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Module 8 · L13 of 20 ~30 min ⚡ +90 XP available

Integrating $\dfrac{1}{\sqrt{a^2-x^2}}$

The inverse sine function hides inside every integral of the form $\displaystyle\int \dfrac{1}{\sqrt{a^2-x^2}}\,dx$. Once you see why the derivative of $\sin^{-1}\!\left(\tfrac{x}{a}\right)$ produces exactly that denominator, the formula becomes something you derive — not just memorise. In this lesson you'll unpack the proof, recognise the pattern at a glance, and handle definite integrals confidently.

Today's hook — Before using any formula, write down what $\dfrac{d}{dx}\!\left[\sin^{-1}\!\left(\dfrac{x}{a}\right)\right]$ equals. If you can recall that, you already know the integral. Jot it now — you'll check it after card 05.

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Recall — your gut answer first

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From last lesson (Chain Rule with Inverse Trig), recall: what is $\dfrac{d}{dx}\!\left[\sin^{-1}\!\left(\dfrac{x}{a}\right)\right]$? Write it below without looking it up. Even a rough guess is valuable.

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The key idea: reversing a derivative

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Integration is the reverse of differentiation. We already know (from Lesson 12) that:

$$\frac{d}{dx}\!\left[\sin^{-1}\!\left(\frac{x}{a}\right)\right] = \frac{1}{\sqrt{a^2 - x^2}}$$

So reversing this derivative immediately gives us the integral:

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

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

$a$ must be positive

The formula requires $a > 0$ and $|x| < a$ so the square root is defined and real.

Special case $a = 1$

When $a = 1$: $\displaystyle\int \dfrac{dx}{\sqrt{1-x^2}} = \sin^{-1} x + C$. Recognise this as a special case.

Don't forget $+C$

Every indefinite integral needs a constant of integration. A missing $+C$ costs a mark in the HSC.

What you'll master

Know

Key facts

$\displaystyle\int \dfrac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\!\!\left(\dfrac{x}{a}\right)+C$ is a standard integral
The formula holds when $a > 0$ and $|x| < a$
This result follows directly from differentiating $\sin^{-1}(x/a)$

Understand

Concepts

Why the denominator $\sqrt{a^2-x^2}$ signals arcsine
How to spot and rewrite integrands to match the standard form
Why the domain restriction $|x| < a$ matters for definite integrals

Can do

Skills

Evaluate indefinite integrals of the form $\dfrac{k}{\sqrt{a^2-x^2}}$
Identify and extract the constant $a$ from the integrand
Compute definite integrals using the arcsine primitive

Key terms

Standard integralA pre-proved integral result listed in the HSC reference sheet. Recognise the form, apply directly — no need to re-derive in an exam.

$\sin^{-1}(x/a)$The inverse sine (arcsine) of $x/a$. Output is an angle in $[-\pi/2, \pi/2]$. Also written $\arcsin(x/a)$.

$a$ (scale parameter)A positive constant that determines the width of the domain. It appears under the square root and in the denominator of the arcsine argument.

Indefinite integralIntegration without limits; always includes a constant of integration $+C$.

Domain restriction$|x| < a$ is required so $a^2 - x^2 > 0$ and the square root is real and positive.

Primitive functionAnother name for antiderivative. If $F'(x) = f(x)$ then $F$ is a primitive of $f$.

The standard integral — proof and form

core concept

Proof from first principles. Apply the chain rule to $F(x) = \sin^{-1}\!\left(\dfrac{x}{a}\right)$:

$$F'(x) = \frac{1}{\sqrt{1 - (x/a)^2}} \cdot \frac{1}{a} = \frac{1}{\sqrt{1 - x^2/a^2}} \cdot \frac{1}{a}$$

Multiply numerator and denominator by $a$ to clear the compound fraction under the root:

$$F'(x) = \frac{1}{a\sqrt{(a^2 - x^2)/a^2}} = \frac{1}{a \cdot \frac{1}{a}\sqrt{a^2 - x^2}} = \frac{1}{\sqrt{a^2 - x^2}}$$

Since $F'(x) = \dfrac{1}{\sqrt{a^2-x^2}}$, reversing gives the standard integral:

$$\boxed{\int \frac{1}{\sqrt{a^2 - x^2}}\,dx = \sin^{-1}\!\left(\frac{x}{a}\right) + C, \quad a > 0}$$

HSC reference sheet. This integral is listed in the standard integrals table you receive in the exam. However, understanding the proof means you can reconstruct it from memory and adapt it — essential when the integrand is scaled or shifted.

$\int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}(x/a)+C$. Proof: $\frac{d}{dx}\sin^{-1}(x/a)=\frac{1}{a}\cdot\frac{1}{\sqrt{1-x^2/a^2}}=\frac{1}{\sqrt{a^2-x^2}}$.

