Chapter 7 · Integrals

Example 31: $\int_{-1}^{1}\sin^5x\cos^4x\,dx$

ExamplesProperties of definite integrals · Section 7.10

The question

Evaluate $\displaystyle\int_{-1}^{1} \sin^5 x\,\cos^4 x\; dx$ ⁠.

Exam answer

Let $I = \displaystyle\int_{-1}^{1}\sin^5 x\cos^4 x\,dx$ ⁠.

Let $f(x) = \sin^5 x\cos^4 x$ ⁠. Then

\begin{aligned} f(-x) &= \sin^5(-x)\cos^4(-x) \\ &= (-\sin x)^5(\cos x)^4 \\ &= -\sin^5 x\cos^4 x = -f(x) \end{aligned}

So $f$ is an odd function.

Using $\displaystyle\int_{-a}^{a} f(x)\,dx = 0$ when $f$ is odd,

$\therefore\ I = 0$ ⁠.

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

The one idea this problem is built on

The interval $[-1, 1]$ is symmetric about 0, and the integrand is an odd function: $f(-x) = -f(x)$ ⁠. Its graph on the left is the graph on the right turned upside down, so the signed area below the axis on $[-1, 0]$ exactly cancels the area above it on $[0, 1]$ ⁠. Property P7 (ii) states this: $\int_{-a}^{a} f(x)\,dx = 0$ for odd $f$ ⁠. No antiderivative is needed.

Given, to find, and the plan

Given: $f(x) = \sin^5 x\cos^4 x$ on $[-1, 1]$ (the limits are $\pm 1$ radian).
To find: $I = \int_{-1}^{1} f(x)\,dx$ ⁠.
Plan: Show $f$ is odd and continuous; apply P7 (ii) with $a = 1$ ⁠; confirm by integrating directly.

Basics you need

1Odd and even functions

$f$ is odd if $f(-x) = -f(x)$ for all $x$ ⁠, even if $f(-x) = f(x)$ ⁠. $\sin$ is odd and $\cos$ is even. An odd power of an odd function is odd; any power of an even function is even; odd times even is odd. Basics: symmetry and mirror images

2Property P7 (ii)

$\int_{-a}^{a} f(x)\,dx = 0$ if $f$ is odd (Section 7.10). Lesson: properties of definite integrals

3Signed area

A definite integral counts area above the x-axis as positive and area below as negative. Basics: area as a sum of thin strips

Solution

Show that $f$ is odd

Use $\sin(-x) = -\sin x$ and $\cos(-x) = \cos x$ ⁠:
$\begin{aligned}f(-x) &= \sin^5(-x)\,\cos^4(-x) \\ &= (-\sin x)^5(\cos x)^4 \\ &= -\sin^5 x\cos^4 x = -f(x).\end{aligned}$
This holds for every real $x$ ⁠, so $f$ is odd. It is also continuous everywhere (products of powers of $\sin$ and $\cos$ ⁠).

The fifth power keeps the minus sign, $(-1)^5 = -1$ ⁠, while the fourth power of $\cos$ is unaffected because $\cos$ is even. One odd factor times one even factor gives an odd product.

$\sin$ is odd, $\cos$ is even. Basics: symmetry
Apply P7 (ii) with $a = 1$

The interval $[-1, 1]$ has the form $[-a, a]$ with $a = 1$ ⁠, and $f$ is odd (Step 1). Therefore
$I = \int_{-1}^{1}\sin^5 x\cos^4 x\,dx = 0.$
The proof of P7 splits at 0 by P2 and substitutes $t = -x$ in the left part, which turns $\int_{-a}^0 f$ into $\int_0^a f(-x)\,dx = -\int_0^a f(x)\,dx$ ⁠. The two halves cancel. The limits here are 1 and $-1$ radian, not $\pm\pi$ ⁠; P7 only needs the interval to be symmetric about 0.

