Chapter 7 · Integrals

Exercise 7.3 · Q5: $\int\sin^3x\cos^3x\,dx$

Exercise 7.3Integration using trigonometric identities · Section 7.3.2

The question

Find the integral of the function

\sin^3 x\cos^3 x.

Exam answer

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

Using $\sin^2 x = 1 - \cos^2 x$ ⁠,

I = \int(1 - \cos^2 x)\cos^3 x\sin x\,dx

Put $\cos x = t$ ⁠. Then $-\sin x\,dx = dt$ ⁠.

\begin{aligned} \therefore\ I &= -\int(1 - t^2)t^3\,dt \\ &= -\int(t^3 - t^5)\,dt \\ &= -\frac{t^4}{4} + \frac{t^6}{6} + C \end{aligned}

$\therefore\ I = \dfrac16\cos^6 x - \dfrac14\cos^4 x + C$ ⁠.

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

The one idea this problem is built on

When sine (or cosine) appears to an odd power, keep one factor of it aside and turn the rest into the other function with $\sin^2 x = 1 - \cos^2 x$ ⁠. The spare factor then becomes the differential of the substitution. Keeping one $\sin x$ aside and putting $t = \cos x$ turns the integrand into the polynomial $-(t^3 - t^5)$ ⁠.

Given, to find, and the plan

Given: $\sin^3 x\cos^3 x$ ⁠, defined for every real $x$ ⁠.
To find: $\displaystyle\int \sin^3 x\cos^3 x\,dx$ ⁠.
Plan: (1) Keep one $\sin x$ aside and convert $\sin^2 x$ into cosines. (2) Substitute $t = \cos x$ ⁠. (3) Integrate the polynomial and return to $x$ ⁠. (4) Differentiate back to check. (5) Show that the other choice, $t = \sin x$ ⁠, gives an equivalent answer.

Basics you need

1The Pythagorean identity

$\sin^2 x + \cos^2 x = 1$ for every $x$ ⁠. Basics: trigonometric identities for integrals

2Substitution

With $t = \cos x$ ⁠, $dt = -\sin x\,dx$ ⁠; with $t = \sin x$ ⁠, $dt = \cos x\,dx$ (Section 7.3.1). Basics: substitution

3The power formula

$\int t^n\,dt = \frac{t^{n+1}}{n+1} + C$ ⁠, $n \ne -1$ (Section 7.2). Basics: powers and the power rule

4Equivalent answers

Two antiderivatives of the same function differ by a constant (Property (II), Section 7.2.1). Basics: integration undoes differentiation

Solution

Keep one $\sin x$ aside
$\begin{aligned} \sin^3 x\cos^3 x &= \sin^2 x\cos^3 x\cdot\sin x \\ &= (1 - \cos^2 x)\cos^3 x\cdot\sin x. \end{aligned}$
Both powers are odd, so either function could give up a factor. Taking a $\sin x$ aside leaves $\sin^2 x$ ⁠, an even power, which becomes $1 - \cos^2 x$ ⁠. Now everything except the spare $\sin x$ is a function of $\cos x$ ⁠: the set-up for $t = \cos x$ ⁠. This choice leads to the textbook's form of the answer.

$\sin^2 x = 1 - \cos^2 x$ ⁠. Basics: trigonometric identities
Substitute $t = \cos x$

Then $dt = -\sin x\,dx$ ⁠, so $\sin x\,dx = -dt$ ⁠, and
$\begin{aligned} &\int \sin^3 x\cos^3 x\,dx \\ &= \int (1 - \cos^2 x)\cos^3 x\,\sin x\,dx \\ &= -\int (1 - t^2)\,t^3\,dt \\ &= -\int (t^3 - t^5)\,dt. \end{aligned}$
The spare $\sin x\,dx$ is exactly $-dt$ ⁠. The minus sign comes from $\frac{d}{dx}\cos x = -\sin x$ and must be carried all the way.

Change of variable, Section 7.3.1 (as in Example 6(i)). Lesson: integration by substitution
Integrate and return to $x$
$\begin{aligned} -\int (t^3 - t^5)\,dt &= -\Big(\frac{t^4}{4} - \frac{t^6}{6}\Big) + C \\ &= \frac{t^6}{6} - \frac{t^4}{4} + C \\ &= \frac16\cos^6 x - \frac14\cos^4 x + C. \end{aligned}$
Power formula on each term, then the minus sign flips both signs, then $t = \cos x$ goes back in.

Power formula, $n = 3$ and $n = 5$ ⁠. Basics: the power rule
Check: differentiate the answer back

With $\frac{d}{dx}\cos^k x = k\cos^{k-1}x\cdot(-\sin x)$ (chain rule):
$\begin{aligned} F'(x) &= \frac16\cdot 6\cos^5 x(-\sin x) \\ &\qquad - \frac14\cdot 4\cos^3 x(-\sin x) \\ &= \sin x\cos^3 x - \sin x\cos^5 x \\ &= \sin x\cos^3 x\,(1 - \cos^2 x) \\ &= \sin^3 x\cos^3 x. \end{aligned}$
The answer is correct for every real $x$ ⁠.

Taking out the common factor $\sin x\cos^3 x$ leaves $1 - \cos^2 x = \sin^2 x$ ⁠, which rebuilds $\sin^3 x$ ⁠.

