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Chapter 7 · Integrals

Example 40: ∫sin⁡2xcos⁡2x9−cos⁡42x dx\int\frac{\sin 2x\cos 2x}{\sqrt{9-\cos^4 2x}}\,dx

ExamplesSubstitution and special integrals · Miscellaneous Examples
The question

Find

∫sin⁡2xcos⁡2x dx9−cos⁡4(2x)\int\frac{\sin 2x\cos 2x\,dx}{\sqrt{9 - \cos^4(2x)}}
Exam answer

Let I=∫sin⁡2xcos⁡2x dx9−cos⁡42xI = \displaystyle\int\frac{\sin 2x\cos 2x\,dx}{\sqrt{9 - \cos^4 2x}}⁠.

Put cos⁡22x=t\cos^2 2x = t⁠. Then −4sin⁡2xcos⁡2x dx=dt-4\sin 2x\cos 2x\,dx = dt⁠.

I=−14∫dt9−t2=−14∫dt32−t2\begin{aligned} I &= -\frac14\int\frac{dt}{\sqrt{9 - t^2}} \\ &= -\frac14\int\frac{dt}{\sqrt{3^2 - t^2}} \end{aligned}

Using ∫dta2−t2=sin⁡−1ta+C\displaystyle\int\frac{dt}{\sqrt{a^2 - t^2}} = \sin^{-1}\frac{t}{a} + C⁠,

I=−14sin⁡−1t3+CI = -\frac14\sin^{-1}\frac{t}{3} + C

∴ I=−14sin⁡−1(13cos⁡22x)+C\therefore\ I = -\dfrac14\sin^{-1}\left(\dfrac13\cos^2 2x\right) + C⁠.

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

The one idea this problem is built on

Under the root, cos⁡42x=(cos⁡22x)2\cos^4 2x = (\cos^2 2x)^2 is the square of cos⁡22x\cos^2 2x⁠. And the numerator sin⁡2xcos⁡2x\sin 2x\cos 2x is, apart from a constant, the derivative of cos⁡22x\cos^2 2x⁠. So with t=cos⁡22xt = \cos^2 2x the integral becomes a constant times ∫dt9−t2\int\frac{dt}{\sqrt{9 - t^2}}⁠, the standard inverse-sine form ∫dta2−t2\int\frac{dt}{\sqrt{a^2 - t^2}} with a=3a = 3⁠.

Given, to find, and the plan
Given
The integrand sin⁡2xcos⁡2x9−cos⁡42x\frac{\sin 2x\cos 2x}{\sqrt{9 - \cos^4 2x}}⁠.
To find
Its indefinite integral.
Plan
(1) Domain. (2) Substitute t=cos⁡22xt = \cos^2 2x and find dtdt⁠. (3) Use ∫dta2−t2=sin⁡−1ta\int\frac{dt}{\sqrt{a^2 - t^2}} = \sin^{-1}\frac{t}{a}⁠. (4) Back to xx⁠. (5) Check by differentiating.

Basics you need

1Chain rule for cos⁡22x\cos^2 2x

cos⁡22x=(cos⁡2x)2\cos^2 2x = (\cos 2x)^2 has three layers: squaring, cos⁡\cos⁠, and 2x2x⁠. Its derivative is 2cos⁡2x⋅(−sin⁡2x)⋅22\cos 2x\cdot(-\sin 2x)\cdot 2⁠. Basics: substitution (the chain rule backwards)

2Section 7.4, formula (5)

∫dta2−t2=sin⁡−1ta+C\int\frac{dt}{\sqrt{a^2 - t^2}} = \sin^{-1}\frac{t}{a} + C⁠, for −a<t<a-a < t < a⁠. Lesson: integrals of particular functions

3Derivative of sin⁡−1\sin^{-1}

ddusin⁡−1u=11−u2\frac{d}{du}\sin^{-1}u = \frac{1}{\sqrt{1 - u^2}} for −1<u<1-1 < u < 1⁠. Basics: integration undoes differentiation

Solution

  1. The integrand is defined for every xx

    Since 0≤cos⁡42x≤10 \le \cos^4 2x \le 1⁠, we have 9−cos⁡42x≥8>09 - \cos^4 2x \ge 8 > 0⁠. So the square root is always defined and never zero, and the integrand is continuous on the whole real line.

