Chapter 7 · Integrals

Exercise 7.3 · Q6: $\int\sin x\sin 2x\sin 3x\,dx$

Exercise 7.3Integration using trigonometric identities · Section 7.3.2

The question

Find the integral of the function

\sin x\sin 2x\sin 3x.

Exam answer

Let $I = \displaystyle\int \sin x\sin 2x\sin 3x\,dx$ ⁠.

Using $2\sin A\sin B$ $= \cos(A - B) - \cos(A + B)$ ⁠,

\begin{aligned} &\sin x\sin 2x\sin 3x \\ &= \frac12(\cos 2x - \cos 4x)\sin 2x \\ &= \frac12\sin 2x\cos 2x \\ &\quad - \frac12\sin 2x\cos 4x \end{aligned}

Using $2\sin A\cos A = \sin 2A$ and $2\sin A\cos B$ $= \sin(A + B) + \sin(A - B)$ ⁠,

\begin{aligned} &\sin x\sin 2x\sin 3x \\ &= \frac14\sin 4x - \frac14(\sin 6x - \sin 2x) \\ &= \frac14(\sin 2x + \sin 4x - \sin 6x) \end{aligned}

\begin{aligned} \therefore\ I &= \frac14\int\bigg(\sin 2x + \sin 4x \\ &\qquad - \sin 6x\bigg)dx \\ &= \frac14\bigg(-\frac{\cos 2x}{2} - \frac{\cos 4x}{4} \\ &\qquad + \frac{\cos 6x}{6}\bigg) + C \end{aligned}

$\therefore\ I = \dfrac14\Big[\dfrac16\cos 6x - \dfrac14\cos 4x$ $- \dfrac12\cos 2x\Big] + C$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Reduce the product two factors at a time. $2\sin A\sin B = \cos(A-B) - \cos(A+B)$ turns $\sin x\sin 3x$ into cosines; multiplying by $\sin 2x$ gives products of a sine and a cosine, which $2\sin A\cos B = \sin(A+B) + \sin(A-B)$ turns into single sines. A sum of single sines integrates term by term.

Given, to find, and the plan

Given: $\sin x\sin 2x\sin 3x$ ⁠, defined for every real $x$ ⁠.
To find: $\displaystyle\int \sin x\sin 2x\sin 3x\,dx$ ⁠.
Plan: (1) Turn $\sin x\sin 3x$ into a difference of cosines. (2) Multiply by $\sin 2x$ and turn each product into sines. (3) Collect: $\frac14[\sin 2x + \sin 4x - \sin 6x]$ ⁠. (4) Integrate. (5) Differentiate back to check.

Basics you need

1Product of two sines

Subtracting $\cos(A+B) = \cos A\cos B - \sin A\sin B$ from $\cos(A-B) = \cos A\cos B + \sin A\sin B$ gives $2\sin A\sin B = \cos(A-B) - \cos(A+B)$ ⁠. Basics: trigonometric identities for integrals

2Sine times cosine

$2\sin A\cos B = \sin(A+B) + \sin(A-B)$ (Example 7(ii), Section 7.3.2), and its special case $2\sin A\cos A = \sin 2A$ ⁠.

3Even and odd

$\cos(-\theta) = \cos\theta$ and $\sin(-\theta) = -\sin\theta$ ⁠.

4Sine of a multiple of $x$

$\int\sin mx\,dx = -\frac1m\cos mx + C$ ⁠, $m \ne 0$ (Example 5(i), Section 7.3.1). Basics: substitution

Solution

Turn $\sin x\sin 3x$ into cosines

Basic 1 with $A = x$ ⁠, $B = 3x$ ⁠: $A - B = -2x$ ⁠, $A + B = 4x$ ⁠, and $\cos(-2x) = \cos 2x$ ⁠, so
$\sin x\sin 3x = \tfrac12\big[\cos 2x - \cos 4x\big].$
Pairing the outer two factors, $x$ and $3x$ ⁠, is a good choice: their sum and difference, $4x$ and $2x$ ⁠, are small even multiples, and the leftover factor $\sin 2x$ will combine neatly with them.

$2\sin A\sin B = \cos(A-B) - \cos(A+B)$ ⁠; cosine is even. Basics: trigonometric identities
Multiply by $\sin 2x$ and turn each product into sines
$\sin x\sin 2x\sin 3x = \tfrac12\big[\sin 2x\cos 2x - \sin 2x\cos 4x\big].$
- $\sin 2x\cos 2x = \frac12\sin 4x$ (double angle, $A = 2x$ ⁠).
- Basic 2 with $A = 2x$ ⁠, $B = 4x$ ⁠:
$\begin{aligned} \sin 2x\cos 4x &= \tfrac12\big[\sin 6x + \sin(-2x)\big] \\ &= \tfrac12\big[\sin 6x - \sin 2x\big]. \end{aligned}$
Each product now has one sine and one cosine, the case of Basic 2. In the second one the angle $A - B = -2x$ is negative; sine is odd, so $\sin(-2x) = -\sin 2x$ ⁠. This sign is the easiest thing to get wrong in the whole question.
Collect the integrand
$\begin{aligned} &\sin x\sin 2x\sin 3x \\ &= \tfrac12\Big[\tfrac12\sin 4x - \tfrac12\sin 6x + \tfrac12\sin 2x\Big] \\ &= \tfrac14\big[\sin 2x + \sin 4x - \sin 6x\big]. \end{aligned}$
Subtracting $\frac12[\sin 6x - \sin 2x]$ gives $-\frac12\sin 6x + \frac12\sin 2x$ ⁠. The product of three sines is now a sum of three single sines.
Integrate term by term
$\begin{aligned} &\int \sin x\sin 2x\sin 3x\,dx \\ &= \tfrac14\Big[-\tfrac12\cos 2x - \tfrac14\cos 4x + \tfrac16\cos 6x\Big] + C \\ &= \tfrac14\Big[\tfrac16\cos 6x - \tfrac14\cos 4x - \tfrac12\cos 2x\Big] + C. \end{aligned}$
$\int\sin mx\,dx = -\frac1m\cos mx$ ⁠: so $\sin 2x \to -\frac12\cos 2x$ ⁠, $\sin 4x \to -\frac14\cos 4x$ ⁠, and $-\sin 6x \to +\frac16\cos 6x$ ⁠. The second line only reorders the terms to match the textbook.

