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Module 15 · L02 of 16 ~40 min ⚡ +90 XP available

Standard Integrals with Inverse Trigonometric Functions

Some integrals look like they'd need a substitution — but the result is already a known function in disguise. The inverse trig family $\sin^{-1}, \tan^{-1}, \sec^{-1}$ are the antiderivatives of three quadratic-denominator patterns you must recognise on sight. This lesson nails the three standard forms, the linear substitution $u = x/a$ that produces them, and how to spot completing-the-square cases.

Today's hook — Differentiate $\tan^{-1}(x/3)$ using the chain rule. Then look at $\displaystyle \int \frac{1}{9 + x^2}\,dx$ — what should the answer be? Now try $\displaystyle \int \frac{1}{4 + x^2}\,dx$ from the same logic. What's the pattern?

0/5QUESTS

Recall — your gut answer first

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From Y12 Advanced you know $\dfrac{d}{dx}\bigl[\tan^{-1} x\bigr] = \dfrac{1}{1 + x^2}$ and $\dfrac{d}{dx}\bigl[\sin^{-1} x\bigr] = \dfrac{1}{\sqrt{1 - x^2}}$. Before checking — what do you predict the derivative of $\tan^{-1}(x/a)$ is, and therefore $\int \dfrac{1}{a^2 + x^2}\,dx$?

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The two moves for inverse-trig integrals

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Two habits handle every inverse-trig integral on the HSC: match the shape of the denominator to one of three templates ($a^2 + x^2$, $\sqrt{a^2 - x^2}$, $x\sqrt{x^2 - a^2}$), then identify $a$ by reading the constant. Anything that doesn't fit instantly usually needs completing the square first.

The shape-and-constant reading: (1) is the denominator a sum, a square-root of a difference, or an $x\sqrt{}$ form? (2) read off $a$ as $\sqrt{\text{constant}}$, (3) apply the matching standard integral verbatim.

$a^2 + x^2 \to \tfrac{1}{a}\tan^{-1}(x/a)$ · $\sqrt{a^2-x^2} \to \sin^{-1}(x/a)$ · $x\sqrt{x^2-a^2} \to \tfrac{1}{a}\sec^{-1}(x/a)$

$\displaystyle \int \frac{dx}{a^2+x^2} = \frac{1}{a}\tan^{-1}\!\frac{x}{a} + C$

Sum of squares → arctan

$a^2 + x^2$ in the denominator (no square root, no $x$ factor) is always arctan. Watch the leading $\tfrac{1}{a}$ — students lose marks by writing $\tan^{-1}(x/a) + C$ without it.

Square-root of a difference → arcsin

$\sqrt{a^2 - x^2}$ in the denominator (square root, difference) is arcsin. No leading $\tfrac{1}{a}$ here — the answer is $\sin^{-1}(x/a) + C$ outright.

$x\sqrt{x^2-a^2}$ → arcsec

An $x$ factor outside the root, with $x^2 - a^2$ inside, gives $\tfrac{1}{a}\sec^{-1}(x/a) + C$. Less common in HSC but worth recognising.

What you'll master

Know

Key facts

$\int \dfrac{dx}{a^2 + x^2} = \tfrac{1}{a}\tan^{-1}(x/a) + C$
$\int \dfrac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}(x/a) + C$ (for $|x| < a$)
$\int \dfrac{dx}{x\sqrt{x^2 - a^2}} = \tfrac{1}{a}\sec^{-1}(x/a) + C$ (for $|x| > a$)
Underlying substitution $u = x/a$ derives each form from the unit-$a$ case

Understand

Concepts

Why a "sum of squares vs difference under a root" diagnostic distinguishes arctan from arcsin
Why the $\tfrac{1}{a}$ factor appears for arctan and arcsec but not arcsin
How completing the square reduces $ax^2 + bx + c$ denominators to one of the three forms

Can do

Skills

Recognise and integrate the three standard inverse-trig forms by inspection
Identify $a$ correctly when the constant is not a perfect square
Set up the substitution $u = x/a$ when asked to derive (not just quote)

Key terms

Standard integralAn integral whose antiderivative is listed in the NESA reference sheet — to be quoted, not re-derived, in exams (unless the question says "show").

Inverse tangent ($\tan^{-1}$)The function with $\dfrac{d}{dx}\tan^{-1} x = \dfrac{1}{1 + x^2}$. Antiderivative pattern for $\dfrac{1}{a^2 + x^2}$.