    No domain restriction is needed, so the final answer will hold for every real xx⁠. It is worth checking anyway: with 1−cos⁡42x1 - \cos^4 2x in place of 9−cos⁡42x9 - \cos^4 2x⁠, the denominator would vanish wherever cos⁡2x=±1\cos 2x = \pm1⁠.

  2. Substitute t=cos⁡22xt = \cos^2 2x

    By the chain rule,

    dtdx=2cos⁡2x⋅(−sin⁡2x)⋅2=−4sin⁡2xcos⁡2x,\frac{dt}{dx} = 2\cos 2x\cdot(-\sin 2x)\cdot 2 = -4\sin 2x\cos 2x ,

    so dt=−4sin⁡2xcos⁡2x dxdt = -4\sin 2x\cos 2x\,dx⁠, that is,

    sin⁡2xcos⁡2x dx=−14 dt.\sin 2x\cos 2x\,dx = -\frac14\,dt .

    Also cos⁡42x=(cos⁡22x)2=t2\cos^4 2x = (\cos^2 2x)^2 = t^2⁠. Therefore

    I=∫−14 dt9−t2=−14∫dt9−t2.I = \int\frac{-\frac14\,dt}{\sqrt{9 - t^2}} = -\frac14\int\frac{dt}{\sqrt{9 - t^2}} .
    The numerator, with dxdx⁠, is exactly −14 dt-\frac14\,dt⁠, and the denominator depends only on tt⁠. So no xx is left: the substitution works. Notice that t=cos⁡22xt = \cos^2 2x lies in [0,1][0, 1]⁠, well inside (−3,3)(-3, 3)⁠, where the standard formula holds.
    dduu2=2u\frac{d}{du}u^2 = 2u⁠; ddxcos⁡2x=−2sin⁡2x\frac{d}{dx}\cos 2x = -2\sin 2x⁠. Lesson: integration by substitution

  3. Use the standard inverse-sine integral

    With a=3a = 3 (9=329 = 3^2⁠):

    −14∫dt32−t2=−14sin⁡−1t3+C.-\frac14\int\frac{dt}{\sqrt{3^2 - t^2}} = -\frac14\sin^{-1}\frac{t}{3} + C .
    Section 7.4, formula (5). The −14-\frac14 is a constant factor and stays in front (Property IV, Section 7.2.1).

  4. Back to xx

    I=−14sin⁡−1(13cos⁡22x)+C.I = -\frac14\sin^{-1}\left(\frac13\cos^2 2x\right) + C .
    Replace tt by cos⁡22x\cos^2 2x⁠. The argument 13cos⁡22x\frac13\cos^2 2x lies in [0,13][0, \frac13]⁠, inside the domain [−1,1][-1, 1] of sin⁡−1\sin^{-1}⁠.

  5. Check: ddx\frac{d}{dx} of our answer

    Let u=13cos⁡22xu = \frac13\cos^2 2x⁠. By Step 2, dudx=13(−4sin⁡2xcos⁡2x)\frac{du}{dx} = \frac13(-4\sin 2x\cos 2x)⁠, that is, dudx=−43sin⁡2xcos⁡2x\frac{du}{dx} = -\frac43\sin 2x\cos 2x⁠. Then

    ddx[−14sin⁡−1u]=−14⋅11−u2⋅(−43sin⁡2xcos⁡2x).\begin{aligned}&\frac{d}{dx}\left[-\frac14\sin^{-1}u\right]\\ &= -\frac14\cdot\frac{1}{\sqrt{1 - u^2}}\cdot\left(-\frac43\sin 2x\cos 2x\right) .\end{aligned}

    Here

    1−u2=1−cos⁡42x9=139−cos⁡42x.\sqrt{1 - u^2} = \sqrt{1 - \frac{\cos^4 2x}{9}} = \frac13\sqrt{9 - \cos^4 2x} .

