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Chapter 7 · Integrals

Miscellaneous · Q10: ∫sin⁡8x−cos⁡8x1−2sin⁡2xcos⁡2x dx\int\frac{\sin^8x-\cos^8x}{1-2\sin^2x\cos^2x}\,dx

Miscellaneous ExerciseSimplify with identities, then integrate · Miscellaneous Exercise on Chapter 7
The question

Integrate the functions in Exercises 1 to 23.

10. sin⁡8x−cos⁡8x1−2sin⁡2xcos⁡2x\dfrac{\sin^8x - \cos^8x}{1 - 2\sin^2x\cos^2x}

(The book prints the numerator as "sin⁡8−cos⁡8x\sin^8 - \cos^8x"; the xx belongs to both terms.)

Exam answer

Let I=∫sin⁡8x−cos⁡8x1−2sin⁡2xcos⁡2x dxI = \displaystyle\int\frac{\sin^8 x - \cos^8 x}{1 - 2\sin^2 x\cos^2 x}\,dx⁠.

Write s=sin⁡xs = \sin x and c=cos⁡xc = \cos x⁠, so s2+c2=1s^2 + c^2 = 1⁠. Then

s8−c8=(s4−c4)(s4+c4)=(s2−c2)(s2+c2)×(s4+c4)=(s2−c2)(s4+c4)\begin{aligned} s^8 - c^8 &= (s^4 - c^4)(s^4 + c^4) \\ &= (s^2 - c^2)(s^2 + c^2) \\ &\quad\times(s^4 + c^4) \\ &= (s^2 - c^2)(s^4 + c^4) \end{aligned} 1−2s2c2=(s2+c2)2−2s2c2=s4+c4\begin{aligned} 1 - 2s^2c^2 &= (s^2 + c^2)^2 - 2s^2c^2 \\ &= s^4 + c^4 \end{aligned}

Cancelling s4+c4s^4 + c^4⁠, and using c2−s2=cos⁡2xc^2 - s^2 = \cos 2x⁠,

∴ I=∫(s2−c2) dx=−∫cos⁡2x dx\begin{aligned} \therefore\ I &= \int(s^2 - c^2)\,dx \\ &= -\int\cos 2x\,dx \end{aligned}

∴ I=−12sin⁡2x+C\therefore\ I = -\dfrac12\sin 2x + C (reading the numerator as sin⁡8x−cos⁡8x\sin^8 x - \cos^8 x⁠).

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

The one idea this problem is built on

The numerator is a difference of squares twice over, and the denominator is the second factor in disguise: 1−2sin⁡2xcos⁡2x1 - 2\sin^2x\cos^2x =sin⁡4x+cos⁡4x= \sin^4x + \cos^4x⁠, because 1=(sin⁡2x+cos⁡2x)21 = (\sin^2x + \cos^2x)^2⁠. After cancelling, the integrand is just sin⁡2x−cos⁡2x=−cos⁡2x\sin^2x - \cos^2x = -\cos 2x⁠. As in Q8, the method is "simplify first".

Given, to find, and the plan
Given
f(x)=sin⁡8x−cos⁡8x1−2sin⁡2xcos⁡2xf(x) = \dfrac{\sin^8x - \cos^8x}{1 - 2\sin^2x\cos^2x}⁠. Step 2 shows the denominator is never 0, so ff is defined for every real xx⁠.
To find
∫f(x) dx\displaystyle\int f(x)\,dx⁠.
Plan
(1) Factorise the numerator. (2) Rewrite the denominator. (3) Cancel and use the double-angle formula. (4) Integrate. (5) Differentiate back.

Basics you need

1Difference of squares

p2−q2=(p−q)(p+q)p^2 - q^2 = (p - q)(p + q)⁠, and (p+q)2=p2+2pq+q2(p + q)^2 = p^2 + 2pq + q^2⁠.

2Pythagorean and double-angle identities

sin⁡2x+cos⁡2x=1\sin^2x + \cos^2x = 1⁠; cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = \cos^2x - \sin^2x⁠. Basics: trigonometric identities

3∫cos⁡kx dx\int\cos kx\,dx

∫cos⁡kx dx=sin⁡kxk+C\int\cos kx\,dx = \frac{\sin kx}{k} + C for k≠0k \ne 0 (substitution t=kxt = kx and standard formula (ii)). Basics: substitution

Solution

  1. Factorise the numerator

    Write s=sin⁡xs = \sin x⁠, c=cos⁡xc = \cos x⁠. Two differences of squares:

    s8−c8=(s4−c4)(s4+c4)=(s2−c2)(s2+c2)(s4+c4)=(s2−c2)(s4+c4).\begin{aligned} s^8 - c^8 &= (s^4 - c^4)(s^4 + c^4) \\ &= (s^2 - c^2)(s^2 + c^2)(s^4 + c^4) \\ &= (s^2 - c^2)(s^4 + c^4). \end{aligned}
    An eighth power is a square of a fourth power, so the difference of squares applies; then s4−c4s^4 - c^4 is again a difference of squares. Finally s2+c2=1s^2 + c^2 = 1⁠. We do this because a high power like sin⁡8x\sin^8x has no direct integral we know, and the factorisation is what lets the denominator cancel.
    Difference of squares (Basic 1); sin⁡2x+cos⁡2x=1\sin^2x + \cos^2x = 1 (Basic 2).

  2. Rewrite the denominator

    1−2s2c2=(s2+c2)2−2s2c2=s4+2s2c2+c4−2s2c2=s4+c4.\begin{aligned} 1 - 2s^2c^2 &= (s^2 + c^2)^2 - 2s^2c^2 \\ &= s^4 + 2s^2c^2 + c^4 - 2s^2c^2 = s^4 + c^4. \end{aligned}

    Also s4+c4=1−2s2c2s^4 + c^4 = 1 - 2s^2c^2 =1−12sin⁡22x≥12= 1 - \frac12\sin^2 2x \ge \frac12⁠, so the denominator is never 0.

