Trigonometric identities in integration | Year 12

01 / LEARN

Understand the idea.

Make the integrand integrable

A trigonometric identity can expose an integrable sum. The power-reduction formulas are ${sin}^{2} x = \frac{1 - cos (2 x)}{2}$ and ${cos}^{2} x = \frac{1 + cos (2 x)}{2}$ . After rewriting, integrate each term and compensate for the doubled angle.

For an odd power of sine, reserve one $sin x$ factor and replace the remaining even power using ${sin}^{2} x = 1 - {cos}^{2} x$ . Then $u = cos x$ absorbs the reserved factor, with a minus sign in $d u$ .

Select an identity for the task

Choose an identity for what it makes simpler. A product such as $sin x cos x$ becomes $\frac{1}{2} sin (2 x)$ ; products with different frequencies can use product-to-sum formulas. An odd cosine power can instead reserve $cos x$ for $u = sin x$ .

An identity must hold throughout the relevant domain. Testing an angle may reveal a wrong sign, but one test does not prove an identity. To verify an antiderivative, differentiate it and use identities to recover the original integrand.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

Find $\int {sin}^{2} x d x$ .

Worked solution

1
Reduce the squared sine.
$\int {sin}^{2} x d x = \frac{1}{2} \int (1 - cos (2 x)) d x$
The double-angle identity turns the squared trigonometric term into a sum.
2
Integrate both terms.
$\frac{x}{2} - \frac{1}{2} \cdot \frac{sin (2 x)}{2} + C = \frac{x}{2} - \frac{sin (2 x)}{4} + C$
Integrating cosine of 2x supplies a further factor of one half.
3
Verify by differentiation.
$\frac{d}{d x} (\frac{x}{2} - \frac{sin (2 x)}{4}) = \frac{1 - cos (2 x)}{2} = {sin}^{2} x$
The original identity confirms the derivative equals the integrand.

Answer

$x / 2 - sin 2 x / 4 + C$ .

EXAMPLE 02

Worked example 2

Find $\int {sin}^{3} x d x$ .

Worked solution

1
Reserve one sine factor.
${sin}^{3} x = (1 - {cos}^{2} x) sin x$
The remaining squared sine can be expressed in terms of cosine.
2
Substitute for cosine.
$u = cos x, d u = - sin x d x$
The reserved sine factor supplies the differential with a negative sign.
3
Integrate the polynomial.
$\int {sin}^{3} x d x = - \int (1 - u^{2}) d u = - u + \frac{u^{3}}{3} + C$
Distribute the minus sign before integrating the two powers.
4
Return to x and check.
$\begin{array}{c} - cos x + \frac{{cos}^{3} x}{3} + C \\ \frac{d}{d x} (- cos x + \frac{{cos}^{3} x}{3}) = sin x (1 - {cos}^{2} x) = {sin}^{3} x \end{array}$
The chain rule and Pythagorean identity recover the original odd power.

Answer

$- cos x + {cos}^{3} x / 3 + C$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

Rewrite sin squared x times cos squared x.

Show answer & reasoning

$(1 - cos 4 x) / 8$ .

First, ${sin}^{2} x {cos}^{2} x = \frac{1}{4} {sin}^{2} (2 x)$ . Then reduce the remaining square to obtain $\frac{1}{8} (1 - cos (4 x))$ .

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Find $\int {cos}^{2} (2 x) d x$ .
2 marks
Find $\int sin x cos x d x$ .
2 marks
Find $\int {sin}^{2} x {cos}^{2} x d x$ .
2 marks
Evaluate $\int_{0}^{\frac{π}{2}} {sin}^{2} x d x$ .
2 marks
Find $\int sin x cos (3 x) d x$ .
2 marks
Find $\int {cos}^{3} x d x$ .
2 marks

Read the worked solutions

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

Find $\int {cos}^{2} (2 x) d x$ .
2 marks
Use $\frac{1 + cos 4 x}{2}$ .
Divide the cosine antiderivative by its frequency 4.
$\frac{x}{2} + \frac{sin 4 x}{8} + C$ .
Find $\int sin x cos x d x$ .
2 marks
Use $u = sin x$ .
Integrate $u d u$ .
$\frac{{sin}^{2} x}{2} + C$ .
Find $\int {sin}^{2} x {cos}^{2} x d x$ .
2 marks
Rewrite as $\frac{1 - cos 4 x}{8}$ .
Integrate each term.
$\frac{x}{8} - \frac{sin 4 x}{32} + C$ .
Evaluate $\int_{0}^{\frac{π}{2}} {sin}^{2} x d x$ .
2 marks
Use $\frac{x}{2} - \frac{sin 2 x}{4}$ .
The sine endpoint terms are zero.
$\frac{π}{4}$ .
Find $\int sin x cos (3 x) d x$ .
2 marks
Use $sin x cos 3 x = \frac{sin 4 x - sin 2 x}{2}$ .
Integrate both terms with their frequencies.
$- \frac{cos 4 x}{8} + \frac{cos 2 x}{4} + C$ .
Find $\int {cos}^{3} x d x$ .
2 marks
Write $(1 - {sin}^{2} x) cos x$ .
Use $u = sin x$ .
$sin x - \frac{{sin}^{3} x}{3} + C$ .

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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.

Make the integrand integrable

Select an identity for the task

See the method in action.

Worked example 1

Reduce the squared sine.

Integrate both terms.

Verify by differentiation.

Worked example 2

Reserve one sine factor.

Substitute for cosine.

Integrate the polynomial.

Return to x and check.

Try it yourself.

Check your understanding

Keep practising