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Chapter 7 · Integrals

Exercise 7.4 · Q20: ∫x+24x−x2 dx\int\frac{x + 2}{\sqrt{4x - x^2}}\,dx

Exercise 7.4Integrals of some particular functions · Section 7.4
The question

Integrate the function

x+24x−x2.\frac{x + 2}{\sqrt{4x - x^2}}.
Exam answer

Let I=∫x+24x−x2 dxI = \displaystyle\int\frac{x + 2}{\sqrt{4x - x^2}}\,dx⁠.

Let x+2=A(4−2x)+Bx + 2 = A(4 - 2x) + B⁠, where 4−2x=ddx(4x−x2)4 - 2x = \frac{d}{dx}(4x - x^2)⁠.

Comparing coefficients: −2A=1-2A = 1 and 4A+B=24A + B = 2⁠. So A=−12A = -\frac12⁠, B=4B = 4⁠.

∴ I=−12∫(4−2x) dx4x−x2+4∫dx4x−x2=−12I1+4I2\begin{aligned} \therefore\ I &= -\frac12\int\frac{(4 - 2x)\,dx}{\sqrt{4x - x^2}} \\ &\quad + 4\int\frac{dx}{\sqrt{4x - x^2}} \\ &= -\frac12 I_1 + 4I_2 \end{aligned}

In I1I_1⁠, put 4x−x2=t4x - x^2 = t⁠. Then (4−2x) dx=dt(4 - 2x)\,dx = dt⁠.

I1=∫dtt=2t=24x−x2\begin{aligned} I_1 &= \int\frac{dt}{\sqrt t} = 2\sqrt t \\ &= 2\sqrt{4x - x^2} \end{aligned}

For I2I_2⁠, complete the square:

4x−x2=−[(x−2)2−4]=22−(x−2)2\begin{aligned} 4x - x^2 &= -\left[(x - 2)^2 - 4\right] \\ &= 2^2 - (x - 2)^2 \end{aligned}

[Using ∫dxa2−x2\displaystyle\int\frac{dx}{\sqrt{a^2 - x^2}} =sin⁡−1xa+C= \sin^{-1}\frac xa + C]

I2=∫dx22−(x−2)2=sin⁡−1x−22\begin{aligned} I_2 &= \int\frac{dx}{\sqrt{2^2 - (x - 2)^2}} \\ &= \sin^{-1}\frac{x - 2}{2} \end{aligned}

∴ I=−4x−x2\therefore\ I = -\sqrt{4x - x^2} +4sin⁡−1(x−22)+C+ 4\sin^{-1}\left(\dfrac{x - 2}{2}\right) + C⁠.

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

The one idea this problem is built on

Type 7.4 (10): split the numerator as A⋅Q′+BA\cdot Q' + B with Q=4x−x2Q = 4x - x^2⁠, Q′=4−2xQ' = 4 - 2x⁠. This gives x+2=−12(4−2x)+4x + 2 = -\frac12(4 - 2x) + 4⁠. The Q′Q' part becomes 2Q2\sqrt Q by substitution. For the constant part, completing the square gives 4x−x2=22−(x−2)24x - x^2 = 2^2 - (x - 2)^2⁠, "constant minus square", so formula 7.4 (5) gives an inverse sine.

Given, to find, and the plan
Given
The integrand x+24x−x2\frac{x + 2}{\sqrt{4x - x^2}}⁠.
To find
∫x+24x−x2 dx\int\frac{x + 2}{\sqrt{4x - x^2}}\,dx⁠, and where the answer is valid.
Plan
(1) Find the domain. (2) Find AA⁠, BB with x+2=A(4−2x)+Bx + 2 = A(4 - 2x) + B⁠. (3) Split into I1I_1 and I2I_2⁠. (4) I1I_1 by substitution. (5) I2I_2 by completing the square and 7.4 (5). (6) Combine. (7) Differentiate back.

Basics you need

1Comparing coefficients

If px+q=A(2ax+b)+Bpx + q = A(2ax + b) + B for all xx⁠, the coefficients of xx and the constants on the two sides are equal. Basics: splitting fractions

2Substitution t=Q(x)t = Q(x)

∫Q′(x)Q(x) dx=2Q(x)+C\int\frac{Q'(x)}{\sqrt{Q(x)}}\,dx = 2\sqrt{Q(x)} + C⁠. Basics: substitution

3Completing the square

x2−4x=(x−2)2−4x^2 - 4x = (x - 2)^2 - 4⁠. Basics: completing the square

4Formula 7.4 (5)

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

Solution

  1. Where the integrand exists

    We need 4x−x2=x(4−x)>04x - x^2 = x(4 - x) > 0⁠, that is 0<x<40 < x < 4⁠.

