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Chapter 7 · Integrals

Miscellaneous · Q9: ∫cos⁡x4−sin⁡2x dx\int\frac{\cos x}{\sqrt{4-\sin^2x}}\,dx

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

Integrate the functions in Exercises 1 to 23.

9. cos⁡x4−sin⁡2x\dfrac{\cos x}{\sqrt{4 - \sin^2x}}

Exam answer

Let I=∫cos⁡x dx4−sin⁡2xI = \displaystyle\int\frac{\cos x\,dx}{\sqrt{4 - \sin^2 x}}⁠.

Put sin⁡x=t\sin x = t⁠. Then cos⁡x dx=dt\cos x\,dx = dt⁠.

I=∫dt22−t2I = \int\frac{dt}{\sqrt{2^2 - t^2}}

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

I=sin⁡−1t2+CI = \sin^{-1}\frac{t}{2} + C

∴ I=sin⁡−1(sin⁡x2)+C\therefore\ I = \sin^{-1}\left(\dfrac{\sin x}{2}\right) + C⁠.

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

The one idea this problem is built on

The numerator cos⁡x\cos x is the derivative of sin⁡x\sin x⁠, which is the quantity squared under the root. So the substitution t=sin⁡xt = \sin x turns the integral into ∫dt4−t2\int\frac{dt}{\sqrt{4 - t^2}}⁠, the standard form ∫dxa2−x2=sin⁡−1xa+C\int\frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac xa + C with a=2a = 2⁠.

Given, to find, and the plan
Given
f(x)=cos⁡x4−sin⁡2xf(x) = \dfrac{\cos x}{\sqrt{4 - \sin^2x}}⁠. Since sin⁡2x≤1\sin^2x \le 1⁠, 4−sin⁡2x≥3>04 - \sin^2x \ge 3 > 0⁠, so ff is defined for every real xx⁠.
To find
∫cos⁡x4−sin⁡2x dx\displaystyle\int\frac{\cos x}{\sqrt{4 - \sin^2x}}\,dx⁠.
Plan
(1) Why t=sin⁡xt = \sin x⁠. (2) Substitute. (3) Use formula (5) of Section 7.4. (4) Differentiate back.

Basics you need

1Substitution

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

2Formula (5) of Section 7.4

∫dxa2−x2=sin⁡−1xa+C\int\frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac xa + C for a>0a > 0⁠, ∣x∣<a|x| < a⁠. Lesson: integrals of particular functions

Solution

  1. Why the substitution t=sin⁡xt = \sin x

    The integrand is a function of sin⁡x\sin x multiplied by cos⁡x\cos x⁠, and cos⁡x=ddxsin⁡x\cos x = \frac{d}{dx}\sin x⁠. That is the pattern h(g(x)) g′(x)h(g(x))\,g'(x) with g(x)=sin⁡xg(x) = \sin x⁠.

    Whenever the derivative of an inner expression stands beside it as a factor, substitution removes both at once. Here it also turns the root into the standard form a2−t2\sqrt{a^2 - t^2}⁠.
    Section 7.3.1. Lesson: integration by substitution

  2. Substitute

    Put t=sin⁡xt = \sin x⁠, so dt=cos⁡x dxdt = \cos x\,dx⁠. Then

    ∫cos⁡x dx4−sin⁡2x=∫dt4−t2=∫dt22−t2.\begin{aligned} \int\frac{\cos x\,dx}{\sqrt{4 - \sin^2x}} &= \int\frac{dt}{\sqrt{4 - t^2}} \\ &= \int\frac{dt}{\sqrt{2^2 - t^2}}. \end{aligned}
    The whole numerator cos⁡x dx\cos x\,dx becomes dtdt⁠. The new variable satisfies −1≤t≤1-1 \le t \le 1⁠, well inside ∣t∣<2|t| < 2⁠, so the root 4−t2\sqrt{4 - t^2} is real and positive.
    Substitution (Basic 1).

