Year 12 Maths Extension 1 (2027) Calculus

Integration giving inverse trig functions

Q: What integral gives arcsin?

The integral of dx over the square root of a squared minus x squared equals arcsin of x over a plus C.

Q: What integral gives arctan?

The integral of dx over a squared plus x squared equals one over a times arctan of x over a plus C.

Q: How do you find a in these integrals?

Take a as the square root of the constant; for nine minus x squared, a is three.

Q: What if there is a coefficient on x squared?

Factor it out first, for example one minus nine x squared equals one minus three x all squared, then substitute u equals three x.

Q: Does the arctan form need a one over a?

Yes, the arctan form always carries the one over a factor; the arcsin form does not.

20 practice questions 2 video lessons Theory + worked examples

Learn integration giving inverse trigonometric functions for NSW Year 12 Mathematics Extension 1. Certain standard forms integrate to arcsin or arctan, reversing the derivatives you already know.

You will learn to recognise the integrand forms that lead to inverse trig results, and to evaluate both indefinite and definite integrals using them — a key addition to your integration toolkit for the Extension 1 course.

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Practice questions

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Integration Resulting in Inverse trigonometric functions Watch
Integration into Inverse trigonometric functions using Substitution Watch

Theory

Two standard integrals reverse the inverse-trig derivatives: $\int \frac{d x}{\sqrt{a^{2} - x^{2}}} = \sin^{- 1} \frac{x}{a} + C$ and $\int \frac{d x}{a^{2} + x^{2}} = \frac{1}{a} \tan^{- 1} \frac{x}{a} + C$ . This NSW Year 12 Mathematics Extension 1 topic (NESA outcome ME1-12-04) matches integrands to these forms.

Two standard integrals reverse the inverse-trig derivatives, producing $\sin^{- 1}$ and $\tan^{- 1}$ .

$\int \frac{d x}{\sqrt{a^{2} - x^{2}}} = \sin^{- 1} \frac{x}{a} + C$ and $\int \frac{d x}{a^{2} + x^{2}} = \frac{1}{a} \tan^{- 1} \frac{x}{a} + C$ (for $a > 0$ ). The $\tan^{- 1}$ form carries an extra $\frac{1}{a}$ ; the $\sin^{- 1}$ form does not.

Match $a^{2}$ to the constant, so $a = \sqrt{constant}$ . A coefficient on $x^{2}$ is factored out first, e.g. $1 - 9 x^{2} = 1 - (3 x)^{2}$ .

NESA link. Part of the Year 12 Calculus topic, outcome ME1-12-04 ("selects and applies differentiation and integration techniques to solve problems") with MAO-WM-01. This is part of the Techniques of integration focus area.

\int_{0}^{2} \frac{d x}{4 + x^{2}} = \frac{π}{8}

— the tangent form.

\int \frac{d x}{\sqrt{a^{2} - x^{2}}} = \sin^{- 1} \frac{x}{a} + C

— the sine form.

\int \frac{d x}{\sqrt{a^{2} - x^{2}}} = \sin^{- 1} \frac{x}{a} + C .

integral dx / sqrt(a^2 - x^2) = arcsin(x/a) + C

\int \frac{d x}{a^{2} + x^{2}} = \frac{1}{a} \tan^{- 1} \frac{x}{a} + C, a > 0.

integral dx / (a^2 + x^2) = (1/a) arctan(x/a) + C

Key checks. The $\tan^{- 1}$ form has the extra $\frac{1}{a}$ ; match $a^{2}$ to the constant (e.g. $9 - x^{2} \Rightarrow a = 3$ ); factor out any coefficient on $x^{2}$ before matching.

How to integrate to an inverse trig function

Identify the form — a root $\sqrt{a^{2} - x^{2}}$ gives $\sin^{- 1}$ ; a sum $a^{2} + x^{2}$ gives $\tan^{- 1}$ .
Read off $a$ as the square root of the constant.
Factor a coefficient on $x^{2}$ first, e.g. $1 - 9 x^{2} = 1 - (3 x)^{2}$ (a substitution $u = 3 x$ tidies it).
Include $\frac{1}{a}$ for the $\tan^{- 1}$ form, then evaluate any limits for an exact value.

