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Chapter 7 · Integrals

Exercise 7.7 · Q2: ∫1−4x2 dx\int\sqrt{1 - 4x^2}\,dx

Exercise 7.7Integrals of some more types · Section 7.6.2
The question

Integrate the function

1−4x2.\sqrt{1 - 4x^2}.
Exam answer

Let I=∫1−4x2 dxI = \displaystyle\int\sqrt{1 - 4x^2}\,dx⁠.

Put 2x=t2x = t⁠. Then 2 dx=dt2\,dx = dt⁠.

I=12∫1−t2 dtI = \frac12\int\sqrt{1 - t^2}\,dt

Using ∫a2−t2 dt=t2a2−t2\displaystyle\int\sqrt{a^2 - t^2}\,dt = \frac{t}{2}\sqrt{a^2 - t^2} +a22sin⁡−1ta+C+ \frac{a^2}{2}\sin^{-1}\frac{t}{a} + C with a=1,{a = 1,}

I=12[t21−t2+12sin⁡−1t]+C=t41−t2+14sin⁡−1t+C\begin{aligned} I &= \frac12\bigg[\frac{t}{2}\sqrt{1 - t^2} \\ &\qquad + \frac12\sin^{-1}t\bigg] + C \\ &= \frac{t}{4}\sqrt{1 - t^2} + \frac14\sin^{-1}t + C \end{aligned}

∴ I=x21−4x2\therefore\ I = \dfrac{x}{2}\sqrt{1 - 4x^2} +14sin⁡−12x+C+ \dfrac14\sin^{-1}2x + C⁠.

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

The one idea this problem is built on

4x2=(2x)24x^2 = (2x)^2⁠, so 1−4x2=12−(2x)21 - 4x^2 = 1^2 - (2x)^2 is "a2a^2 minus a square" with a=1a = 1⁠. The substitution t=2xt = 2x turns the integral into 12∫1−t2 dt\frac12\int\sqrt{1 - t^2}\,dt⁠, which formula 7.6.2 (iii) handles. The only new thing compared with Q1 is the factor 12\frac12 that comes from dx=12 dtdx = \frac12\,dt⁠.

Given, to find, and the plan
Given
The integrand 1−4x2\sqrt{1 - 4x^2}⁠.
To find
∫1−4x2 dx\int\sqrt{1 - 4x^2}\,dx⁠, and where the answer is valid.
Plan
(1) Find where the root is real. (2) Substitute t=2xt = 2x⁠. (3) Use 7.6.2 (iii) with a=1a = 1⁠. (4) Return to xx⁠. (5) Differentiate back.

Basics you need

1Substitution

If t=g(x)t = g(x)⁠, then dt=g′(x) dxdt = g'(x)\,dx⁠; every xx and the dxdx must be replaced. For t=2xt = 2x⁠: dt=2 dxdt = 2\,dx⁠, so dx=12 dtdx = \frac12\,dt⁠. Basics: substitution

2Formula 7.6.2 (iii)

∫a2−t2 dt=t2a2−t2+a22sin⁡−1ta+C\int\sqrt{a^2 - t^2}\,dt = \frac{t}{2}\sqrt{a^2 - t^2} + \frac{a^2}{2}\sin^{-1}\frac{t}{a} + C⁠, a>0a > 0⁠. It is derived by parts in Exercise 7.7 Q1. Lesson: integration by parts

3Derivative of sin⁡−1\sin^{-1}

ddxsin⁡−1u=11−u2⋅dudx\frac{d}{dx}\sin^{-1}u = \frac{1}{\sqrt{1 - u^2}}\cdot\frac{du}{dx} for ∣u∣<1|u| < 1 (standard formula (viii), Section 7.2, with the chain rule). Basics: integration undoes differentiation

Solution

  1. Where the square root is real

    We need 1−4x2≥01 - 4x^2 \ge 0⁠, that is x2≤14x^2 \le \frac14⁠, that is

    −12≤x≤12.-\tfrac12 \le x \le \tfrac12.
    4x2≤1  ⟺  x2≤14  ⟺  ∣x∣≤124x^2 \le 1 \iff x^2 \le \frac14 \iff |x| \le \frac12⁠. Outside this interval the integrand is not a real number.

