Chapter 7 · Integrals

Exercise 7.3 · Q3: $\int\cos 2x\cos 4x\cos 6x\,dx$

Exercise 7.3Integration using trigonometric identities · Section 7.3.2

The question

Find the integral of the function

\cos 2x\cos 4x\cos 6x.

Exam answer

Let $I = \displaystyle\int \cos 2x\cos 4x\cos 6x\,dx$ ⁠.

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

\begin{aligned} &\cos 2x\cos 4x\cos 6x \\ &= \frac12(\cos 6x + \cos 2x)\cos 6x \\ &= \frac12\cos^2 6x + \frac12\cos 2x\cos 6x \end{aligned}

Using $\cos^2\theta = \frac{1 + \cos 2\theta}{2}$ and the same identity again,

\begin{aligned} &\cos 2x\cos 4x\cos 6x \\ &= \frac14(1 + \cos 12x) \\ &\quad + \frac14(\cos 8x + \cos 4x) \end{aligned}

\begin{aligned} \therefore\ I &= \frac14\int\bigg(1 + \cos 4x + \cos 8x \\ &\qquad + \cos 12x\bigg)dx \\ &= \frac14\bigg(x + \frac{\sin 4x}{4} + \frac{\sin 8x}{8} \\ &\qquad + \frac{\sin 12x}{12}\bigg) + C \end{aligned}

$\therefore\ I = \dfrac14\Big[x + \dfrac14\sin 4x + \dfrac18\sin 8x$ $+ \dfrac{1}{12}\sin 12x\Big] + C$ ⁠.

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

The one idea this problem is built on

A product of three cosines is handled two at a time. The identity $2\cos A\cos B = \cos(A+B) + \cos(A-B)$ turns one pair into a sum; multiplying that sum by the third cosine gives new products of two cosines, and the identity (or $\cos^2\theta = \frac{1+\cos 2\theta}{2}$ ⁠) is used again. In the end the integrand is a constant plus single cosines, each of which integrates directly.

Given, to find, and the plan

Given: $\cos 2x\cos 4x\cos 6x$ ⁠, defined for every real $x$ ⁠.
To find: $\displaystyle\int \cos 2x\cos 4x\cos 6x\,dx$ ⁠.
Plan: (1) Turn $\cos 2x\cos 4x$ into a sum. (2) Multiply by $\cos 6x$ and reduce the two new products. (3) Collect the integrand as $\frac14[1 + \cos 4x + \cos 8x + \cos 12x]$ ⁠. (4) Integrate term by term. (5) Differentiate back to check.

Basics you need

1Product of two cosines

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

2Square of a cosine

$\cos^2\theta = \frac{1+\cos 2\theta}{2}$ (Example 7(i), Section 7.3.2). It is the special case $A = B = \theta$ of Basic 1.

3Cosine of a multiple of $x$

$\int\cos mx\,dx = \frac1m\sin mx + C$ for $m \ne 0$ (substitution $t = mx$ ⁠, Section 7.3.1). Basics: substitution

4Checking an integral

$F(x) + C$ is right exactly when $F'(x)$ is the integrand (Property (I), Section 7.2.1). Basics: integration undoes differentiation

Solution

Turn the first pair into a sum

Use $2\cos A\cos B = \cos(A+B) + \cos(A-B)$ with $A = 4x$ ⁠, $B = 2x$ ⁠:
$\cos 2x\cos 4x = \tfrac12\big[\cos 6x + \cos 2x\big].$
We pick a pair and remove the product. Any pair would do (see Common mistakes); the pair $2x, 4x$ is convenient because the sum $6x$ matches the third factor, which will give a square that the power-lowering identity handles.

Product-to-sum identity for two cosines. Basics: trigonometric identities
Multiply by the third cosine
$\cos 2x\cos 4x\cos 6x = \tfrac12\big[\cos^2 6x + \cos 2x\cos 6x\big].$
Reduce each new product:
- $\cos^2 6x = \frac12\big[1 + \cos 12x\big]$ (power lowering with $\theta = 6x$ ⁠).
- $\cos 2x\cos 6x = \frac12\big[\cos 8x + \cos 4x\big]$ (Basic 1 with $A = 6x$ ⁠, $B = 2x$ ⁠).
Multiplying out creates two products of two cosines. Each is the same kind of problem as Step 1, so the same identity finishes them. Taking $A$ as the larger angle keeps $A - B$ positive; with the other order you would get $\cos(-4x)$ ⁠, which is the same thing because cosine is even.
Collect the integrand
$\begin{aligned} &\cos 2x\cos 4x\cos 6x \\ &= \tfrac12\Big[\tfrac12(1 + \cos 12x) \\ &\qquad + \tfrac12(\cos 8x + \cos 4x)\Big] \\ &= \tfrac14\big[1 + \cos 4x + \cos 8x + \cos 12x\big]. \end{aligned}$
The triple product is now a constant plus three single cosines. No products and no powers are left, so every term has a standard integral.
Integrate term by term
$\begin{aligned} &\int \cos 2x\cos 4x\cos 6x\,dx \\ &= \tfrac14\Big[x + \tfrac14\sin 4x + \tfrac18\sin 8x \\ &\qquad + \tfrac{1}{12}\sin 12x\Big] + C. \end{aligned}$
$\int 1\,dx = x$ ⁠, and each $\int\cos mx\,dx = \frac1m\sin mx$ by the substitution $t = mx$ ⁠. The factor $\frac14$ in front applies to every term.

