Chapter 7 · Integrals

Exercise 7.3 · Q11: $\int\cos^4 2x\,dx$

Exercise 7.3Integration using trigonometric identities · Section 7.3.2

The question

Find the integral of the function

\cos^4 2x.

Exam answer

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

Using $\cos^2\theta = \frac{1 + \cos 2\theta}{2}$ ⁠,

\begin{aligned} &\cos^4 2x = \left(\frac{1 + \cos 4x}{2}\right)^2 \\ &= \frac14\left(1 + 2\cos 4x + \cos^2 4x\right) \\ &= \frac14\bigg(1 + 2\cos 4x \\ &\qquad + \frac{1 + \cos 8x}{2}\bigg) \\ &= \frac38 + \frac12\cos 4x + \frac18\cos 8x \end{aligned}

\begin{aligned} \therefore\ I &= \int\bigg(\frac38 + \frac12\cos 4x \\ &\qquad + \frac18\cos 8x\bigg)dx \\ &= \frac38x + \frac12\cdot\frac{\sin 4x}{4} \\ &\quad + \frac18\cdot\frac{\sin 8x}{8} + C \end{aligned}

$\therefore\ I = \dfrac{3x}{8} + \dfrac18\sin 4x$ $+ \dfrac{1}{64}\sin 8x + C$ ⁠.

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

The one idea this problem is built on

The same double power-lowering as Question 10, now for cosine and with the angle $2x$ ⁠: $\cos^2\theta = \frac{1+\cos 2\theta}{2}$ applied twice turns $\cos^4 2x$ into a constant plus $\cos 4x$ and $\cos 8x$ ⁠. Each angle doubles at each step, so keep careful track of it.

Given, to find, and the plan

Given: $\cos^4 2x$ ⁠, meaning $(\cos 2x)^4$ ⁠, defined for every real $x$ ⁠.
To find: $\displaystyle\int \cos^4 2x\,dx$ ⁠.
Plan: (1) Write $\cos^4 2x = (\cos^2 2x)^2$ and lower the power. (2) Expand the square. (3) Lower the power of $\cos^2 4x$ ⁠. (4) Integrate term by term. (5) Differentiate back to check.

Basics you need

1Power lowering for cosine

$\cos^2\theta = \frac{1 + \cos 2\theta}{2}$ for every $\theta$ (Example 7(i), Section 7.3.2). Basics: trigonometric identities for integrals

2Squaring a binomial

$(a + b)^2 = a^2 + 2ab + b^2$ ⁠.

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

Lower the power once

With $\theta = 2x$ ⁠, so $2\theta = 4x$ ⁠:
$\cos^4 2x = (\cos^2 2x)^2 = \Big(\frac{1 + \cos 4x}{2}\Big)^2.$
The fourth power is the square of a square, and the identity handles squares. The angle doubles from $2x$ to $4x$ ⁠.

$\cos^2\theta = \frac{1 + \cos 2\theta}{2}$ ⁠. Basics: trigonometric identities
Expand the square
$\Big(\frac{1 + \cos 4x}{2}\Big)^2 = \frac14\big(1 + 2\cos 4x + \cos^2 4x\big).$
$(a + b)^2 = a^2 + 2ab + b^2$ with $a = 1$ ⁠, $b = \cos 4x$ ⁠. The last term is still a square, so one more power-lowering is needed.
Lower the power of $\cos^2 4x$