  2. Substitute t=2xt = 2x

    Put t=2xt = 2x⁠. Then dt=2 dxdt = 2\,dx⁠, so dx=12 dtdx = \frac12\,dt⁠, and 1−4x2=1−t21 - 4x^2 = 1 - t^2⁠:

    ∫1−4x2 dx=12∫1−t2 dt.\int\sqrt{1 - 4x^2}\,dx = \frac12\int\sqrt{1 - t^2}\,dt.

    As xx runs over [−12,12]\left[-\frac12, \frac12\right]⁠, tt runs over [−1,1][-1, 1]⁠.

    The substitution makes the square inside the root a plain t2t^2⁠, which is the shape in the formula. Replacing dxdx is essential: dxdx is half of dtdt⁠, not equal to it, and that 12\frac12 stays in front.
    Integration by substitution, Section 7.3.1. Basics: substitution

  3. Use formula 7.6.2 (iii) with a=1a = 1

    12∫1−t2 dt=12[t21−t2+12sin⁡−1t]+C=t41−t2+14sin⁡−1t+C.\begin{aligned} \frac12\int\sqrt{1 - t^2}\,dt &= \frac12\left[\frac{t}{2}\sqrt{1 - t^2} + \frac12\sin^{-1}t\right] \\ &\quad + C \\ &= \frac{t}{4}\sqrt{1 - t^2} + \frac14\sin^{-1}t + C. \end{aligned}
    With a=1a = 1⁠: a22=12\frac{a^2}{2} = \frac12 and ta=t\frac{t}{a} = t⁠. Multiplying the bracket by 12\frac12 halves both terms. A constant times an arbitrary constant is still an arbitrary constant, so we keep calling it CC⁠.
    Formula 7.6.2 (iii); Property (IV), Section 7.2.1. Lesson: integration by parts

  4. Return to xx

    Put t=2xt = 2x back: t4=2x4=x2\frac{t}{4} = \frac{2x}{4} = \frac{x}{2} and 1−t2=1−4x21 - t^2 = 1 - 4x^2⁠:

    ∫1−4x2 dx=x21−4x2+14sin⁡−12x+C.\begin{aligned} \int\sqrt{1 - 4x^2}\,dx &= \frac{x}{2}\sqrt{1 - 4x^2} \\ &\quad + \frac14\sin^{-1}2x + C. \end{aligned}
    An indefinite integral found in tt must be written back in the original variable. Another route gives the same thing: 1−4x2=214−x2\sqrt{1 - 4x^2} = 2\sqrt{\frac14 - x^2}⁠, and 7.6.2 (iii) with a=12a = \frac12 gives 2[x214−x2+18sin⁡−12x]2\left[\frac{x}{2}\sqrt{\frac14 - x^2} + \frac18\sin^{-1}2x\right]⁠, which is the same expression since x14−x2=x21−4x2x\sqrt{\frac14 - x^2} = \frac{x}{2}\sqrt{1 - 4x^2}⁠.

  5. Check: differentiate the answer

    For −12<x<12-\frac12 < x < \frac12⁠, write r=1−4x2r = \sqrt{1 - 4x^2}⁠, so drdx=−8x2r=−4xr\frac{dr}{dx} = \frac{-8x}{2r} = \frac{-4x}{r}⁠. Then

    ddx[x2 r]=r2+x2⋅−4xr=r2−2x2r,ddx[14sin⁡−12x]=14⋅21−4x2=12r.\begin{aligned} \frac{d}{dx}\left[\frac{x}{2}\,r\right] &= \frac{r}{2} + \frac{x}{2}\cdot\frac{-4x}{r} = \frac{r}{2} - \frac{2x^2}{r}, \\ \frac{d}{dx}\left[\frac14\sin^{-1}2x\right] &= \frac14\cdot\frac{2}{\sqrt{1 - 4x^2}} = \frac{1}{2r}. \end{aligned}