    Replacing 1 by (s2+c2)2(s^2 + c^2)^2 reveals that the denominator is exactly the factor s4+c4s^4 + c^4 from Step 1. Using 2sc=sin⁡2x2sc = \sin 2x⁠, 2s2c2=12sin⁡22x≤122s^2c^2 = \frac12\sin^2 2x \le \frac12⁠, so the denominator stays at least 12\frac12⁠: no point of R\R is excluded.
    (p+q)2(p + q)^2 expansion (Basic 1); sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x⁠. Basics: trigonometric identities

  3. Cancel and use the double-angle formula

    f(x)=(s2−c2)(s4+c4)s4+c4=sin⁡2x−cos⁡2x=−cos⁡2x.\begin{aligned} f(x) &= \frac{(s^2 - c^2)(s^4 + c^4)}{s^4 + c^4} \\ &= \sin^2x - \cos^2x = -\cos 2x. \end{aligned}
    The common factor s4+c4s^4 + c^4 is never 0 (Step 2), so cancelling is valid for every xx⁠. Then sin⁡2x−cos⁡2x\sin^2x - \cos^2x =−(cos⁡2x−sin⁡2x)=−cos⁡2x= -(\cos^2x - \sin^2x) = -\cos 2x⁠.
    cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = \cos^2x - \sin^2x (Basic 2).

  4. Integrate

    ∫f(x) dx=−∫cos⁡2x dx=−12sin⁡2x+C.\int f(x)\,dx = -\int\cos 2x\,dx = -\frac12\sin 2x + C.
    With t=2xt = 2x⁠, dt=2 dxdt = 2\,dx⁠: ∫cos⁡2x dx=12∫cos⁡t dt=12sin⁡t\int\cos 2x\,dx = \frac12\int\cos t\,dt = \frac12\sin t⁠. This is the textbook's answer, valid for every real xx⁠.
    Standard formula (ii), Section 7.2; substitution, Section 7.3.1.

  5. Check: differentiate the answer

    ddx(−12sin⁡2x)\dfrac{d}{dx}\Big(-\dfrac12\sin 2x\Big) =−12⋅2cos⁡2x=−cos⁡2x= -\dfrac12\cdot 2\cos 2x = -\cos 2x⁠, which is f(x)f(x) by Step 3. ✓

    The chain rule gives the inner derivative 2, cancelling the 12\frac12⁠.
    Chain rule, Chapter 5. Basics: integration undoes differentiation
Result

For every real xx⁠:

∫sin⁡8x−cos⁡8x1−2sin⁡2xcos⁡2x dx=−12sin⁡2x+C.\int\frac{\sin^8x - \cos^8x}{1 - 2\sin^2x\cos^2x}\,dx = -\frac12\sin 2x + C.

(Textbook answer: −12sin⁡2x+C-\frac12\sin 2x + C⁠, the same.)

Check it with numbers

At x=π6x = \frac\pi6⁠: sin⁡8x=1256\sin^8x = \frac{1}{256}⁠, cos⁡8x=81256\cos^8x = \frac{81}{256}⁠, so the numerator is −80256=−516-\frac{80}{256} = -\frac{5}{16}⁠. The denominator is 1−2⋅14⋅34=581 - 2\cdot\frac14\cdot\frac34 = \frac58⁠. The fraction is −516⋅85=−12-\frac{5}{16}\cdot\frac85 = -\frac12⁠, and −cos⁡π3=−12-\cos\frac\pi3 = -\frac12⁠. ✓

Area check on [0,π4][0, \frac\pi4]⁠: −12sin⁡π2+12sin⁡0=−0.5-\frac12\sin\frac\pi2 + \frac12\sin 0 = -0.5⁠. Numerical integration of the original fraction gives −0.5000000-0.5000000⁠.

Common mistakes
  1. Expanding sin⁡8x\sin^8x with power-reduction formulas. It works in principle but is very long; factorising is the intended route.
  2. Stopping at s4−c4s^4 - c^4 without factorising again, so the cancellation is missed.
  3. ∫cos⁡2x dx=sin⁡2x\int\cos 2x\,dx = \sin 2x: the factor 12\frac12 is needed.
What to take away
  1. sin⁡4x+cos⁡4x\sin^4x + \cos^4x =1−2sin⁡2xcos⁡2x= 1 - 2\sin^2x\cos^2x and sin⁡4x−cos⁡4x=sin⁡2x−cos⁡2x\sin^4x - \cos^4x = \sin^2x - \cos^2x⁠: both come from sin⁡2x+cos⁡2x=1\sin^2x + \cos^2x = 1⁠.
  2. When an integrand looks too complicated, factorise and simplify before choosing a method.

Every fact used, and where it comes from

FactUsed inSource
p2−q2=(p−q)(p+q)p^2 - q^2 = (p-q)(p+q)Step 1Algebra
sin⁡2x+cos⁡2x=1\sin^2x + \cos^2x = 1Steps 1, 2Trigonometry (Basics)
sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x⁠, cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = \cos^2x - \sin^2xSteps 2, 3Trigonometry (Class XI)
∫cos⁡x dx=sin⁡x+C\int\cos x\,dx = \sin x + CStep 4Standard formula (ii), Section 7.2
Substitution t=2xt = 2xStep 4Section 7.3.1
Chain ruleStep 5Chapter 5
All decimalsCheckRecomputed in the site's verification script
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