Chapter 7 · Integrals

Exercise 7.3 · Q7: $\int\sin 4x\sin 8x\,dx$

Exercise 7.3Integration using trigonometric identities · Section 7.3.2

The question

Find the integral of the function

\sin 4x\sin 8x.

Exam answer

Let $I = \displaystyle\int \sin 4x\sin 8x\,dx$ ⁠.

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

\begin{aligned} &\sin 4x\sin 8x \\ &= \frac12\big(\cos(-4x) - \cos 12x\big) \\ &= \frac12(\cos 4x - \cos 12x) \end{aligned}

\begin{aligned} \therefore\ I &= \frac12\int(\cos 4x - \cos 12x)\,dx \\ &= \frac12\left(\frac{\sin 4x}{4} - \frac{\sin 12x}{12}\right) + C \end{aligned}

$\therefore\ I = \dfrac12\Big[\dfrac14\sin 4x - \dfrac{1}{12}\sin 12x\Big]$ $+ C$ ⁠.

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

The one idea this problem is built on

A product of two sines becomes a difference of cosines: $2\sin A\sin B = \cos(A-B) - \cos(A+B)$ ⁠. Single cosines of multiples of $x$ integrate straight away.

Given, to find, and the plan

Given: $\sin 4x\sin 8x$ ⁠, defined for every real $x$ ⁠.
To find: $\displaystyle\int \sin 4x\sin 8x\,dx$ ⁠.
Plan: (1) Derive and apply the product-to-sum identity for two sines. (2) Integrate the two cosines. (3) Differentiate back to check.

Basics you need

1Compound-angle formulas for cosine

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

2Cosine is even

$\cos(-\theta) = \cos\theta$ for every $\theta$ ⁠.

3Cosine of a multiple of $x$

$\int\cos mx\,dx = \frac1m\sin mx + C$ ⁠, $m \ne 0$ (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

The identity for a product of two sines

Subtract the two formulas of Basic 1; the $\cos A\cos B$ terms cancel:
$\cos(A-B) - \cos(A+B) = 2\sin A\sin B.$
With $A = 4x$ ⁠, $B = 8x$ ⁠: $A - B = -4x$ ⁠, $A + B = 12x$ ⁠, and $\cos(-4x) = \cos 4x$ ⁠. So
$\sin 4x\sin 8x = \tfrac12\big[\cos 4x - \cos 12x\big].$
The difference angle is negative here, but cosine is even, so the sign of the angle does not matter. (That is why the order of $A$ and $B$ never matters in this identity, unlike the sine-cosine one in Question 2.) The minus sign between the cosines does matter: it comes from subtracting the formulas.

Compound-angle formulas (Class XI trigonometry). Basics: trigonometric identities
Integrate the two cosines
$\begin{aligned} &\int \sin 4x\sin 8x\,dx \\ &= \frac12\int\cos 4x\,dx - \frac12\int\cos 12x\,dx \\ &= \frac12\Big[\frac14\sin 4x - \frac{1}{12}\sin 12x\Big] + C. \end{aligned}$
Each $\int\cos mx\,dx = \frac1m\sin mx$ ⁠: the substitution $t = mx$ makes $dx = \frac{dt}{m}$ ⁠. Here $m = 4$ and $m = 12$ ⁠.

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

Let $F(x) = \frac18\sin 4x - \frac{1}{24}\sin 12x$ (the answer with the $\frac12$ multiplied in). By the chain rule,
$\begin{aligned} F'(x) &= \tfrac12\big[\cos 4x - \cos 12x\big] \\ &= \sin 4x\sin 8x, \end{aligned}$
by Step 1. The answer is correct for every real $x$ ⁠.

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

Result

\int \sin 4x\sin 8x\,dx = \frac12\Big[\frac14\sin 4x - \frac{1}{12}\sin 12x\Big] + C,

that is $\frac18\sin 4x - \frac{1}{24}\sin 12x + C$ ⁠, valid for all real $x$ ⁠. This is exactly the textbook answer.

Check it with numbers

The identity at $x = 1$ ⁠: $\sin 4\sin 8$ $\approx (-0.7568025)(0.9893582)$ $\approx -0.7487488$ ⁠, and $\frac12[\cos 4 - \cos 12]$ $\approx \frac12[-0.6536436 - 0.8438540]$ $\approx -0.7487488$ ⁠.

The derivative at $x = 1$ ⁠, $h = 0.001$ ⁠: $F(1.001) \approx -0.0729926977$ ⁠, $F(0.999) \approx -0.0714952221$ ⁠, slope $\approx -0.748738$ ⁠, against the integrand $-0.748749$ ⁠: agreement to four significant figures (the gap is the error of the difference quotient; $\sin 12x$ varies quickly).

Common mistakes

Getting the order of the cosines wrong. $2\sin A\sin B = \cos(A-B) - \cos(A+B)$ ⁠. Test: with $A = B$ it gives $2\sin^2 A = 1 - \cos 2A$ ⁠, which is true; the reversed order would give $2\sin^2 A = \cos 2A - 1$ ⁠, which is never positive, so it cannot be right.
Writing $\cos(-4x) = -\cos 4x$ ⁠. Cosine is even.
Forgetting to divide by 12 in $\int\cos 12x\,dx$ ⁠.

What to take away

$\sin\cdot\sin$ and $\cos\cdot\cos$ become sums of cosines; $\sin\cdot\cos$ becomes a sum of sines.
Check any product-to-sum identity with $A = B$ ⁠.
Question 6 uses the same identity as its first move.

Every fact used, and where it comes from

Fact	Used in	Source
$\cos(A\mp B) = \cos A\cos B \pm \sin A\sin B$	Step 1	Compound-angle formulas (Class XI trigonometry)
$2\sin A\sin B = \cos(A-B) - \cos(A+B)$	Steps 1, 3	Derived in Step 1
$\cos(-\theta) = \cos\theta$	Step 1	Cosine is even (Class XI)
$\int\cos mx\,dx = \frac1m\sin mx + C$	Step 2	Standard formula for $\cos x$ (Section 7.2) with $t = mx$ (Section 7.3.1)
Chain rule; differentiation undoes integration	Step 3	Theorem 4, Section 5.3.1; Property (I), Section 7.2.1

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Exercise 7.3 · Q7: ∫sin⁡4xsin⁡8x dx\int\sin 4x\sin 8x\,dx∫sin4xsin8xdx

Basics you need

Solution

The identity for a product of two sines

Integrate the two cosines

Check: differentiate the answer back

Every fact used, and where it comes from

Exercise 7.3 · Q7: $\int\sin 4x\sin 8x\,dx$