    Over the common denominator 2r2r⁠:

    r2−4x2+12r=2−8x22r=1−4x2r=r. ✓\frac{r^2 - 4x^2 + 1}{2r} = \frac{2 - 8x^2}{2r} = \frac{1 - 4x^2}{r} = r.\ \checkmark
    r2=r22r\frac{r}{2} = \frac{r^2}{2r} and r2=1−4x2r^2 = 1 - 4x^2⁠, so the numerator is (1−4x2)−4x2+1=2(1−4x2)(1 - 4x^2) - 4x^2 + 1 = 2(1 - 4x^2)⁠. Dividing r2r^2 by rr gives r=1−4x2r = \sqrt{1 - 4x^2}⁠, the integrand. At x=±12x = \pm\frac12 the integrand is 0 and the answer's one-sided slope is 0 too, so the answer is good on the whole closed interval.
    Product rule; chain rule; standard formula (viii), Section 7.2. Basics: integration undoes differentiation
Result ∫1−4x2 dx=x21−4x2+14sin⁡−12x+C,\begin{aligned} \int\sqrt{1 - 4x^2}\,dx &= \frac{x}{2}\sqrt{1 - 4x^2} \\ &\quad + \frac14\sin^{-1}2x + C, \end{aligned}

valid for −12≤x≤12-\frac12 \le x \le \frac12⁠. (Textbook answer: 14sin⁡−12x+12x1−4x2+C\frac14\sin^{-1}2x + \frac12 x\sqrt{1 - 4x^2} + C⁠, the same two terms in the other order.)

Check it with numbers

By Section 7.8, ∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a) for our answer FF⁠, on any interval where the integrand is continuous. The middle column is a computer's sum of very many thin strips, found without our answer.

IntervalSum of thin stripsF(b)−F(a)F(b) - F(a)
[0,0.4][0, 0.4]0.35182380.35182380.35182380.3518238
[−0.3,0.2][-0.3, 0.2]0.47540600.47540600.47540600.4754060
Common mistakes
  1. Losing the 12\frac12 from dx=12 dtdx = \frac12\,dt⁠. That gives x1−4x2+12sin⁡−12xx\sqrt{1 - 4x^2} + \frac12\sin^{-1}2x⁠, exactly twice the right answer; its derivative is 21−4x22\sqrt{1 - 4x^2}⁠.
  2. Taking a=2a = 2⁠. The number outside the square is a2=1a^2 = 1⁠; the 4 belongs to (2x)2(2x)^2⁠.
  3. Writing sin⁡−1x\sin^{-1}x instead of sin⁡−12x\sin^{-1}2x⁠. After the substitution the angle is sin⁡−1t\sin^{-1}t with t=2xt = 2x⁠.
What to take away
  1. a2−(kx)2\sqrt{a^2 - (kx)^2}⁠: substitute t=kxt = kx⁠, and remember the factor 1k\frac1k from dx=1k dtdx = \frac1k\,dt⁠.
  2. Or take the coefficient out of the root first: 1−4x2=214−x2\sqrt{1 - 4x^2} = 2\sqrt{\frac14 - x^2}⁠. Both routes agree.

Every fact used, and where it comes from

FactUsed inSource
Substitution t=2xt = 2x⁠, dx=12 dtdx = \frac12\,dtStep 2Section 7.3.1 (Basics: substitution)
∫a2−t2 dt\int\sqrt{a^2 - t^2}\,dt formulaStep 3Formula 7.6.2 (iii), Section 7.6.2
∫kf(x) dx=k∫f(x) dx\int k f(x)\,dx = k\int f(x)\,dxStep 3Property (IV), Section 7.2.1
ddxsin⁡−1x=11−x2\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^2}}Step 5Standard formula (viii), Section 7.2
∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a)Numerical checkSection 7.8 (Lesson: definite integrals)
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