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Chapter 7 · Integrals

Exercise 7.10 · Q13: ∫−π/2π/2sin⁡7x dx\int_{-\pi/2}^{\pi/2}\sin^7x\,dx

Exercise 7.10Properties of definite integrals (P7) · Section 7.10
The question

By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

13. ∫−π/2π/2sin⁡7x dx\displaystyle\int_{-\pi/2}^{\pi/2}\sin^7 x\,dx

Exam answer

Let I=∫−π/2π/2sin⁡7x dxI = \displaystyle\int_{-\pi/2}^{\pi/2}\sin^7 x\,dx⁠.

Let f(x)=sin⁡7xf(x) = \sin^7 x⁠. Then

f(−x)=(sin⁡(−x))7=(−sin⁡x)7=−sin⁡7x=−f(x)\begin{aligned} f(-x) &= \left(\sin(-x)\right)^7 \\ &= (-\sin x)^7 = -\sin^7 x \\ &= -f(x) \end{aligned}

So ff is an odd function.

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

∴ I=0\therefore\ I = 0⁠.

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

The one idea this problem is built on

The interval [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}] is symmetric about 0, and sin⁡7x\sin^7 x is an odd function: changing xx to −x-x changes its sign. So the graph on the left is the graph on the right turned upside down, and the area below the axis on the left cancels the area above it on the right. Property P7 (ii) says exactly this, and no antiderivative is needed.

Given, to find, and the plan
Given
f(x)=sin⁡7xf(x) = \sin^7 x on [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]⁠.
To find
I=∫−π/2π/2sin⁡7x dxI = \displaystyle\int_{-\pi/2}^{\pi/2}\sin^7 x\,dx⁠.
Plan
(1) Continuity. (2) Show f(−x)=−f(x)f(-x) = -f(x)⁠. (3) Apply P7 (ii) with a=π2a = \frac{\pi}{2}⁠. (4) Confirm by computing each half directly.

Basics you need

1Odd and even functions

ff is odd if f(−x)=−f(x)f(-x) = -f(x) for every xx⁠, and even if f(−x)=f(x)f(-x) = f(x)⁠. sin⁡\sin is odd; an odd power of an odd function is odd. Basics: symmetry and mirror images

2Property P7 (Section 7.10)

(i) ∫−aaf(x) dx=2∫0af(x) dx\int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx if ff is even; (ii) ∫−aaf(x) dx=0\int_{-a}^{a} f(x)\,dx = 0 if ff is odd. Lesson: properties of definite integrals

3Signed area

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

4Substitution in a definite integral (Section 7.9)

Used only in the check of Step 4. Basics: substitution

Solution

  1. The integrand is continuous

    sin⁡x\sin x is continuous for every real xx⁠, and so is any power of it. So f(x)=sin⁡7xf(x) = \sin^7 x is continuous on [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}] and the integral is an ordinary definite integral.

    P7, like every property in Section 7.10, is stated for functions continuous on the closed interval. Here nothing can go wrong, but it is a habit worth keeping: in Q20 the same interval carries a function that is not continuous there.
    Products (hence powers) of continuous functions are continuous (Chapter 5, Section 5.2.1).

  2. Show that ff is odd

    For every real xx⁠, using sin⁡(−x)=−sin⁡x\sin(-x) = -\sin x⁠:

    f(−x)=(sin⁡(−x))7=(−sin⁡x)7=(−1)7sin⁡7x=−sin⁡7x=−f(x).\begin{aligned}f(-x) &= \big(\sin(-x)\big)^7 = (-\sin x)^7 \\ &= (-1)^7\sin^7 x = -\sin^7 x = -f(x) .\end{aligned}
    This is the condition of P7 (ii). It has to hold for every xx in the interval, and it does, because the calculation used nothing about xx⁠. The exponent 7 is odd, so (−1)7=−1(-1)^7 = -1 keeps the minus sign; with an even power such as sin⁡6x\sin^6 x the sign would disappear and the function would be even instead.
    sin⁡(−x)=−sin⁡x\sin(-x) = -\sin x⁠: the point at angle −x-x on the unit circle is the reflection in the horizontal axis of the point at angle xx⁠. Basics: symmetry

  3. Apply P7 (ii) with a=π2a = \frac{\pi}{2}

    The interval has the form [−a,a][-a, a] with a=π2a = \frac{\pi}{2}⁠, and ff is odd (Step 2) and continuous (Step 1). By P7 (ii):

