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Chapter 7 · Integrals

Exercise 7.4 · Q12: ∫dx7−6x−x2\int\frac{dx}{\sqrt{7-6x-x^2}}

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

Integrate the function (Exercise 7.4, "Integrate the functions in Exercises 1 to 23"):

17−6x−x2\frac{1}{\sqrt{7 - 6x - x^2}}
Exam answer

Let I=∫dx7−6x−x2I = \displaystyle\int\frac{dx}{\sqrt{7 - 6x - x^2}}⁠.

7−6x−x2=7−[(x+3)2−9]=42−(x+3)2\begin{aligned} 7 - 6x - x^2 &= 7 - \left[(x + 3)^2 - 9\right] \\ &= 4^2 - (x + 3)^2 \end{aligned}

Put x+3=tx + 3 = t⁠. Then dx=dtdx = dt⁠.

I=∫dt42−t2I = \int\frac{dt}{\sqrt{4^2 - t^2}}

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

I=sin⁡−1t4+CI = \sin^{-1}\frac t4 + C

∴ I=sin⁡−1(x+34)+C\therefore\ I = \sin^{-1}\left(\dfrac{x + 3}{4}\right) + C⁠.

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

The one idea this problem is built on

The x2x^2 term is negative, so completing the square produces "a constant minus a square": 7−6x−x2=16−(x+3)2=42−(x+3)27 - 6x - x^2 = 16 - (x + 3)^2 = 4^2 - (x + 3)^2⁠. Under a root, that is the shape of formula 7.4 (5), ∫dxa2−x2=sin⁡−1xa+C\int\frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac xa + C⁠.

Given, to find, and the plan
Given
f(x)=17−6x−x2f(x) = \dfrac{1}{\sqrt{7 - 6x - x^2}}⁠.
To find
∫f(x) dx\int f(x)\,dx⁠, and where the answer is valid.
Plan
(1) Complete the square, keeping track of the minus sign. (2) Read off the domain. (3) Shift t=x+3t = x + 3 and use formula 7.4 (5) with a=4a = 4⁠. (4) Return to xx⁠. (5) Differentiate back.

Basics you need

1Completing the square with a minus sign

Bracket the xx terms with the minus outside: −x2−6x=−(x2+6x)-x^2 - 6x = -(x^2 + 6x)⁠, then complete the square inside. Basics: completing the square

2Formula 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

3Derivative of sin⁡−1\sin^{-1}

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

Solution

  1. Complete the square

    7−6x−x2=7−(x2+6x)=7−[(x+3)2−9]=16−(x+3)2=42−(x+3)2.\begin{aligned} 7 - 6x - x^2 &= 7 - (x^2 + 6x) \\ &= 7 - \left[(x + 3)^2 - 9\right] \\ &= 16 - (x + 3)^2 \\ &= 4^2 - (x + 3)^2. \end{aligned}
    Inside the bracket, half of 6 is 3, so x2+6x=(x+3)2−9x^2 + 6x = (x + 3)^2 - 9⁠. The minus sign in front of the bracket turns −9-9 into +9+9⁠, and 7+9=167 + 9 = 16⁠. The result is "constant minus square", the shape that leads to sin⁡−1\sin^{-1}⁠.
    Basics: completing the square

  2. Where is the integrand defined?

    We need 16−(x+3)2>016 - (x + 3)^2 > 0⁠, that is, ∣x+3∣<4|x + 3| < 4⁠, that is,

    −7<x<1.-7 < x < 1.
    x y −7 −5 −3 −1 1 8 16 −8 vertex (−3, 16) y = 7 − 6x − x²
    y=7−6x−x2=16−(x+3)2y = 7 - 6x - x^2 = 16 - (x + 3)^2 is a downward parabola with vertex (−3,16)(-3, 16)⁠. It is positive (green) only between its roots x=−7x = -7 and x=1x = 1⁠; that open interval, marked on the xx-axis, is where the integrand exists.
    The root sits in a denominator, so the quantity under it must be strictly positive. The answer is only claimed on this interval. (Check: 7−6x−x2=−(x+7)(x−1)7 - 6x - x^2 = -(x + 7)(x - 1)⁠, which vanishes at −7-7 and 11⁠.)

