Inverse-trigonometric derivatives and integrals | Year 12

01 / LEARN

Understand the idea.

Recognise the two basic patterns

For $a > 0$ , a denominator $\sqrt{a^{2} - x^{2}}$ suggests $arcsin (\frac{x}{a})$ ; a denominator $a^{2} + x^{2}$ suggests $\frac{1}{a} arctan (\frac{x}{a})$ . Factor constants or complete the square before comparing with either pattern.

These rules use radians. The arcsine antiderivative is differentiable only for $- a < x < a$ , where the integrand is real and finite. The arctangent pattern has no real denominator zeros, so its domain is all real numbers.

Check scaling with a derivative

Scaling inside an inverse trigonometric function changes its derivative. For instance, $\frac{d}{d x} arctan (2 x) = \frac{2}{1 + 4 x^{2}}$ . Match the numerator as well as the denominator; a substitution makes any required factor explicit.

The word inverse refers to undoing a function, not taking a reciprocal. The value $arcsin x$ is an angle on the principal range; $\frac{1}{sin x}$ is a different function. Differentiate your proposed antiderivative to distinguish them.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

Find $\int 1 / \sqrt{9 - x^{2}} d x$ .

Worked solution

1
Identify the arcsine pattern.
$9 - x^{2} = 3^{2} - x^{2}, a = 3$
The denominator is a square root of a difference of squares.
2
Apply the scaled formula.
$\int \frac{1}{\sqrt{9 - x^{2}}} d x = arcsin (\frac{x}{3}) + C$
The standard arcsine rule already includes the scaling of its input.
3
Check the derivative and domain.
$\frac{d}{d x} arcsin (\frac{x}{3}) = \frac{1 / 3}{\sqrt{1 - x^{2} / 9}} = \frac{1}{\sqrt{9 - x^{2}}}, - 3 < x < 3$
Inside this interval the square root is positive, and the simplification uses 3 greater than zero.

Answer

$arcsin (x / 3) + C$ , -3<x<3.

EXAMPLE 02

Worked example 2

Find $\int 1 / (x^{2} + 4) d x$ .

Worked solution

1
Match the sum of squares.
$x^{2} + 4 = x^{2} + 2^{2}$
A sum of squares without a square root matches the arctangent rule.
2
Include the outside scale factor.
$\int \frac{1}{x^{2} + 4} d x = \frac{1}{2} arctan (\frac{x}{2}) + C$
The rule has both an input divisor and an outside factor of one over 2.
3
Differentiate to verify.
$\frac{d}{d x} [\frac{1}{2} arctan (\frac{x}{2})] = \frac{1 / 4}{1 + x^{2} / 4} = \frac{1}{x^{2} + 4}$
The two factors of one half yield the required numerator; the denominator is positive for every real input.

Answer

$\frac{1}{2} arctan (x / 2) + C$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

Differentiate arctan(3x).

Show answer & reasoning

$3 / (1 + 9 x^{2})$ .

Using $u = 3 x$ , the arctangent rule gives $\frac{u^{'}}{1 + u^{2}} = \frac{3}{1 + 9 x^{2}}$ . The numerator 3 comes from the chain rule.

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Find $\int \frac{3}{9 + x^{2}} d x$ .
2 marks
Find $\int \frac{2}{\sqrt{1 - 4 x^{2}}} d x$ .
2 marks
Find $\int \frac{1}{1 + 9 x^{2}} d x$ .
2 marks
Evaluate $\int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{1 - x^{2}}} d x$ .
2 marks
Find $\int \frac{1}{x^{2} + 16} d x$ .
2 marks
Differentiate $arccos (\frac{x}{2})$ on $- 2 < x < 2$ .
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find $\int \frac{3}{9 + x^{2}} d x$ .
2 marks
Use the arctangent rule with $a = 3$ .
The factor 3 cancels the outside $\frac{1}{3}$ .
$arctan (\frac{x}{3}) + C$ .
Find $\int \frac{2}{\sqrt{1 - 4 x^{2}}} d x$ .
2 marks
Use $u = 2 x$ and $d u = 2 d x$ .
The transformed integral is $\int {(1 - u^{2})}^{- \frac{1}{2}} d u$ .
$arcsin (2 x) + C$ , $∣ x ∣ < \frac{1}{2}$ .
Find $\int \frac{1}{1 + 9 x^{2}} d x$ .
2 marks
Differentiate $arctan (3 x)$ to see the extra factor 3.
Divide by 3.
$\frac{1}{3} arctan (3 x) + C$ .
Evaluate $\int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{1 - x^{2}}} d x$ .
2 marks
The antiderivative is $arcsin x$ .
Evaluate $arcsin (\frac{1}{2}) - arcsin (0)$ .
$\frac{π}{6}$ .
Find $\int \frac{1}{x^{2} + 16} d x$ .
2 marks
Use $a = 4$ in the arctangent pattern.
Check the chain factor.
$\frac{1}{4} arctan (\frac{x}{4}) + C$ .
Differentiate $arccos (\frac{x}{2})$ on $- 2 < x < 2$ .
2 marks
Use derivative $- \frac{u^{'}}{\sqrt{1 - u^{2}}}$ .
Simplify with $u = \frac{x}{2}$ .
$- \frac{1}{\sqrt{4 - x^{2}}}$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics integration techniques and applications, including inverse-trigonometric patterns, substitution, parts, simple partial fractions, areas and solids of revolution.

Official curriculum source ↗

In this lesson

Understand the idea.

Recognise the two basic patterns

Check scaling with a derivative

See the method in action.

Worked example 1

Identify the arcsine pattern.

Apply the scaled formula.

Check the derivative and domain.

Worked example 2

Match the sum of squares.

Include the outside scale factor.

Differentiate to verify.

Try it yourself.

Check your understanding

Keep practising