Properties (III), (IV), Section 7.2.1; Example 5(i). Lesson: integration by substitution
Check: differentiate the answer back

Since $\frac{d}{dx}\big(-\frac1m\cos mx\big) = \sin mx$ ⁠, differentiating the bracket term by term gives
$F'(x) = \tfrac14\big[-\sin 6x + \sin 4x + \sin 2x\big],$
which is the collected form of Step 3, hence equal to $\sin x\sin 2x\sin 3x$ ⁠. The answer is valid for every real $x$ ⁠.

Chain rule, Theorem 4, Section 5.3.1. Basics: integration undoes differentiation

Result

\begin{aligned} &\int \sin x\sin 2x\sin 3x\,dx \\ &= \frac14\Big[\frac16\cos 6x - \frac14\cos 4x - \frac12\cos 2x\Big] + C, \end{aligned}

valid for all real $x$ ⁠. This is exactly the textbook answer.

Check it with numbers

The collected form at $x = 1$ ⁠: $\sin 1\sin 2\sin 3$ $\approx 0.8414710\times 0.9092974\times 0.1411200$ $\approx 0.1079776$ ⁠, and $\frac14[\sin 2 + \sin 4 - \sin 6]$ $\approx \frac14[0.9092974 - 0.7568025 + 0.2794155]$ $\approx 0.1079776$ ⁠.

The derivative at $x = 1$ ⁠, $h = 0.001$ ⁠: $F(1.001) \approx 0.132985002707$ ⁠, $F(0.999) \approx 0.132769047625$ ⁠, slope $\approx 0.1079775$ ⁠, matching the integrand to six significant figures.

Common mistakes

Wrong sign in $2\sin A\sin B$ ⁠. It is $\cos(A-B) - \cos(A+B)$ ⁠: the difference angle comes first. Check with $A = B$ ⁠: $2\sin^2 A = 1 - \cos 2A$ ⁠, which is right.
$\sin(-2x) = \sin 2x$ ⁠. No: sine is odd, $\sin(-2x) = -\sin 2x$ ⁠.
Forgetting a $\frac12$ ⁠. Two identities each bring a $\frac12$ ⁠, giving $\frac14$ overall.
Sign of $\int\sin mx\,dx$ ⁠. It is $-\frac1m\cos mx$ ⁠; so $\int(-\sin 6x)\,dx = +\frac16\cos 6x$ ⁠.

What to take away

Three-factor products: two factors at a time, multiply out, repeat (as in Question 3).
Sanity-check any product-to-sum identity by setting $A = B$ or $B = 0$ ⁠.
Different pairings give the same collected form; always reach "sum of single sines or cosines" before integrating.

Every fact used, and where it comes from

Fact	Used in	Source
$2\sin A\sin B = \cos(A-B) - \cos(A+B)$	Step 1	From the compound-angle formulas for $\cos(A\pm B)$ (Class XI trigonometry)
$2\sin A\cos B = \sin(A+B) + \sin(A-B)$ ⁠; $2\sin A\cos A = \sin 2A$	Step 2	Example 7(ii), Section 7.3.2; double-angle formula
$\cos(-\theta) = \cos\theta$ ⁠, $\sin(-\theta) = -\sin\theta$	Steps 1, 2	Even and odd functions (Class XI)
$\int\sin mx\,dx = -\frac1m\cos mx + C$	Step 4	Example 5(i), Section 7.3.1
Chain rule; differentiation undoes integration	Step 5	Theorem 4, Section 5.3.1; Property (I), Section 7.2.1

Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →

Exercise 7.3 · Q6: ∫sin⁡xsin⁡2xsin⁡3x dx\int\sin x\sin 2x\sin 3x\,dx∫sinxsin2xsin3xdx

Basics you need

Solution

Turn sin⁡xsin⁡3x\sin x\sin 3xsinxsin3x into cosines

Multiply by sin⁡2x\sin 2xsin2x and turn each product into sines

Collect the integrand

Integrate term by term

Check: differentiate the answer back

Every fact used, and where it comes from

Exercise 7.3 · Q6: $\int\sin x\sin 2x\sin 3x\,dx$

Turn $\sin x\sin 3x$ into cosines

Multiply by $\sin 2x$ and turn each product into sines