Chapter 7 · Integrals

Miscellaneous · Q34: $\int_{-1}^{1}x^{17}\cos^4x\,dx=0$

Miscellaneous ExerciseOdd functions on a symmetric interval (P7) · Miscellaneous Exercise

The question

Prove that

\int_{-1}^{1} x^{17}\cos^4 x\,dx = 0

Exam answer

Let $I = \displaystyle\int_{-1}^{1}x^{17}\cos^4 x\,dx$ and $f(x) = x^{17}\cos^4 x$ ⁠.

\begin{aligned} f(-x) &= (-x)^{17}\cos^4(-x) \\ &= -x^{17}\cos^4 x = -f(x) \end{aligned}

So $f$ is an odd function.

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

$\therefore\ I = 0 = \text{RHS}$ ⁠. Hence proved.

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

The one idea this problem is built on

Finding an antiderivative of $x^{17}\cos^4 x$ would take seventeen rounds of integration by parts. We do not need one. The interval $[-1, 1]$ is symmetric about 0, and the integrand is an odd function: replacing $x$ by $-x$ changes its sign. Property P7 (ii) says the integral of an odd function over $[-a, a]$ is 0, because the area to the right of the axis is cancelled exactly by an equal area, below the axis, on the left.

Given, to show, and the plan

Given: $f(x) = x^{17}\cos^4 x$ on $[-1, 1]$ ⁠.
To show: $\int_{-1}^{1} f(x)\,dx = 0$ ⁠.
Plan: (1) Continuity. (2) Show $f(-x) = -f(x)$ ⁠. (3) Apply P7 (ii) with $a = 1$ ⁠, and see why it works. (4) Picture and numerical check.

Basics you need

1Odd and even functions

$f$ is even if $f(-x) = f(x)$ and odd if $f(-x) = -f(x)$ ⁠, for every $x$ in a domain symmetric about 0. Odd powers of $x$ are odd; $\cos x$ and all its powers are even. Basics: symmetry and mirror images

2Property P7 (Section 7.10)

(i) $\int_{-a}^{a} f = 2\int_0^a f$ if $f$ is even; (ii) $\int_{-a}^{a} f = 0$ if $f$ is odd. Lesson: properties of definite integrals

3Signed area

Area below the $x$ -axis counts as negative in a definite integral. Basics: area as a sum of thin strips

Solution

The integrand is continuous on $[-1, 1]$

$x^{17}$ is a polynomial and $\cos^4 x = (\cos x)^4$ is a power of a continuous function, so their product is continuous on all of $\R$ ⁠, in particular on $[-1, 1]$ ⁠.

P7 is proved in Section 7.10 for integrals that exist in the sense of Theorem 2, so continuity on the closed interval is the condition to check first.

Lesson: definite integrals and the fundamental theorem
Show the integrand is odd

For every real $x$ ⁠,
$\begin{aligned} f(-x) &= (-x)^{17}\cos^4(-x) \\ &= (-x^{17})(\cos x)^4 \\ &= -x^{17}\cos^4 x = -f(x) . \end{aligned}$
So $f$ is an odd function.

Two facts: $(-x)^{17} = (-1)^{17}x^{17} = -x^{17}$ because 17 is odd; and $\cos(-x) = \cos x$ ⁠, so $\cos^4(-x) = \cos^4 x$ ⁠. An odd function times an even function is odd. The domain $[-1, 1]$ is symmetric about 0, which P7 also needs.

$\cos(-x) = \cos x$ (cosine is even). Basics: symmetry and mirror images
Apply P7 (ii) with $a = 1$
$\int_{-1}^{1} x^{17}\cos^4 x\,dx = 0 .$
Why P7 (ii) holds here: by P2, split at 0, and in the left part put $x = -t$ ⁠, $dx = -dt$ (limits $-1 \to 1$ ⁠, $0 \to 0$ ⁠):
$\begin{aligned} \int_{-1}^{0} f(x)\,dx &= \int_{1}^{0} f(-t)(-dt) \\ &= \int_0^1 f(-t)\,dt \\ &= -\int_0^1 f(t)\,dt . \end{aligned}$
So the left part is exactly the negative of the right part $\int_0^1 f$ ⁠, and the two add to 0. This is the required result.

