Chapter 7 · Integrals

Exercise 7.3 · Q10: $\int\sin^4x\,dx$

Exercise 7.3Integration using trigonometric identities · Section 7.3.2

The question

Find the integral of the function

\sin^4 x.

Exam answer

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

Using $\sin^2 x = \frac{1 - \cos 2x}{2}$ ⁠,

\begin{aligned} \sin^4 x &= \left(\frac{1 - \cos 2x}{2}\right)^2 \\ &= \frac14\left(1 - 2\cos 2x + \cos^2 2x\right) \end{aligned}

Using $\cos^2 2x = \frac{1 + \cos 4x}{2}$ ⁠,

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

\begin{aligned} \therefore\ I &= \int\bigg(\frac38 - \frac12\cos 2x \\ &\qquad + \frac18\cos 4x\bigg)dx \\ &= \frac38x - \frac12\cdot\frac{\sin 2x}{2} \\ &\quad + \frac18\cdot\frac{\sin 4x}{4} + C \end{aligned}

$\therefore\ I = \dfrac{3x}{8} - \dfrac14\sin 2x$ $+ \dfrac{1}{32}\sin 4x + C$ ⁠.

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

The one idea this problem is built on

A fourth power is the square of a square. Lower the power once with $\sin^2 x = \frac{1-\cos 2x}{2}$ ⁠; squaring that produces a $\cos^2 2x$ ⁠, so lower the power a second time with $\cos^2\theta = \frac{1+\cos 2\theta}{2}$ ⁠. What is left is a constant plus single cosines.

Given, to find, and the plan

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

Basics you need

1Power-lowering identities

$\sin^2\theta = \frac{1 - \cos 2\theta}{2}$ and $\cos^2\theta = \frac{1 + \cos 2\theta}{2}$ ⁠, both from the double-angle formula; the second is used in 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
$\sin^4 x = (\sin^2 x)^2 = \Big(\frac{1 - \cos 2x}{2}\Big)^2.$
The fourth power has no direct formula. Writing it as the square of $\sin^2 x$ lets us use the one identity we have for squares. One application halves the power from 4 to 2 (now in terms of $\cos 2x$ ⁠).

$\sin^2\theta = \frac{1 - \cos 2\theta}{2}$ ⁠. Basics: trigonometric identities
Expand the square
$\Big(\frac{1 - \cos 2x}{2}\Big)^2 = \frac14\big(1 - 2\cos 2x + \cos^2 2x\big).$
$(a - b)^2 = a^2 - 2ab + b^2$ with $a = 1$ ⁠, $b = \cos 2x$ ⁠, and $2^2 = 4$ in the denominator. The middle term is a plain cosine (fine), but the last term is still a square.
Lower the power of $\cos^2 2x$

Use $\cos^2\theta = \frac{1 + \cos 2\theta}{2}$ with $\theta = 2x$ ⁠:
$\cos^2 2x = \frac{1 + \cos 4x}{2}.$
Substituting and collecting:
$\begin{aligned} \sin^4 x &= \frac14\Big(1 - 2\cos 2x + \frac12 + \frac12\cos 4x\Big) \\ &= \frac14\Big(\frac32 - 2\cos 2x + \frac12\cos 4x\Big) \\ &= \frac38 - \frac12\cos 2x + \frac18\cos 4x. \end{aligned}$
This is the second power-lowering. The constant parts $1 + \frac12 = \frac32$ combine before multiplying by $\frac14$ ⁠. Now there are no powers at all: a constant and two single cosines.

$\cos^2\theta = \frac{1 + \cos 2\theta}{2}$ (Example 7(i), Section 7.3.2).
Integrate term by term
$\begin{aligned} &\int \sin^4 x\,dx \\ &= \frac38 x - \frac12\cdot\frac12\sin 2x + \frac18\cdot\frac14\sin 4x + C \\ &= \frac{3x}{8} - \frac14\sin 2x + \frac{1}{32}\sin 4x + C. \end{aligned}$
$\int\cos 2x\,dx = \frac12\sin 2x$ and $\int\cos 4x\,dx = \frac14\sin 4x$ ⁠. The constant term $\frac38$ gives $\frac{3x}{8}$ ⁠: $\frac38$ is the average value of $\sin^4 x$ ⁠, as the figure shows.

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

Let
$F(x) = \frac{3x}{8} - \frac14\sin 2x + \frac{1}{32}\sin 4x.$
Then
$\begin{aligned} F'(x) &= \frac38 - \frac14\cdot 2\cos 2x + \frac{1}{32}\cdot 4\cos 4x \\ &= \frac38 - \frac12\cos 2x + \frac18\cos 4x = \sin^4 x, \end{aligned}$
by Step 3. The answer is valid for every real $x$ ⁠.

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

y = \sin^4 x

0 \le x \le 2\pi

⁠. It touches 0 at multiples of

\pi

and 1 at odd multiples of

\frac{\pi}{2}

⁠, and its average value is

\frac38

(dashed line), which is why the answer begins with

\frac{3x}{8}

⁠.

Result

\int \sin^4 x\,dx = \frac{3x}{8} - \frac14\sin 2x + \frac{1}{32}\sin 4x + C,

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

Check it with numbers

The identity of Step 3 at $x = 1$ ⁠: $\sin^4 1 \approx 0.5013680$ ⁠, and $\frac38 - \frac12\cos 2 + \frac18\cos 4$ $\approx 0.375 + 0.20807342 - 0.08170545$ $\approx 0.5013680$ ⁠.

The derivative at $x = 1$ ⁠, $h = 0.001$ ⁠: $F(1.001) \approx 0.124527577209$ ⁠, $F(0.999) \approx 0.123524841119$ ⁠, slope $\approx 0.5013680$ ⁠, matching the integrand to seven significant figures.

Common mistakes

Stopping after one power-lowering. $\cos^2 2x$ is still a square and must be lowered again.
Squaring the fraction wrongly. $\big(\frac{1 - \cos 2x}{2}\big)^2$ has denominator 4, and the middle term is $-2\cos 2x$ ⁠.
Doubling the angle wrongly in the second step. $\cos^2 2x = \frac{1 + \cos 4x}{2}$ ⁠, not $\frac{1 + \cos 2x}{2}$ ⁠.
Dividing by the wrong number. $\int\frac18\cos 4x\,dx = \frac{1}{32}\sin 4x$ ⁠.

What to take away

Even powers of sine or cosine: lower the power repeatedly until only single cosines remain (Question 11 is the cosine version).
$\sin^4 x = \frac38 - \frac12\cos 2x + \frac18\cos 4x$ is worth knowing; its constant $\frac38$ is the average value of $\sin^4 x$ ⁠.
Collect constants before multiplying through to keep the arithmetic clean.

Every fact used, and where it comes from

Fact	Used in	Source
$\sin^2\theta = \frac{1 - \cos 2\theta}{2}$	Step 1	Double-angle formula (Class XI trigonometry)
$\cos^2\theta = \frac{1 + \cos 2\theta}{2}$	Step 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 · Q10: ∫sin⁡4x dx\int\sin^4x\,dx∫sin4xdx

Basics you need

Solution

Lower the power once

Expand the square

Lower the power of cos⁡22x\cos^2 2xcos22x

Integrate term by term

Check: differentiate the answer back

Every fact used, and where it comes from

Exercise 7.3 · Q10: $\int\sin^4x\,dx$

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