    So the derivative is

    13sin⁡2xcos⁡2x⋅39−cos⁡42x=sin⁡2xcos⁡2x9−cos⁡42x,\frac{1}{3}\sin 2x\cos 2x\cdot\frac{3}{\sqrt{9 - \cos^4 2x}} = \frac{\sin 2x\cos 2x}{\sqrt{9 - \cos^4 2x}} ,

    the integrand.

    The two minus signs cancel, and 14⋅43=13\frac14\cdot\frac43 = \frac13 is cancelled by the 13\frac13 that comes out of the square root.
Result

∫sin⁡2xcos⁡2x9−cos⁡42x dx=−14sin⁡−1(13cos⁡22x)+C\begin{aligned}&\int\frac{\sin 2x\cos 2x}{\sqrt{9 - \cos^4 2x}}\,dx\\ &= -\frac14\sin^{-1}\left(\frac13\cos^2 2x\right) + C\end{aligned} for every real xx⁠. This is the textbook's answer.

Check it with numbers

At x=π8x = \frac{\pi}{8}⁠: sin⁡π4cos⁡π4=12\sin\frac{\pi}{4}\cos\frac{\pi}{4} = \frac12 and cos⁡4π4=14\cos^4\frac{\pi}{4} = \frac14⁠, so the integrand is 1/235/4=135≈0.169031\frac{1/2}{\sqrt{35/4}} = \frac{1}{\sqrt{35}} \approx 0.169031⁠; a numerical derivative of the answer gives 0.1690310.169031⁠. At x=1x = 1⁠: integrand ≈−0.126344\approx -0.126344⁠, numerical derivative −0.126344-0.126344⁠.

Common mistakes
  1. Missing a factor 2 in dtdt⁠. There are two: one from the square and one from the 2x2x inside cosine, so dt=−4sin⁡2xcos⁡2x dxdt = -4\sin 2x\cos 2x\,dx⁠.
  2. Sign of ddxcos⁡2x\frac{d}{dx}\cos 2x⁠. It is −2sin⁡2x-2\sin 2x⁠; losing the minus gives the answer with the wrong sign.
  3. Stopping halfway with t=cos⁡2xt = \cos 2x⁠. That choice is not wrong, only longer: it leaves −12∫t dt9−t4-\frac12\int\frac{t\,dt}{\sqrt{9 - t^4}}⁠, which still needs the substitution u=t2u = t^2 before the standard form appears.
What to take away
  1. A fourth power under a root, a2−(something)4\sqrt{a^2 - (\text{something})^4}⁠, suggests calling the square of that something tt⁠.
  2. sin⁡θcos⁡θ\sin\theta\cos\theta is the derivative of 12sin⁡2θ\frac12\sin^2\theta and of −12cos⁡2θ-\frac12\cos^2\theta⁠: either can be the substitution.
  3. Always count every constant produced by the chain rule; the check by differentiation confirms them.

Every fact used, and where it comes from

FactUsed inSource
Chain ruleSteps 2, 5Chapter 5, Section 5.3.1
Substitution t=cos⁡22xt = \cos^2 2xStep 2Section 7.3.1 (Lesson: substitution)
∫dta2−t2=sin⁡−1ta+C\int\frac{dt}{\sqrt{a^2 - t^2}} = \sin^{-1}\frac{t}{a} + CStep 3Section 7.4, formula (5) (Lesson: particular functions)
ddusin⁡−1u=11−u2\frac{d}{du}\sin^{-1}u = \frac{1}{\sqrt{1-u^2}}Step 5Chapter 5, Section 5.3.3
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