    The quadratic is under a square root in a denominator, so it must be strictly positive. The product x(4−x)x(4 - x) is positive exactly between its roots 0 and 4.

  2. Split the numerator as A⋅Q′+BA\cdot Q' + B

    Q=4x−x2Q = 4x - x^2 has Q′=4−2xQ' = 4 - 2x⁠. Write

    x+2=A(4−2x)+B.x + 2 = A(4 - 2x) + B.

    Coefficients of xx⁠: −2A=1-2A = 1⁠. Constants: 4A+B=24A + B = 2⁠. So A=−12A = -\frac12⁠, B=2+2=4B = 2 + 2 = 4⁠.

    This is type 7.4 (10). Check: −12(4−2x)+4=−2+x+4=x+2-\frac12(4 - 2x) + 4 = -2 + x + 4 = x + 2⁠.
    Type 7.4 (10). Basics: splitting fractions

  3. Split the integral into I1I_1 and I2I_2

    ∫x+24x−x2 dx=−12∫(4−2x) dx4x−x2+4∫dx4x−x2=−12I1+4I2.\begin{aligned} \int\frac{x + 2}{\sqrt{4x - x^2}}\,dx &= -\frac12\int\frac{(4 - 2x)\,dx}{\sqrt{4x - x^2}} \\ &\quad + 4\int\frac{dx}{\sqrt{4x - x^2}} \\ &= -\frac12 I_1 + 4I_2. \end{aligned}
    Dividing −12(4−2x)+4-\frac12(4 - 2x) + 4 by the denominator term by term gives two fractions; Property (V) of Section 7.2.1 splits the integral.
    Property (V), Section 7.2.1. Lesson: integrals of particular functions

  4. I1I_1 by substitution

    Put t=4x−x2t = 4x - x^2 (so t>0t > 0⁠), dt=(4−2x) dxdt = (4 - 2x)\,dx⁠:

    I1=∫dtt=2t+C1=24x−x2+C1.I_1 = \int\frac{dt}{\sqrt t} = 2\sqrt t + C_1 = 2\sqrt{4x - x^2} + C_1.
    Power formula with n=−12n = -\frac12⁠: ∫t−1/2 dt=t1/21/2=2t\int t^{-1/2}\,dt = \frac{t^{1/2}}{1/2} = 2\sqrt t⁠.
    Substitution, Section 7.3.1; standard formula (i), Section 7.2. Basics: the power rule

  5. I2I_2 by completing the square

    4x−x2=−(x2−4x)=−[(x−2)2−4]=22−(x−2)2.\begin{aligned} 4x - x^2 &= -\left(x^2 - 4x\right) = -\left[(x - 2)^2 - 4\right] \\ &= 2^2 - (x - 2)^2. \end{aligned}

    Put t=x−2t = x - 2⁠, dt=dxdt = dx⁠; on the domain −2<t<2-2 < t < 2⁠. By 7.4 (5) with a=2a = 2⁠:

    I2=∫dt22−t2=sin⁡−1t2+C2=sin⁡−1x−22+C2.I_2 = \int\frac{dt}{\sqrt{2^2 - t^2}} = \sin^{-1}\frac t2 + C_2 = \sin^{-1}\frac{x - 2}{2} + C_2.
    Half of −4-4 is −2-2⁠, and (−2)2=4(-2)^2 = 4⁠. "Constant minus square" under a root is the inverse sine shape. On the domain 0<x<40 < x < 4⁠, t=x−2t = x - 2 runs over (−2,2)(-2, 2)⁠, exactly where 7.4 (5) holds.
    Type 7.4 (8); formula 7.4 (5). Basics: completing the square

  6. Combine

    ∫x+24x−x2 dx=−4x−x2+4sin⁡−1x−22+C,\begin{aligned} \int\frac{x + 2}{\sqrt{4x - x^2}}\,dx &= -\sqrt{4x - x^2} \\ &\quad + 4\sin^{-1}\frac{x - 2}{2} + C, \end{aligned}

    with C=4C2−C12C = 4C_2 - \frac{C_1}{2}⁠.