Chain rule, Theorem 4, Section 5.3.1. Basics: integration undoes differentiation
The other choice, $t = \sin x$ ⁠, and why both are right

Keep a $\cos x$ aside instead: $\sin^3 x\cos^3 x = \sin^3 x(1 - \sin^2 x)\cos x$ ⁠. With $u = \sin x$ ⁠, $du = \cos x\,dx$ ⁠:
$\int (u^3 - u^5)\,du = \frac14\sin^4 x - \frac16\sin^6 x + C'.$
To compare, write $s = \sin^2 x$ ⁠, so $\cos^2 x = 1 - s$ ⁠:
$\begin{aligned} \frac{(1-s)^3}{6} - \frac{(1-s)^2}{4} &= \frac{1 - 3s + 3s^2 - s^3}{6} - \frac{1 - 2s + s^2}{4} \\ &= -\frac{1}{12} + \frac{s^2}{4} - \frac{s^3}{6}. \end{aligned}$
Since $s^2 = \sin^4 x$ and $s^3 = \sin^6 x$ ⁠, this says
$\begin{aligned} &\frac16\cos^6 x - \frac14\cos^4 x \\ &= \frac14\sin^4 x - \frac16\sin^6 x - \frac{1}{12}. \end{aligned}$
The two answers differ by the constant $\frac{1}{12}$ ⁠, which the arbitrary constant absorbs.

In the expansion the $s$ terms cancel ( $-\frac{3s}{6} + \frac{2s}{4} = 0$ ⁠) and the constants give $\frac16 - \frac14 = -\frac{1}{12}$ ⁠. Different substitutions often produce answers that look unrelated; Property (II) guarantees they differ by a constant, and this computation exhibits it.

Property (II), Section 7.2.1. Basics: integration undoes differentiation

Result

\int \sin^3 x\cos^3 x\,dx = \frac16\cos^6 x - \frac14\cos^4 x + C,

valid for all real $x$ ⁠; equivalently $\frac14\sin^4 x - \frac16\sin^6 x + C$ ⁠. (Textbook answer: $\dfrac16\cos^6 x - \dfrac14\cos^4 x + C$ ⁠.)

Check it with numbers

At $x = 1$ with $h = 0.001$ ⁠: $F(1.001) \approx -0.017065047564$ and $F(0.999) \approx -0.017253004082$ ⁠, so the slope is about $\dfrac{0.000187956518}{0.002} \approx 0.0939783$ ⁠. The integrand is $\sin^3 1\cos^3 1 \approx 0.0939784$ ⁠. They agree to five significant figures (the gap is the error of the difference quotient).

The constant difference at two points: at $x = 0$ the two forms are $\frac16 - \frac14 = -\frac{1}{12}$ and $0$ ⁠; at $x = \frac{\pi}{2}$ they are $0$ and $\frac14 - \frac16 = \frac{1}{12}$ ⁠. In both cases the second minus the first is $\frac{1}{12}$ ⁠.

Common mistakes

Substituting without a spare factor. Putting $t = \cos x$ needs $\sin x\,dx$ available; that is why one $\sin x$ is kept aside first.
Dropping the minus sign from $dt = -\sin x\,dx$ ⁠, which flips the whole answer.
Calling the $t = \sin x$ answer wrong. It differs from the textbook form only by $\frac{1}{12}$ ⁠.
Writing $\sin^3 x\cos^3 x = \sin^3 2x$ ⁠. In fact $\sin x\cos x = \frac12\sin 2x$ ⁠, so $\sin^3 x\cos^3 x = \frac18\sin^3 2x$ ⁠. (That gives a third route: the odd-power method of Question 4.)

What to take away

$\sin^m x\cos^n x$ with an odd power: keep one factor of the odd one aside, convert the rest, substitute the other function.
When both powers are odd, either choice works; the answers differ by a constant.
To compare two answers, write both in one variable (here $s = \sin^2 x$ ⁠) and subtract.

Every fact used, and where it comes from

Fact	Used in	Source
$\sin^2 x + \cos^2 x = 1$	Steps 1, 4, 5	Pythagorean identity (Class XI trigonometry)
Substitutions $t = \cos x$ ⁠, $u = \sin x$	Steps 2, 5	Section 7.3.1; Example 6(i)
$\int t^n\,dt = \frac{t^{n+1}}{n+1} + C$	Steps 3, 5	Standard formulae, Section 7.2
Chain rule	Step 4	Theorem 4, Section 5.3.1
Antiderivatives of one function differ by a constant	Step 5	Property (II), Section 7.2.1
$\sin x\cos x = \frac12\sin 2x$	Common mistakes	Double-angle formula (Class XI)

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 · Q5: ∫sin⁡3xcos⁡3x dx\int\sin^3x\cos^3x\,dx∫sin3xcos3xdx

Basics you need

Solution

Keep one sin⁡x\sin xsinx aside

Substitute t=cos⁡xt = \cos xt=cosx

Integrate and return to xxx

Check: differentiate the answer back

The other choice, t=sin⁡xt = \sin xt=sinx⁠, and why both are right

Every fact used, and where it comes from

Exercise 7.3 · Q5: $\int\sin^3x\cos^3x\,dx$

Keep one $\sin x$ aside

Substitute $t = \cos x$

Integrate and return to $x$

The other choice, $t = \sin x$ ⁠, and why both are right