi}{2} \leq \theta \leq \tfrac{\pi}{2}$ ).

Change the limits: $x = 0 \Rightarrow \theta = 0$ ; $x = 1 \Rightarrow \sin\theta = \tfrac{1}{2} \Rightarrow \theta = \tfrac{\pi}{6}$ .

\int_0^{1}\sqrt{4 - x^2}\,dx = \int_0^{\pi/6} 2\cos\theta\cdot 2\cos\theta\,d\theta = 4\int_0^{\pi/6}\cos^2\theta\,d\theta.

Use

\cos^2\theta = \tfrac{1}{2}(1 + \cos 2\theta)

= 4\cdot\tfrac{1}{2}\left[\theta + \tfrac{1}{2}\sin 2\theta\right]_0^{\pi/6} = 2\left(\tfrac{\pi}{6} + \tfrac{1}{2}\sin\tfrac{\pi}{3}\right) = 2\left(\tfrac{\pi}{6} + \tfrac{\sqrt{3}}{4}\right) = \tfrac{\pi}{3} + \tfrac{\sqrt{3}}{2}.

Markers reward the substitution and $\sqrt{4 - x^2} = 2\cos\theta$ , changing the limits, the double-angle reduction, and the exact value $\tfrac{\pi}{3} + \tfrac{\sqrt{3}}{2}$ .

What this dot point is asking

The answer

Standard integrals giving inverse trig

Completing the square

Trigonometric substitutions

Hyperbolic substitutions

Integrals leading to logarithms

Examples in context

Try this

Exam-style practice questions

Related dot points