With $\theta = 4x$ ⁠: $\cos^2 4x = \frac{1 + \cos 8x}{2}$ ⁠. Then
$\begin{aligned} \cos^4 2x &= \frac14\Big(1 + 2\cos 4x + \frac12 + \frac12\cos 8x\Big) \\ &= \frac14\Big(\frac32 + 2\cos 4x + \frac12\cos 8x\Big) \\ &= \frac38 + \frac12\cos 4x + \frac18\cos 8x. \end{aligned}$
The angle doubles again, $4x \to 8x$ ⁠. Constants $1 + \frac12 = \frac32$ combine first, then everything is multiplied by $\frac14$ ⁠. Compare Question 10: the same numbers $\frac38, \frac12, \frac18$ appear, with plus signs here because cosine's identity has a plus.
Integrate term by term
$\begin{aligned} &\int \cos^4 2x\,dx \\ &= \frac38 x + \frac12\cdot\frac14\sin 4x + \frac18\cdot\frac18\sin 8x + C \\ &= \frac{3x}{8} + \frac18\sin 4x + \frac{1}{64}\sin 8x + C. \end{aligned}$
$\int\cos 4x\,dx = \frac14\sin 4x$ and $\int\cos 8x\,dx = \frac18\sin 8x$ ⁠, by the substitution $t = mx$ ⁠.

Properties (III), (IV), Section 7.2.1; substitution, Section 7.3.1. Lesson: integration by substitution
Check: differentiate the answer back

Let
$F(x) = \frac{3x}{8} + \frac18\sin 4x + \frac{1}{64}\sin 8x.$
Then
$\begin{aligned} F'(x) &= \frac38 + \frac18\cdot 4\cos 4x + \frac{1}{64}\cdot 8\cos 8x \\ &= \frac38 + \frac12\cos 4x + \frac18\cos 8x = \cos^4 2x, \end{aligned}$
by Step 3. The answer is valid for every real $x$ ⁠.

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

Result

\int \cos^4 2x\,dx = \frac{3x}{8} + \frac18\sin 4x + \frac{1}{64}\sin 8x + C,

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

Check it with numbers

The identity of Step 3 at $x = 1$ ⁠: $\cos^4 2$ $\approx (-0.4161468)^4$ $\approx 0.0299907$ ⁠, and $\frac38 + \frac12\cos 4 + \frac18\cos 8$ $\approx 0.375 - 0.3268218 - 0.0181875$ $\approx 0.0299907$ ⁠.

The derivative at $x = 1$ ⁠, $h = 0.001$ ⁠: $F(1.001) \approx 0.295888664566$ ⁠, $F(0.999) \approx 0.295828681064$ ⁠, slope $\approx 0.02999175$ ⁠, against the integrand $0.0299907$ ⁠: agreement to four significant figures (the gap is the error of the difference quotient).

Common mistakes

Not doubling the angle each time. $\cos^2 2x$ lowers to $\cos 4x$ ⁠, and $\cos^2 4x$ lowers to $\cos 8x$ ⁠.
Reading $\cos^4 2x$ as $\cos(2x^4)$ or $\cos^4 x\cdot 2$ ⁠. It means $(\cos 2x)^4$ ⁠.
Dividing by the wrong number. $\int\frac18\cos 8x\,dx = \frac{1}{64}\sin 8x$ ⁠.

What to take away

$\cos^4\theta = \frac38 + \frac12\cos 2\theta + \frac18\cos 4\theta$ and $\sin^4\theta = \frac38 - \frac12\cos 2\theta + \frac18\cos 4\theta$ ⁠.
Every power-lowering doubles the angle.
Check the arithmetic by differentiating; the factors from the chain rule cancel the divisions.

Every fact used, and where it comes from

Fact	Used in	Source
$\cos^2\theta = \frac{1 + \cos 2\theta}{2}$	Steps 1, 3	Example 7(i), Section 7.3.2
$(a + b)^2 = a^2 + 2ab + b^2$	Step 2	Algebra
$\int\cos mx\,dx = \frac1m\sin mx + C$	Step 4	Standard formula for $\cos x$ (Section 7.2) with $t = mx$ (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 · Q11: ∫cos⁡42x dx\int\cos^4 2x\,dx∫cos42xdx

Basics you need

Solution

Lower the power once

Expand the square

Lower the power of cos⁡24x\cos^2 4xcos24x

Integrate term by term

Check: differentiate the answer back

Every fact used, and where it comes from

Exercise 7.3 · Q11: $\int\cos^4 2x\,dx$

Lower the power of $\cos^2 4x$