Pause — copy the standard integral $\int\frac{1}{\sqrt{a^2-x^2}}\,dx=\sin^{-1}(x/a)+C$ with the proof via $\frac{d}{dx}\sin^{-1}(x/a)=\frac{1}{\sqrt{a^2-x^2}}$ into your book.

Quick check: What is $\displaystyle\int \dfrac{1}{\sqrt{9 - x^2}}\,dx$?

Spotting the form — reading the integrand

core concept

We just saw the proof: $\frac{d}{dx}\sin^{-1}(x/a)=\frac{1}{\sqrt{a^2-x^2}}$, so $\int\frac{1}{\sqrt{a^2-x^2}}\,dx=\sin^{-1}(x/a)+C$. That raises a question: before applying this formula, how do you recognise whether an integrand is in the standard form $\frac{1}{\sqrt{a^2-x^2}}$, and what algebraic manipulation reveals the value of $a$? This card answers it → factor the radicand as $a^2-x^2$; take $a=\sqrt{\text{constant term}}$; check $x^2$ has coefficient $-1$ after factoring.

The standard form is $\dfrac{1}{\sqrt{a^2 - x^2}}$. To apply it you need to:

Identify $a^2$: The constant term under the root is $a^2$. Extract $a = \sqrt{\text{constant}}$.
Check the coefficient of $x^2$: It must be $-1$ (i.e., $-x^2$, not $-4x^2$). If it is not, factor it out first.
Handle a scalar multiplier: A constant $k$ in front of the fraction comes straight out: $\displaystyle\int \dfrac{k}{\sqrt{a^2-x^2}}\,dx = k\sin^{-1}\!\!\left(\dfrac{x}{a}\right)+C$.

Example — reading $a$: For $\displaystyle\int \dfrac{1}{\sqrt{16 - x^2}}\,dx$, we have $a^2 = 16$ so $a = 4$. Answer: $\sin^{-1}\!\!\left(\dfrac{x}{4}\right)+C$.

Example — scalar multiple: $\displaystyle\int \dfrac{3}{\sqrt{25 - x^2}}\,dx = 3\sin^{-1}\!\!\left(\dfrac{x}{5}\right)+C$.

Pattern recognition tip. The denominator always contains a difference of a constant squared and $x^2$ under a square root. If you see a sum ($a^2 + x^2$) instead, that signals $\tan^{-1}$ — not arcsine.

The standard form is $\dfrac{1}{\sqrt{a^2 - x^2}}$. To apply it you need to:

Pause — copy the form-recognition steps: (1) check for $\sqrt{a^2-x^2}$ in denominator; (2) read off $a$; (3) apply $\sin^{-1}(x/a)+C$ into your book.

Did you get this? True or false: $\displaystyle\int \dfrac{5}{\sqrt{4 - x^2}}\,dx = 5\sin^{-1}\!\!\left(\dfrac{x}{2}\right)+C$.

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

PROBLEM 1 · INDEFINITE INTEGRAL

Find $\displaystyle\int \dfrac{1}{\sqrt{25 - x^2}}\,dx$.

Identify the form: $\sqrt{a^2 - x^2}$ with $a^2 = 25$, so $a = 5$.

The integrand matches the standard form $\dfrac{1}{\sqrt{a^2 - x^2}}$ with $a = 5 > 0$.

PROBLEM 2 · SCALAR MULTIPLE

Find $\displaystyle\int \dfrac{4}{\sqrt{7 - x^2}}\,dx$.

Factor the constant: $\displaystyle\int \dfrac{4}{\sqrt{7-x^2}}\,dx = 4\int \dfrac{1}{\sqrt{7-x^2}}\,dx$.

Constants factor out of integrals. This isolates the standard form with $a^2 = 7$, $a = \sqrt{7}$.

PROBLEM 3 · DEFINITE INTEGRAL

Evaluate $\displaystyle\int_{0}^{3} \dfrac{1}{\sqrt{9 - x^2}}\,dx$.

Identify $a = 3$. Find the primitive: $F(x) = \sin^{-1}\!\!\left(\dfrac{x}{3}\right)$.

Standard form with $a^2 = 9$. For a definite integral, find the antiderivative first, then apply limits.

Fill the gap: $\displaystyle\int \dfrac{1}{\sqrt{16-x^2}}\,dx = \sin^{-1}\!\!\left(\dfrac{x}{\,\,}\right)+C$. The denominator inside the arcsine should be .

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Writing $\sin^{-1}(ax)$ instead of $\sin^{-1}(x/a)$

The argument of arcsine is $x/a$, not $ax$. For $\sqrt{9-x^2}$: $a = 3$, so the answer is $\sin^{-1}(x/3)$. Writing $\sin^{-1}(3x)$ is wrong — always check by differentiating your answer.