This repeats the book's proof of P7 for our $f$ ⁠, so you can see that nothing more than oddness and a symmetric interval is used. The minus from $dx = -dt$ is undone by swapping the limits back (P1), and the minus from $f(-t) = -f(t)$ is what produces the cancellation.

Section 7.10: P2 (splitting), P1 (swapping limits), P0 (renaming the variable), P7 (ii). Lesson: properties of definite integrals
Picture: two equal and opposite areas

$y = x^{17}\cos^4 x$ on $[-1, 1]$ (vertical scale stretched 6 times compared with the horizontal). The curve is almost flat near 0 because $x^{17}$ is tiny there, and it is symmetric about the origin. The area on the right, about 0.006608, is above the axis; the matching area on the left is below it and counts as $-0.006608$ ⁠. They cancel.

An odd function's graph is unchanged by a half-turn about the origin. A half-turn carries the region above the axis on $[0, 1]$ onto a region of the same size below the axis on $[-1, 0]$ ⁠, and signed areas below the axis are negative.

Basics: area as a sum of thin strips

Result

$f(x) = x^{17}\cos^4 x$ is continuous and odd, and $[-1, 1]$ is symmetric about 0, so by P7 (ii) $\int_{-1}^{1} x^{17}\cos^4 x\,dx = 0,$ as was to be proved.

Check it with numbers

Numerical integration (adaptive quadrature) gives $\int_0^1 x^{17}\cos^4 x\,dx \approx 0.0066078909$ and $\int_{-1}^{0} x^{17}\cos^4 x\,dx \approx -0.0066078909$ ⁠; the whole integral is 0 to all digits computed.

Oddness at a sample point: $0.9^{17} \approx 0.166772$ and $\cos^4 0.9 \approx 0.149304$ ⁠, so $f(0.9) \approx 0.024900$ ⁠, while $f(-0.9)$ $\approx (-0.166772)(0.149304)$ $\approx -0.024900$ ✓.

Common mistakes

Calling the integrand even because $\cos^4 x$ is even. The factor $x^{17}$ is odd, and odd times even is odd.
Using P7 on a non-symmetric interval, such as $[0, 1]$ or $[-1, 2]$ ⁠. The cancellation needs $[-a, a]$ ⁠.
Attempting seventeen integrations by parts. Always look for symmetry before integrating.

What to take away

On $[-a, a]$ ⁠, test $f(-x)$ first: odd means the integral is 0; even means it is $2\int_0^a f$ ⁠.
Odd $\times$ even = odd; odd $\times$ odd = even; even $\times$ even = even (just like the signs $-\times+$ ⁠, $-\times-$ ⁠, $+\times+$ ⁠).

Every fact used, and where it comes from

Fact	Used in	Source
Continuity of polynomials, $\cos$ ⁠, products	Step 1	Chapter 5, Section 5.2
$\cos(-x) = \cos x$ ⁠; $(-x)^{17} = -x^{17}$	Step 2	Trigonometric functions (Class XI); laws of exponents
P7 (ii): $\int_{-a}^{a} f = 0$ for odd $f$	Step 3	Section 7.10, property P7 (ii) (Lesson: properties)
P0, P1, P2 and substitution, used in the proof of P7	Step 3	Section 7.10 (Proof of P7); Section 7.9
Signed area	Step 4	Section 7.8.1 (Basics: area)
All decimals	Step 4, Check	Recomputed (mpmath quadrature, 30 digits) in the site's verification script

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Miscellaneous · Q34: ∫−11x17cos⁡4x dx=0\int_{-1}^{1}x^{17}\cos^4x\,dx=0∫−11​x17cos4xdx=0

Basics you need

Solution

The integrand is continuous on [−1,1][-1, 1][−1,1]

Show the integrand is odd

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

Picture: two equal and opposite areas

Every fact used, and where it comes from

Miscellaneous · Q34: $\int_{-1}^{1}x^{17}\cos^4x\,dx=0$

The integrand is continuous on $[-1, 1]$

Apply P7 (ii) with $a = 1$