Inverse sine ($\sin^{-1}$)The function with $\dfrac{d}{dx}\sin^{-1} x = \dfrac{1}{\sqrt{1 - x^2}}$. Antiderivative pattern for $\dfrac{1}{\sqrt{a^2 - x^2}}$.

Inverse secant ($\sec^{-1}$)The function with $\dfrac{d}{dx}\sec^{-1} x = \dfrac{1}{|x|\sqrt{x^2 - 1}}$. Antiderivative pattern for $\dfrac{1}{x\sqrt{x^2 - a^2}}$.

Completing the squareRewriting $ax^2 + bx + c$ as $a\bigl((x + \tfrac{b}{2a})^2 + k\bigr)$ to expose a standard inverse-trig form.

Linear substitution $u = x/a$The substitution that reduces $a^2 \pm x^2$ to $a^2(1 \pm u^2)$, exposing the unit-$a$ standard form. Source of the $\tfrac{1}{a}$ factor.

MEX-C1NESA outcome (Further Integration): integrates functions using the standard integrals listed in the reference sheet, including inverse trigonometric forms.

The three standard inverse-trig integrals

core concept

Three quadratic-denominator shapes have inverse-trig antiderivatives. Memorise them by feature, not by formula alone:

$$\int \frac{dx}{a^2 + x^2} = \frac{1}{a}\tan^{-1}\!\frac{x}{a} + C$$

$$\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\!\frac{x}{a} + C \quad (|x| < a)$$

$$\int \frac{dx}{x\sqrt{x^2 - a^2}} = \frac{1}{a}\sec^{-1}\!\frac{x}{a} + C \quad (|x| > a)$$

Why $u = x/a$ derives all three. Take the arctan case. Set $u = x/a$, so $x = au$ and $dx = a\,du$. Then $a^2 + x^2 = a^2(1 + u^2)$, and

$\displaystyle \int \frac{dx}{a^2 + x^2} = \int \frac{a\,du}{a^2(1 + u^2)} = \frac{1}{a}\int \frac{du}{1 + u^2} = \frac{1}{a}\tan^{-1} u + C = \frac{1}{a}\tan^{-1}\!\frac{x}{a} + C$.

The same substitution gives the arcsin form (with one less $\tfrac{1}{a}$ because $\sqrt{a^2} = a$ cancels the $a\,du$).

Where the $\tfrac{1}{a}$ comes from. Arctan: $\tfrac{a\,du}{a^2(1+u^2)} = \tfrac{1}{a} \cdot \tfrac{du}{1+u^2}$. Arcsin: $\tfrac{a\,du}{\sqrt{a^2(1-u^2)}} = \tfrac{a\,du}{a\sqrt{1-u^2}} = \tfrac{du}{\sqrt{1-u^2}}$ — the $a$'s cancel, no leading factor.

$\int \tfrac{dx}{a^2+x^2} = \tfrac{1}{a}\tan^{-1}(x/a) + C$ · $\int \tfrac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}(x/a) + C$ · $\int \tfrac{dx}{x\sqrt{x^2-a^2}} = \tfrac{1}{a}\sec^{-1}(x/a) + C$ · Substitution $u = x/a$ derives all three; $a$-factors give the leading $\tfrac{1}{a}$ for arctan and arcsec

Pause — copy the three standard inverse-trig integrals (arctan, arcsin, arcsec forms) with the $1/a$ factor and the derivation sketch ($u = x/a$) into your book.

Quick check: Which standard integral does $\displaystyle \int \frac{dx}{\sqrt{25 - x^2}}$ match?

Spotting the form — completing the square

core concept

We just saw the three standard inverse-trig integrals: $\int dx/(a^2+x^2) = \tfrac{1}{a}\tan^{-1}(x/a)+C$, $\int dx/\sqrt{a^2-x^2} = \sin^{-1}(x/a)+C$, and the arcsec form, all derived by $u = x/a$. That raises a question: what if the quadratic in the denominator has a linear term? This card answers it → complete the square first: $x^2+bx+c = (x+b/2)^2 + (c-b^2/4)$, then substitute $w = x+b/2$ to match the standard arctan form.

Sometimes the denominator hides a standard form behind a general quadratic. The fix: complete the square, then identify $a$.