Properties (III), (IV), Section 7.2.1; $\int\cos x\,dx = \sin x + C$ (Section 7.2) with $t = mx$ (Section 7.3.1). Lesson: integration by substitution
Check: differentiate the answer back

Let
$F(x) = \frac14\Big[x + \frac14\sin 4x + \frac18\sin 8x + \frac{1}{12}\sin 12x\Big].$
Since $\frac{d}{dx}\big(\frac1m\sin mx\big) = \cos mx$ ⁠,
$F'(x) = \tfrac14\big[1 + \cos 4x + \cos 8x + \cos 12x\big],$
which is $\cos 2x\cos 4x\cos 6x$ by Step 3. The answer is correct for every real $x$ ⁠.

The derivative lands on the "collected" form, and Steps 1 to 3 show that form equals the original product. So the check is complete once those steps are right; the numerical check below tests them independently.

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

Result

\begin{aligned} &\int \cos 2x\cos 4x\cos 6x\,dx \\ &= \frac14\Big[x + \frac14\sin 4x + \frac18\sin 8x \\ &\qquad + \frac{1}{12}\sin 12x\Big] + C, \end{aligned}

valid for all real $x$ ⁠. Textbook answer:

\begin{aligned} &\frac14\Big[\frac{1}{12}\sin 12x + x \\ &\qquad + \frac18\sin 8x + \frac14\sin 4x\Big] + C, \end{aligned}

the same four terms in a different order.

Check it with numbers

The collected form at $x = 1$ ⁠: $\cos 2\cos 4\cos 6$ $\approx (-0.41614684)(-0.65364362)$ $\times\, 0.96017029$ $\approx 0.2611776$ ⁠, and $\frac14[1 + \cos 4 + \cos 8 + \cos 12]$ $\approx \frac14[1 - 0.6536436 - 0.1455000 + 0.8438540]$ $\approx 0.2611776$ ⁠.

The derivative at $x = 1$ with $h = 0.001$ ⁠: $F(1.001) \approx 0.2227000540$ ⁠, $F(0.999) \approx 0.2221777074$ ⁠, slope $\approx 0.261173$ ⁠, against the integrand $0.261178$ ⁠: agreement to four significant figures (the gap is the error of the difference quotient, which is larger here because $\cos 12x$ changes quickly).

Common mistakes

Applying the identity to all three factors at once. $2\cos A\cos B$ handles exactly two factors; multiply out before using it again.
Dropping a factor $\frac12$ ⁠. Two identities each bring a $\frac12$ ⁠, so the collected form has $\frac14$ in front.
Worrying about the pairing. Pairing $\cos 4x\cos 6x$ first gives $\frac12[\cos 10x + \cos 2x]$ ⁠, then $\frac14[\cos 12x + \cos 8x] + \frac14[1 + \cos 4x]$ ⁠: exactly the same collected form.
Forgetting to divide by $m$ in $\int\cos mx\,dx = \frac1m\sin mx$ ⁠.

What to take away

Longer products: apply the product-to-sum identity to two factors, multiply out, repeat.
A product $\cos\theta\cos\theta$ that appears along the way is just $\cos^2\theta = \frac{1+\cos 2\theta}{2}$ ⁠.
Answers to these questions can look different (order, pairing) and still be equal; check by differentiating.

Every fact used, and where it comes from

Fact	Used in	Source
$2\cos A\cos B = \cos(A+B) + \cos(A-B)$	Steps 1, 2	From the compound-angle formulas for $\cos(A\pm B)$ (Class XI trigonometry)
$\cos^2\theta = \frac{1+\cos 2\theta}{2}$	Step 2	Example 7(i), Section 7.3.2
Sum and constant-multiple rules	Step 4	Properties (III), (IV), Section 7.2.1
$\int\cos mx\,dx = \frac1m\sin mx + C$	Step 4	Standard formula for $\cos x$ (Section 7.2) with substitution $t = mx$ (Section 7.3.1)
Chain rule; differentiation undoes integration	Step 5	Theorem 4, Section 5.3.1; Property (I), Section 7.2.1

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Exercise 7.3 · Q3: ∫cos⁡2xcos⁡4xcos⁡6x dx\int\cos 2x\cos 4x\cos 6x\,dx∫cos2xcos4xcos6xdx

Basics you need

Solution

Turn the first pair into a sum

Multiply by the third cosine

Collect the integrand

Integrate term by term

Check: differentiate the answer back

Every fact used, and where it comes from

Exercise 7.3 · Q3: $\int\cos 2x\cos 4x\cos 6x\,dx$