    −12⋅2Q=−Q-\frac12\cdot 2\sqrt Q = -\sqrt Q⁠. The constants merge into one. This is exactly the textbook's form.

  7. Check: differentiate the answer

    ddx(−4x−x2)=−4−2x24x−x2=x−24x−x2,ddx 4sin⁡−1x−22=4⋅121−(x−2)24=44−(x−2)2.\begin{aligned} \frac{d}{dx}\left(-\sqrt{4x - x^2}\right) &= -\frac{4 - 2x}{2\sqrt{4x - x^2}} = \frac{x - 2}{\sqrt{4x - x^2}}, \\ \frac{d}{dx}\,4\sin^{-1}\frac{x - 2}{2} &= \frac{4\cdot\frac12}{\sqrt{1 - \frac{(x - 2)^2}{4}}} = \frac{4}{\sqrt{4 - (x - 2)^2}}. \end{aligned}

    Since 4−(x−2)2=4x−x24 - (x - 2)^2 = 4x - x^2⁠, the sum is x−2+44x−x2=x+24x−x2\frac{x - 2 + 4}{\sqrt{4x - x^2}} = \frac{x + 2}{\sqrt{4x - x^2}}⁠. ✓\checkmark

    Both lines use the chain rule. In the second, multiplying the top and bottom by 2 turns 21−(x−2)2/4\frac{2}{\sqrt{1 - (x - 2)^2/4}} into 44−(x−2)2\frac{4}{\sqrt{4 - (x - 2)^2}}⁠. Valid for 0<x<40 < x < 4⁠.
    Chain rule (Theorem 4, Section 5.3.1); derivative (viii), Section 7.2. Basics: integration undoes differentiation
Result ∫x+24x−x2 dx=−4x−x2+4sin⁡−1x−22+C,\begin{aligned} \int\frac{x + 2}{\sqrt{4x - x^2}}\,dx &= -\sqrt{4x - x^2} \\ &\quad + 4\sin^{-1}\frac{x - 2}{2} + C, \end{aligned}

valid for 0<x<40 < x < 4⁠. (Textbook answer: −4x−x2+4sin⁡−1(x−22)+C-\sqrt{4x - x^2} + 4\sin^{-1}\left(\frac{x - 2}{2}\right) + C⁠.)

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. 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)
[1,3][1, 3]4.18879024.18879024π3=4.1887902\frac{4\pi}{3} = 4.1887902

The root terms cancel because 4x−x2=34x - x^2 = 3 at both ends, and 4[sin⁡−112−sin⁡−1(−12)]=4⋅π34\left[\sin^{-1}\frac12 - \sin^{-1}\left(-\frac12\right)\right] = 4\cdot\frac{\pi}{3}⁠.

Common mistakes
  1. Sign of AA⁠. Q′=4−2xQ' = 4 - 2x has a negative xx-coefficient, so A=−12A = -\frac12⁠.
  2. Wrong BB⁠. B=2−4A=2+2=4B = 2 - 4A = 2 + 2 = 4⁠, not 22⁠.
  3. Completing the square with the wrong sign. 4x−x2=4−(x−2)24x - x^2 = 4 - (x - 2)^2⁠, not (x−2)2−4(x - 2)^2 - 4⁠; the second would lead to the wrong (log) formula.
What to take away
  1. Numerator =A⋅Q′+B= A\cdot Q' + B⁠, even when Q′Q' has a negative xx-coefficient.
  2. A quadratic with negative x2x^2 term under a root completes to a2−(x−h)2a^2 - (x - h)^2 and gives sin⁡−1x−ha\sin^{-1}\frac{x - h}{a}⁠.

Every fact used, and where it comes from

FactUsed inSource
px+q=Addx(ax2+bx+c)+Bpx + q = A\frac{d}{dx}(ax^2 + bx + c) + BStep 2Type 7.4 (10) (Lesson: integrals of particular functions)
Integral of a sum of constant multiplesStep 3Property (V), Section 7.2.1
∫t−1/2 dt=2t+C\int t^{-1/2}\,dt = 2\sqrt t + CStep 4Standard formula (i), Section 7.2
SubstitutionSteps 4, 5Section 7.3.1 (Basics: substitution)
∫dxa2−x2=sin⁡−1xa+C\int\frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac xa + CStep 5Formula 7.4 (5)
ddxsin⁡−1x=11−x2\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^2}}⁠; chain ruleStep 7Derivative (viii), Section 7.2; Theorem 4, Section 5.3.1
∫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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