Example 1 — Sine form

Find

\int \frac{d x}{\sqrt{25 - x^{2}}}

Solution

$a^{2} = 25$ , so $a = 5$ .

\int \frac{d x}{\sqrt{25 - x^{2}}}

=

\sin^{- 1} \frac{x}{5} + C

integral dx / sqrt(25 - x^2) = arcsin(x/5) + C

So the integral is $\sin^{- 1} \frac{x}{5} + C$ .

Example 2 — Tangent form

Find

\int \frac{d x}{x^{2} + 9}

Solution

$a^{2} = 9$ , so $a = 3$ ; include the $\frac{1}{a}$ .

\int \frac{d x}{x^{2} + 9}

=

\frac{1}{3} \tan^{- 1} \frac{x}{3} + C

integral dx / (x^2 + 9) = (1/3) arctan(x/3) + C

So the integral is $\frac{1}{3} \tan^{- 1} \frac{x}{3} + C$ .

Example 3 — Coefficient inside the root

Find

\int \frac{d x}{\sqrt{1 - 9 x^{2}}}

Solution

Write $1 - 9 x^{2} = 1 - (3 x)^{2}$ ; let $u = 3 x$ , $d u = 3 d x$ .

$\int \frac{d x}{\sqrt{1 - 9 x^{2}}}$	$=$	$\frac{1}{3} \int \frac{d u}{\sqrt{1 - u^{2}}}$
	$=$	$\frac{1}{3} \sin^{- 1} (3 x) + C$

integral dx / sqrt(1 - 9x^2) = (1/3) arcsin(3x) + C

So the integral is $\frac{1}{3} \sin^{- 1} (3 x) + C$ .

Example 4 — A definite integral

Evaluate

\int_{0}^{2} \frac{d x}{4 + x^{2}}

Solution

$a = 2$ , primitive $\frac{1}{2} \tan^{- 1} \frac{x}{2}$ .

$\int_{0}^{2} \frac{d x}{4 + x^{2}}$	$=$	$\frac{1}{2} [\tan^{- 1} \frac{x}{2}]_{0}^{2}$
	$=$	$\frac{1}{2} \cdot \frac{π}{4} = \frac{π}{8}$

integral 0 to 2 dx / (4 + x^2) = pi/8

Value: $\frac{π}{8}$ .

Common pitfalls

The extra factor. The

\tan^{- 1}

form has a

\frac{1}{a}

factor; the

\sin^{- 1}

form does not.

Reading $a$ . Match

a^{2}

to the constant, so

a = \sqrt{constant}

(e.g.

9 - x^{2} \Rightarrow a = 3

Coefficient on $x^{2}$ . Factor it out before matching the form, e.g.

1 - 9 x^{2} = 1 - (3 x)^{2}

Mixing the forms. A root in the denominator gives

\sin^{- 1}

; a plain sum gives

\tan^{- 1}

Frequently asked questions

What integral gives arcsin?

$\int \frac{d x}{\sqrt{a^{2} - x^{2}}} = \sin^{- 1} \frac{x}{a} + C$ .

What integral gives arctan?

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

How do you find a in these integrals?

Take $a = \sqrt{constant}$ ; for $9 - x^{2}$ , $a = 3$ .

What if there is a coefficient on x squared?

Factor it out first, e.g. $1 - 9 x^{2} = 1 - (3 x)^{2}$ , then substitute $u = 3 x$ .

Does the arctan form need a one over a?

Yes — the $\tan^{- 1}$ form always carries the $\frac{1}{a}$ factor; the $\sin^{- 1}$ form does not.

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Derivatives of inverse trig functions

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Integration by substitution

Integration giving inverse trig functions

📖 Theory

How to integrate to an inverse trig function

Common pitfalls

Frequently asked questions

Theory