    I=∫−π/2π/2sin⁡7x dx=0.I = \int_{-\pi/2}^{\pi/2}\sin^7 x\,dx = 0 .
    x y −π/2 π/2 1 −1 +16/35 −16/35 blue: y = sin⁷x, an odd function green area above the axis = rose area below it total signed area: 16/35 − 16/35 = 0
    y=sin⁡7xy = \sin^7 x on [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]⁠. The rose region on the left is the green region on the right turned upside down, so their signed areas +1635+\frac{16}{35} and −1635-\frac{16}{35} cancel.
    The book proves P7 by splitting at 0 with P2 and putting t=−xt = -x in the left half; for an odd function the left half then becomes minus the right half. The figure shows the same thing: every strip at xx above the axis is matched by a strip of the same size below the axis at −x-x⁠.
    P7 (ii), Section 7.10. Lesson: properties of definite integrals

  4. Check: compute each half directly

    On [0,π2][0, \frac{\pi}{2}] write sin⁡7x=(1−cos⁡2x)3sin⁡x\sin^7 x = (1 - \cos^2 x)^3\sin x and put t=cos⁡xt = \cos x⁠, so dt=−sin⁡x dxdt = -\sin x\,dx⁠. When x=0x = 0⁠, t=1t = 1⁠; when x=π2x = \frac{\pi}{2}⁠, t=0t = 0⁠. Then

    ∫0π/2sin⁡7x dx=∫01(1−t2)3 dt=∫01(1−3t2+3t4−t6) dt=1−1+35−17=1635.\begin{aligned}\int_0^{\pi/2}\sin^7 x\,dx &= \int_0^1 (1 - t^2)^3\,dt \\ &= \int_0^1 (1 - 3t^2 + 3t^4 - t^6)\,dt \\ &= 1 - 1 + \frac{3}{5} - \frac{1}{7} = \frac{16}{35} .\end{aligned}

    By Step 2 and the substitution x→−xx \to -x⁠, the left half is the negative of this, −1635-\frac{16}{35}⁠. The two halves add to 0, as P7 (ii) said.

    This confirms the answer by a route that does not use P7. The substitution is valid on the whole interval: t=cos⁡xt = \cos x is continuous and decreasing on [0,π2][0, \frac{\pi}{2}]⁠. The sign change dt=−sin⁡x dxdt = -\sin x\,dx is absorbed by swapping the limits (P1).
    sin⁡2x=1−cos⁡2x\sin^2 x = 1 - \cos^2 x⁠; (1−u)3=1−3u+3u2−u3(1 - u)^3 = 1 - 3u + 3u^2 - u^3 with u=t2u = t^2⁠; power rule. Basics: powers of x
Result

∫−π/2π/2sin⁡7x dx=0\int_{-\pi/2}^{\pi/2}\sin^7 x\,dx = 0 (Textbook answer: 0.)

Check it with numbers

Numerical integration over [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}] gives 0 (to more than 25 decimal places). Over the right half it gives 0.457142857…0.457142857\ldots⁠, and 1635=0.457142857…\frac{16}{35} = 0.457142857\ldots⁠.

Oddness at one point: sin⁡7(0.5)≈0.005822\sin^7(0.5) \approx 0.005822 and sin⁡7(−0.5)≈−0.005822\sin^7(-0.5) \approx -0.005822⁠.

Common mistakes
  1. Using P7 (i) instead of (ii), giving 2×1635=32352 \times \frac{16}{35} = \frac{32}{35}⁠. Check whether the function is odd or even before choosing.
  2. Thinking an odd power makes an even function. Odd power of an odd function: odd. Even power of an odd function: even.
  3. Applying P7 on an interval that is not symmetric, such as [0,π][0, \pi]⁠. P7 needs limits −a-a and aa⁠.
What to take away
  1. Before integrating over [−a,a][-a, a]⁠, test the integrand for oddness: f(−x)=−f(x)f(-x) = -f(x) gives 0 at once.
  2. The same reasoning appears in Example 31 (sin⁡5xcos⁡4x\sin^5 x\cos^4 x⁠), and in Q20, where only part of the integrand is odd.

Every fact used, and where it comes from

FactUsed inSource
Powers of continuous functions are continuousStep 1Chapter 5, Section 5.2.1
sin⁡(−x)=−sin⁡x\sin(-x) = -\sin xStep 2Class XI trigonometry
P7 (ii): ∫−aaf(x) dx=0\int_{-a}^{a} f(x)\,dx = 0 for odd ffStep 3Section 7.10, property P7 (Lesson: properties)
Substitution in a definite integral; P1Step 4Section 7.9; Section 7.10, property P1
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