  3. Use formula (5) with a=2a = 2

    ∫dt22−t2=sin⁡−1t2+C=sin⁡−1(sin⁡x2)+C.\begin{aligned} \int\frac{dt}{\sqrt{2^2 - t^2}} &= \sin^{-1}\frac t2 + C \\ &= \sin^{-1}\Big(\frac{\sin x}{2}\Big) + C. \end{aligned}
    Formula (5) needs ∣t∣<a=2|t| < a = 2⁠, which holds since ∣t∣=∣sin⁡x∣≤1|t| = |\sin x| \le 1⁠. The argument sin⁡x2\frac{\sin x}{2} lies in [−12,12][-\frac12, \frac12]⁠, inside the domain [−1,1][-1, 1] of sin⁡−1\sin^{-1}⁠. This is exactly the textbook's answer, valid for every real xx⁠.
    Formula (5), Section 7.4 (Basic 2).

  4. Check: differentiate the answer

    Chain rule with ddusin⁡−1u=11−u2\frac{d}{du}\sin^{-1}u = \frac{1}{\sqrt{1 - u^2}} and u=sin⁡x2u = \frac{\sin x}{2}⁠, u′=cos⁡x2u' = \frac{\cos x}{2}⁠:

    ddxsin⁡−1(sin⁡x2)=11−sin⁡2x4⋅cos⁡x2=cos⁡x2⋅124−sin⁡2x=cos⁡x4−sin⁡2x. ✓\begin{aligned} \frac{d}{dx}\sin^{-1}\Big(\frac{\sin x}{2}\Big) &= \frac{1}{\sqrt{1 - \frac{\sin^2x}{4}}}\cdot\frac{\cos x}{2} \\ &= \frac{\cos x}{2\cdot\frac12\sqrt{4 - \sin^2x}} = \frac{\cos x}{\sqrt{4 - \sin^2x}}. \ \checkmark \end{aligned}
    1−s24=4−s24=124−s2\sqrt{1 - \frac{s^2}{4}} = \sqrt{\frac{4 - s^2}{4}} = \frac12\sqrt{4 - s^2}⁠. The 12\frac12 cancels the 2, returning the integrand for every xx⁠.
    Derivative of sin⁡−1\sin^{-1}⁠, Chapter 5. Basics: integration undoes differentiation
Result

For every real xx⁠:

∫cos⁡x4−sin⁡2x dx=sin⁡−1(sin⁡x2)+C.\int\frac{\cos x}{\sqrt{4 - \sin^2x}}\,dx = \sin^{-1}\Big(\frac{\sin x}{2}\Big) + C.

(Textbook answer: sin⁡−1(sin⁡x2)+C\sin^{-1}\big(\frac{\sin x}{2}\big) + C⁠, the same.)

Check it with numbers

Area check on [0,π2][0, \frac\pi2]⁠: sin⁡−112−sin⁡−10=π6≈0.5235988\sin^{-1}\frac12 - \sin^{-1}0 = \frac\pi6 \approx 0.5235988⁠. Numerical integration of cos⁡x4−sin⁡2x\frac{\cos x}{\sqrt{4 - \sin^2x}} from 0 to π2\frac\pi2 also gives 0.52359880.5235988⁠.

Common mistakes
  1. Writing sin⁡−1(sin⁡x)\sin^{-1}(\sin x) (forgetting the a=2a = 2⁠): the formula gives sin⁡−1ta\sin^{-1}\frac ta⁠.
  2. Leaving cos⁡x\cos x in the integral after substituting; the whole cos⁡x dx\cos x\,dx is dtdt⁠.
What to take away
  1. Look for an inner function whose derivative is a factor of the integrand: cos⁡x\cos x with sin⁡x\sin x⁠, x3x^3 with x4x^4⁠, exe^x with exe^x⁠.
  2. After substitution, compare with the six formulae of Section 7.4; here it was 1a2−t2\frac{1}{\sqrt{a^2 - t^2}}⁠.

Every fact used, and where it comes from

FactUsed inSource
sin⁡2x≤1\sin^2x \le 1Given, Steps 2, 3Trigonometry (Basics)
Substitution t=sin⁡xt = \sin x⁠, dt=cos⁡x dxdt = \cos x\,dxSteps 1, 2Section 7.3.1
∫dxa2−x2=sin⁡−1xa+C\int\frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac xa + CStep 3Formula (5), Section 7.4
Derivative of sin⁡−1u\sin^{-1}u⁠; chain ruleStep 4Chapter 5
All decimalsCheckRecomputed in the site's verification script
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