P7 (ii), Section 7.10. Lesson: properties of definite integrals
The picture

$y = \sin^5 x\cos^4 x$ on $[-1, 1]$ ⁠, with the vertical scale stretched. The blue region on $[0, 1]$ (area A) and the rose region on $[-1, 0]$ (signed area $-A$ ⁠) are mirror images through the origin, so their sum is 0.

"Odd" means symmetric through the origin: rotate the right half by a half-turn about $(0, 0)$ and it lands on the left half. Area above the axis becomes area below, so the signed areas cancel.

Signed area: regions below the axis count as negative. Basics: area
Check by integrating directly

Write $\sin^5 x = \sin x\left(1 - \cos^2 x\right)^2$ and put $u = \cos x$ ⁠, $du = -\sin x\,dx$ ⁠:
$\int\sin^5 x\cos^4 x\,dx = -\int(1 - u^2)^2u^4\,du = -\int\left(u^4 - 2u^6 + u^8\right)du.$
So $F(x) = -\frac{\cos^5 x}{5} + \frac{2\cos^7 x}{7} - \frac{\cos^9 x}{9}$ is an antiderivative. It contains only powers of $\cos x$ ⁠, which is even, so $F(-1) = F(1)$ and
$I = F(1) - F(-1) = 0.$
This confirms P7 (ii) independently. The antiderivative of an odd function can be chosen even, and an even function has the same value at $1$ and $-1$ ⁠.

Substitution, Section 7.3.1, and $\sin^2 x = 1 - \cos^2 x$ ⁠. Basics: trigonometric identities

Result

$\int_{-1}^{1}\sin^5 x\cos^4 x\,dx = 0,$ the same as the book's answer.

Check it with numbers

Numerically, $\int_0^1 \sin^5 x\cos^4 x\,dx \approx 0.0195923$ and $\int_{-1}^0 \sin^5 x\cos^4 x\,dx \approx -0.0195923$ ⁠; their sum is 0. At a sample point, $f(0.5) \approx 0.0150230$ and $f(-0.5) \approx -0.0150230$ ⁠.

Common mistakes

Thinking the answer 0 means there is no region. There are two regions; they have equal size and opposite signs.
Calling $f$ even because $\cos^4 x$ is even. One odd factor ( $\sin^5 x$ ⁠) makes the whole product odd.
Using P7 on a non-symmetric interval. P7 needs limits $-a$ and $a$ ⁠; on $[0, 1]$ the integral is not 0.

What to take away

Before integrating over $[-a, a]$ ⁠, test the parity of the integrand: odd gives 0 at once.
Parity rules: odd $\times$ even = odd, odd $\times$ odd = even, even $\times$ even = even.
A zero integral is a statement about signed area, not about the graph being flat.

Every fact used, and where it comes from

Fact	Used in	Source
$\sin(-x) = -\sin x$ ⁠, $\cos(-x) = \cos x$	Step 1	Trigonometry (Basics)
P7 (ii): $\int_{-a}^a f = 0$ for odd $f$	Step 2	Section 7.10 (Lesson)
Signed area	Step 3	Area meaning of the integral (the book's area function in Section 7.8.1 assumes $f > 0$ and says the results hold for other functions too); parts below the axis count as negative (Basics)
$u = \cos x$ ⁠, $\sin^2 x = 1 - \cos^2 x$	Step 4	Section 7.3.1 (Lesson)

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 →

Example 31: ∫−11sin⁡5xcos⁡4x dx\int_{-1}^{1}\sin^5x\cos^4x\,dx∫−11​sin5xcos4xdx

Basics you need

Solution

Show that fff is odd

Apply P7 (ii) with a=1a = 1a=1

The picture

Check by integrating directly

Every fact used, and where it comes from

Example 31: $\int_{-1}^{1}\sin^5x\cos^4x\,dx$

Show that $f$ is odd

Apply P7 (ii) with $a = 1$