Example. $\displaystyle \int \frac{dx}{x^2 + 4x + 13}$. The denominator is $x^2 + 4x + 13 = (x+2)^2 + 9$. With the substitution $w = x + 2$ (so $dw = dx$),

$\displaystyle \int \frac{dw}{9 + w^2} = \tfrac{1}{3}\tan^{-1}(w/3) + C = \tfrac{1}{3}\tan^{-1}\!\Bigl(\tfrac{x+2}{3}\Bigr) + C$.

The diagnostic: $b^2 - 4ac < 0$ in the quadratic denominator (no real roots) signals arctan. A denominator $\sqrt{-(x-h)^2 + k^2}$ signals arcsin.

$$x^2 + bx + c \;=\; \Bigl(x + \tfrac{b}{2}\Bigr)^2 + \Bigl(c - \tfrac{b^2}{4}\Bigr).$$

Watch the sign. If $c - \tfrac{b^2}{4} > 0$ the expression is a sum of squares — arctan territory. If $c - \tfrac{b^2}{4} < 0$ the quadratic factors over the reals — that's partial fractions (next lesson), not arctan.

Complete the square: $x^2 + bx + c = (x + b/2)^2 + (c - b^2/4)$ · Sum of squares + no real roots → arctan form after $w = x + b/2$ · Real roots ($b^2 - 4c > 0$) → not inverse-trig; use partial fractions · Example: $\int dx/(x^2+4x+13) = \tfrac{1}{3}\tan^{-1}((x+2)/3) + C$

Pause — copy the completing-the-square formula $(x+b/2)^2 + (c-b^2/4)$, the sum-of-squares → arctan rule, the real-roots → partial-fractions rule, and the worked example $\int dx/(x^2+4x+13) = \tfrac{1}{3}\tan^{-1}((x+2)/3)+C$ into your book.

Did you get this? True or false: $\displaystyle \int \frac{dx}{x^2 + 2x + 5}$ requires partial fractions because the denominator is a general quadratic.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · ARCTAN FORM

Evaluate $\displaystyle \int \frac{dx}{9 + x^2}$.

Identify the shape: $9 + x^2 = 3^2 + x^2$. Sum of squares → arctan form with $a = 3$.

No square root, no $x$ factor — this is the cleanest arctan case. Reading $a^2 = 9 \Rightarrow a = 3$ takes one second.

PROBLEM 2 · ARCSIN FORM

Evaluate $\displaystyle \int \frac{dx}{\sqrt{16 - x^2}}$ for $|x| < 4$.

Identify the shape: $\sqrt{16 - x^2} = \sqrt{4^2 - x^2}$. Square root of a difference → arcsin form with $a = 4$.

Reading the constant: $a^2 = 16 \Rightarrow a = 4$. The domain $|x| < a = 4$ ensures the radicand is positive.

PROBLEM 3 · ARCSEC FORM

Evaluate $\displaystyle \int \frac{dx}{x\sqrt{x^2 - 4}}$ for $x > 2$.

Identify the shape: an $x$ factor outside, with $x^2 - 4 = x^2 - 2^2$ inside the root → arcsec form with $a = 2$.

The diagnostic feature is the $x$ multiplying the radical. Reading $a^2 = 4 \Rightarrow a = 2$. Domain $x > 2$ keeps the radicand positive.

Fill the gap: $\displaystyle \int \frac{dx}{49 + x^2} = $ $\tan^{-1}\!\left(\dfrac{x}{\,}\right.$$\!\!\left.\right) + C$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Confusing arctan and arcsin shapes

$a^2 + x^2$ (no root, no $x$ factor) is arctan. $\sqrt{a^2 - x^2}$ (root, difference) is arcsin. Mixing them up writes the wrong function and gets zero. Always check: is there a square root over the denominator?

Trap 02

Forgetting the $\tfrac{1}{a}$ factor

Arctan and arcsec both carry a leading $\tfrac{1}{a}$ outside. Writing $\tan^{-1}(x/3) + C$ instead of $\tfrac{1}{3}\tan^{-1}(x/3) + C$ loses a mark. Differentiate your answer to check it gives back the integrand.

Trap 03

Misreading $a$ from the constant

In $\int \tfrac{dx}{\sqrt{25 - x^2}}$, $a = 5$ (not 25). The constant is $a^2$, so take its square root. Students who write $\sin^{-1}(x/25)$ have skipped this step.

Did you get this? True or false: $\displaystyle \int \frac{dx}{4 + x^2} = \tan^{-1}(x/2) + C$ (without a leading $\tfrac{1}{2}$).

