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Chapter 7 · Integrals

Miscellaneous · Q12: ∫x31−x8 dx\int\frac{x^3}{\sqrt{1-x^8}}\,dx

Miscellaneous ExerciseSubstitution t = x⁴, then the sin⁻¹ formula · Miscellaneous Exercise on Chapter 7
The question

Integrate the functions in Exercises 1 to 23.

12. x31−x8\dfrac{x^3}{\sqrt{1 - x^8}}

Exam answer

Let I=∫x3 dx1−x8I = \displaystyle\int\frac{x^3\,dx}{\sqrt{1 - x^8}}⁠.

Put x4=tx^4 = t⁠. Then 4x3 dx=dt4x^3\,dx = dt⁠.

I=14∫dt1−t2=14sin⁡−1t+CI = \frac14\int\frac{dt}{\sqrt{1 - t^2}} = \frac14\sin^{-1}t + C

∴ I=14sin⁡−1(x4)+C\therefore\ I = \dfrac14\sin^{-1}(x^4) + C⁠.

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

The one idea this problem is built on

See x8x^8 as (x4)2(x^4)^2⁠. The numerator x3x^3 is, up to the factor 4, the derivative of x4x^4⁠. So the substitution t=x4t = x^4 turns the integral into a multiple of the standard form ∫dt1−t2=sin⁡−1t+C\int\frac{dt}{\sqrt{1 - t^2}} = \sin^{-1}t + C⁠.

Given, to find, and the plan
Given
f(x)=x31−x8f(x) = \dfrac{x^3}{\sqrt{1 - x^8}}⁠, defined where 1−x8>01 - x^8 > 0⁠, that is −1<x<1-1 < x < 1⁠.
To find
∫x31−x8 dx\displaystyle\int\frac{x^3}{\sqrt{1 - x^8}}\,dx⁠.
Plan
(1) Domain and why t=x4t = x^4⁠. (2) Substitute. (3) Integrate with standard formula (viii). (4) Differentiate back.

Basics you need

1Substitution

If t=g(x)t = g(x) then dt=g′(x) dxdt = g'(x)\,dx (Section 7.3.1). Basics: substitution

2Standard formula (viii)

∫dx1−x2=sin⁡−1x+C\int\frac{dx}{\sqrt{1 - x^2}} = \sin^{-1}x + C for −1<x<1-1 < x < 1⁠. Basics: integration undoes differentiation

Solution

  1. Domain, and why t=x4t = x^4

    The root needs 1−x8>01 - x^8 > 0 (strictly, since it is a denominator), i.e. x8<1x^8 < 1⁠, i.e. −1<x<1-1 < x < 1⁠. Because x8=(x4)2x^8 = (x^4)^2 and ddxx4=4x3\frac{d}{dx}x^4 = 4x^3⁠, the integrand has the form 14⋅g′(x)1−g(x)2\frac14\cdot\frac{g'(x)}{\sqrt{1 - g(x)^2}} with g(x)=x4g(x) = x^4⁠.

    The power 3 outside is one less than the power 4 that is squared inside the root: the tell-tale sign of a substitution. The substitution does not need to be one-one here; we only use dt=4x3 dxdt = 4x^3\,dx and then return to xx (the "t=g(x)t = g(x)" direction of Section 7.3.1).
    Section 7.3.1. Lesson: integration by substitution

  2. Substitute

    Put t=x4t = x^4⁠, dt=4x3 dxdt = 4x^3\,dx⁠, so x3 dx=14 dtx^3\,dx = \frac14\,dt⁠:

    ∫x3 dx1−x8=14∫dt1−t2.\int\frac{x^3\,dx}{\sqrt{1 - x^8}} = \frac14\int\frac{dt}{\sqrt{1 - t^2}}.
    For −1<x<1-1 < x < 1⁠, t=x4t = x^4 lies in [0,1)[0, 1)⁠, where 1−t2\sqrt{1 - t^2} is real and positive, so the new integral makes sense.
    Substitution (Basic 1).

  3. Integrate with standard formula (viii)

    14∫dt1−t2=14sin⁡−1t+C=14sin⁡−1(x4)+C.\begin{aligned} \frac14\int\frac{dt}{\sqrt{1 - t^2}} &= \frac14\sin^{-1}t + C \\ &= \frac14\sin^{-1}(x^4) + C. \end{aligned}
    Standard formula (viii), then t=x4t = x^4⁠. This is exactly the textbook's answer, valid on −1<x<1-1 < x < 1⁠. (Formula (5) of Section 7.4 with a=1a = 1 says the same.)
    Standard formula (viii), Section 7.2 (Basic 2).

  4. Check: differentiate the answer

    Chain rule with u=x4u = x^4⁠, u′=4x3u' = 4x^3⁠:

    ddx(14sin⁡−1(x4))=14⋅4x31−(x4)2=x31−x8. ✓\begin{aligned} \frac{d}{dx}\Big(\frac14\sin^{-1}(x^4)\Big) &= \frac14\cdot\frac{4x^3}{\sqrt{1 - (x^4)^2}} \\ &= \frac{x^3}{\sqrt{1 - x^8}}. \ \checkmark \end{aligned}
    The 4 from the inner derivative cancels the 14\frac14⁠, and (x4)2=x8(x^4)^2 = x^8 restores the integrand on the whole interval −1<x<1-1 < x < 1⁠.
    Derivative of sin⁡−1u\sin^{-1}u⁠, Chapter 5.
Result

For −1<x<1-1 < x < 1⁠:

∫x31−x8 dx=14sin⁡−1(x4)+C.\int\frac{x^3}{\sqrt{1 - x^8}}\,dx = \frac14\sin^{-1}(x^4) + C.

(Textbook answer: 14sin⁡−1(x4)+C\frac14\sin^{-1}(x^4) + C⁠, the same.)

Check it with numbers

Integrate from 0 to 2−1/4≈0.84089642^{-1/4} \approx 0.8408964⁠, where x4=12x^4 = \frac12⁠: 14(sin⁡−112−sin⁡−10)\frac14\big(\sin^{-1}\frac12 - \sin^{-1}0\big) =π24≈0.1308997= \frac{\pi}{24} \approx 0.1308997⁠. Numerical integration over the same interval also gives 0.13089970.1308997⁠.

Common mistakes
  1. Trying t=x8t = x^8: its derivative 8x78x^7 is not in the integrand.
  2. Forgetting the 14\frac14 from dt=4x3 dxdt = 4x^3\,dx⁠.
  3. Writing sin⁡−1(x8)\sin^{-1}(x^8): the square under the root is (x4)2(x^4)^2⁠, so the answer involves x4x^4⁠.
What to take away
  1. x2nx^{2n} under a root can be read as (xn)2(x^n)^2⁠; if xn−1x^{n-1} stands outside, put t=xnt = x^n⁠.
  2. After substituting, match the result with the standard list: here 11−t2\frac{1}{\sqrt{1 - t^2}}⁠.

Every fact used, and where it comes from

FactUsed inSource
x8<1  ⟺  −1<x<1x^8 < 1 \iff -1 < x < 1Step 1Algebra
Substitution t=x4t = x^4⁠, dt=4x3 dxdt = 4x^3\,dxSteps 1, 2Section 7.3.1
∫dx1−x2=sin⁡−1x+C\int\frac{dx}{\sqrt{1 - x^2}} = \sin^{-1}x + CStep 3Standard formula (viii), Section 7.2
Derivative of sin⁡−1u\sin^{-1}u⁠; chain ruleStep 4Chapter 5
All decimalsCheckRecomputed in the site's verification script
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