Trap 02

Confusing $\sqrt{a^2 - x^2}$ with $\sqrt{a^2 + x^2}$

A difference under the root ($a^2 - x^2$) means arcsine. A sum ($a^2 + x^2$) in the denominator (without a square root) signals arctangent — a completely different integral covered in Lesson 14.

Trap 03

Forgetting to check domain for definite integrals

For $\displaystyle\int_a^b \dfrac{dx}{\sqrt{a^2-x^2}}$, the integrand is only defined for $|x| < a$. If either limit equals $\pm a$ the integral is improper (but still convergent for this form). Always verify the limits are valid.

Did you get this? True or false: for $\displaystyle\int \dfrac{dx}{\sqrt{36 - x^2}}$, the correct primitive is $\sin^{-1}\!\!\left(\dfrac{x}{6}\right)+C$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Find $\displaystyle\int \dfrac{1}{\sqrt{49 - x^2}}\,dx$.

Find $\displaystyle\int \dfrac{2}{\sqrt{1 - x^2}}\,dx$.

Evaluate $\displaystyle\int_{0}^{2} \dfrac{1}{\sqrt{4 - x^2}}\,dx$. Give an exact answer.

Verify that $\dfrac{d}{dx}\sin^{-1}\!\!\left(\dfrac{x}{5}\right) = \dfrac{1}{\sqrt{25-x^2}}$ by differentiating using the chain rule.

Find $\displaystyle\int \dfrac{3}{\sqrt{a^2 - x^2}}\,dx$ in terms of $a$. What is the domain restriction on $x$?

Odd one out: Three of these integrals evaluate to an arcsine expression. Which one does NOT?

Revisit your thinking

Earlier you wrote down what you thought $\dfrac{d}{dx}\sin^{-1}(x/a)$ equals.

The exact answer is $\dfrac{1}{\sqrt{a^2-x^2}}$. This is the integrand we've been working with all lesson — so once you know this derivative, the integral follows immediately by reversal. The key insight: the denominator signals the inverse function. A difference under a square root means arcsine; a sum without a root means arctangent (next lesson).

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Multiple choice

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Short answer

ApplyBand 31 mark

Q1. Find $\displaystyle\int \dfrac{1}{\sqrt{9-x^2}}\,dx$. (1 mark)

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ApplyBand 42 marks

Q2. Evaluate $\displaystyle\int_{0}^{4} \dfrac{1}{\sqrt{16-x^2}}\,dx$. Give an exact answer. (2 marks)

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AnalyseBand 52 marks

Q3. Show that $\dfrac{d}{dx}\sin^{-1}\!\!\left(\dfrac{x}{a}\right) = \dfrac{1}{\sqrt{a^2-x^2}}$ using the chain rule. Hence write down $\displaystyle\int \dfrac{3}{\sqrt{a^2-x^2}}\,dx$. (2 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $\sin^{-1}(x/7)+C$ · 2. $2\sin^{-1}(x)+C$ (since $a=1$) · 3. $[\sin^{-1}(x/2)]_0^2 = \sin^{-1}(1) - \sin^{-1}(0) = \pi/2$ · 4. Chain rule: $\dfrac{1/5}{\sqrt{1-(x/5)^2}} = \dfrac{1}{\sqrt{25-x^2}}$ · 5. $3\sin^{-1}(x/a)+C$, domain $|x| < a$

Q1 (1 mark): $a = 3$; $\displaystyle\int \dfrac{1}{\sqrt{9-x^2}}\,dx = \sin^{-1}\!\!\left(\dfrac{x}{3}\right)+C$ [1].

Q2 (2 marks): $a = 4$; primitive $= \sin^{-1}(x/4)$ [1]. $[\sin^{-1}(x/4)]_0^4 = \sin^{-1}(1) - \sin^{-1}(0) = \dfrac{\pi}{2} - 0 = \dfrac{\pi}{2}$ [1].

Q3 (2 marks): Let $u = x/a$; $\dfrac{d}{dx}\sin^{-1}(u) = \dfrac{1}{\sqrt{1-u^2}}\cdot\dfrac{1}{a} = \dfrac{1}{a\sqrt{1-x^2/a^2}} = \dfrac{1}{\sqrt{a^2-x^2}}$ [1]. Hence $\displaystyle\int \dfrac{3}{\sqrt{a^2-x^2}}\,dx = 3\sin^{-1}\!\!\left(\dfrac{x}{a}\right)+C$ [1].

Boss battle · The Integral Master

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Science Jump · platform challenge

Climb platforms by answering arcsine integration questions. Lighter alternative to the boss.

Mark lesson as complete

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← Lesson 12 · Chain Rule with Inverse Trig Lesson 14 · Integrating a/(a²+x²) →

Module overview · Extension 1 · Checkpoint 3