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Evaluate $\displaystyle \int \frac{dx}{4 + x^2}$.

Evaluate $\displaystyle \int \frac{dx}{\sqrt{9 - x^2}}$ for $|x| < 3$.

Evaluate $\displaystyle \int \frac{dx}{x\sqrt{x^2 - 9}}$ for $x > 3$.

Evaluate $\displaystyle \int_0^1 \frac{dx}{1 + x^2}$ using the standard integral.

Complete the square to evaluate $\displaystyle \int \frac{dx}{x^2 + 2x + 5}$.

Odd one out: Three of these integrals reduce to a single inverse-trig standard form. Which one does NOT?

Revisit your thinking

Earlier you differentiated $\tan^{-1}(x/3)$ and predicted $\int \tfrac{dx}{9+x^2}$ and $\int \tfrac{dx}{4+x^2}$.

The first is $\tfrac{1}{3}\tan^{-1}(x/3) + C$; the second is $\tfrac{1}{2}\tan^{-1}(x/2) + C$. The pattern: for $\int \tfrac{dx}{a^2 + x^2}$ the constant under the antiderivative is $\tfrac{1}{a}$ — coming from the chain rule's $\tfrac{1}{a}$ multiplier when you differentiate $\tan^{-1}(x/a)$. Two checks every time: (1) the constant inside is $a$, not $a^2$, and (2) the leading $\tfrac{1}{a}$ outside is mandatory for arctan and arcsec, absent for arcsin.

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Multiple choice

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Short answer

ApplyBand 32 marks

Q1. Evaluate $\displaystyle \int \frac{dx}{25 + x^2}$. (2 marks)

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ApplyBand 43 marks

Q2. Evaluate $\displaystyle \int_0^{\sqrt{3}} \frac{dx}{\sqrt{4 - x^2}}$. (3 marks)

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AnalyseBand 53 marks

Q3. By completing the square, evaluate $\displaystyle \int \frac{dx}{x^2 + 6x + 13}$. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $a^2 = 4 \Rightarrow a = 2$. $\int \tfrac{dx}{4+x^2} = \tfrac{1}{2}\tan^{-1}(x/2) + C$.

2. $a^2 = 9 \Rightarrow a = 3$. $\int \tfrac{dx}{\sqrt{9-x^2}} = \sin^{-1}(x/3) + C$ (no leading $\tfrac{1}{a}$).

3. $a^2 = 9 \Rightarrow a = 3$. $\int \tfrac{dx}{x\sqrt{x^2-9}} = \tfrac{1}{3}\sec^{-1}(x/3) + C$.

4. $\int_0^1 \tfrac{dx}{1+x^2} = [\tan^{-1} x]_0^1 = \tan^{-1}(1) - \tan^{-1}(0) = \tfrac{\pi}{4} - 0 = \tfrac{\pi}{4}$.

5. $x^2 + 2x + 5 = (x+1)^2 + 4$. With $w = x+1$: $\int \tfrac{dw}{4+w^2} = \tfrac{1}{2}\tan^{-1}(w/2) + C = \tfrac{1}{2}\tan^{-1}\!\bigl(\tfrac{x+1}{2}\bigr) + C$.

Q1 (2 marks): Identify arctan with $a = 5$ [1]. Apply: $\tfrac{1}{5}\tan^{-1}(x/5) + C$ [1].

Q2 (3 marks): Arcsin form with $a = 2$ [1]. Antiderivative: $\sin^{-1}(x/2)$ [1]. Evaluate: $\sin^{-1}(\sqrt{3}/2) - \sin^{-1}(0) = \tfrac{\pi}{3}$ [1].

Q3 (3 marks): Complete the square: $x^2 + 6x + 13 = (x+3)^2 + 4$ [1]. Substitute $w = x + 3$: $\int \tfrac{dw}{4 + w^2}$ [1]. Apply arctan with $a = 2$: $\tfrac{1}{2}\tan^{-1}\!\bigl(\tfrac{x+3}{2}\bigr) + C$ [1].

Boss battle · The Inverse Trig Inquisitor

earn bronze · silver · gold

Five timed questions on shape-spotting, $a$-reading, and completing-the-square forms. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms by answering quick inverse-trig recognition questions. Lighter alternative to the boss.

Mark lesson as complete

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Module overview · Extension 2