  3. Shift t=x+3t = x + 3 and use formula 7.4 (5)

    Put t=x+3t = x + 3⁠, dt=dxdt = dx⁠. For −7<x<1-7 < x < 1 we have −4<t<4-4 < t < 4⁠, so

    ∫dx42−(x+3)2=∫dt42−t2=sin⁡−1t4+C.\begin{aligned} \int\frac{dx}{\sqrt{4^2 - (x + 3)^2}} &= \int\frac{dt}{\sqrt{4^2 - t^2}} \\ &= \sin^{-1}\frac t4 + C. \end{aligned}
    This is formula 7.4 (5) with a=4a = 4⁠, and tt lies exactly in its range −4<t<4-4 < t < 4⁠. The book proves 7.4 (5) by putting x=asin⁡θx = a\sin\theta⁠; here that is t=4sin⁡θt = 4\sin\theta⁠, and taking −π2<θ<π2-\frac\pi2 < \theta < \frac\pi2 (the book does not state the range) gives cos⁡θ>0\cos\theta > 0⁠, so 16−16sin⁡2θ=4cos⁡θ\sqrt{16 - 16\sin^2\theta} = 4\cos\theta and θ=sin⁡−1t4\theta = \sin^{-1}\frac t4⁠.
    Formula 7.4 (5); type (8), Section 7.4. Lesson: integrals of particular functions

  4. Return to xx

    ∫dx7−6x−x2=sin⁡−1(x+34)+C.\int\frac{dx}{\sqrt{7 - 6x - x^2}} = \sin^{-1}\left(\frac{x + 3}{4}\right) + C.

    On the domain, −1<x+34<1-1 < \frac{x + 3}{4} < 1⁠, so sin⁡−1\sin^{-1} is defined there.

    Replace tt by x+3x + 3⁠. The domain of the integrand is exactly the set where the argument of sin⁡−1\sin^{-1} lies strictly between −1-1 and 11⁠.

  5. Check: differentiate the answer back

    Let F(x)=sin⁡−1uF(x) = \sin^{-1}u⁠, u=x+34u = \frac{x + 3}{4}⁠, u′=14u' = \frac14⁠. Then

    F′(x)=11−(x+3)216⋅14=116−(x+3)2=17−6x−x2. ✓\begin{aligned} F'(x) &= \frac{1}{\sqrt{1 - \frac{(x + 3)^2}{16}}}\cdot\frac14 \\ &= \frac{1}{\sqrt{16 - (x + 3)^2}} \\ &= \frac{1}{\sqrt{7 - 6x - x^2}}. \ \checkmark \end{aligned}
    1−(x+3)216=1416−(x+3)2\sqrt{1 - \frac{(x + 3)^2}{16}} = \frac14\sqrt{16 - (x + 3)^2}⁠, and the 14\frac14 cancels the inner derivative. The last line is Step 1 read backwards.
    Standard formula (viii), Section 7.2, with the chain rule. Basics: integration undoes differentiation
Result

∫dx7−6x−x2=sin⁡−1(x+34)+C,\int\frac{dx}{\sqrt{7 - 6x - x^2}} = \sin^{-1}\left(\frac{x + 3}{4}\right) + C, valid for −7<x<1-7 < x < 1⁠. (Textbook answer: sin⁡−1(x+34)+C\sin^{-1}\left(\frac{x + 3}{4}\right) + C⁠, the same.)

Check it with numbers

Take x=0x = 0 (inside the domain) and h=0.001h = 0.001⁠:

  • F(0.001)=0.84844012F(0.001) = 0.84844012 and F(−0.001)=0.84768420F(-0.001) = 0.84768420⁠,
  • slope ≈0.84844012−0.847684200.002=0.3780\approx \frac{0.84844012 - 0.84768420}{0.002} = 0.3780⁠,
  • integrand: 17=0.3780\frac{1}{\sqrt7} = 0.3780⁠. ✓
Common mistakes
  1. Sign slip in completing the square: writing 7−(x+3)2−9=−2−(x+3)27 - (x + 3)^2 - 9 = -2 - (x + 3)^2⁠. The −9-9 is inside the bracket that is subtracted, so it becomes +9+9⁠.
  2. Using a logarithm formula. "Constant minus square" under a root gives sin⁡−1\sin^{-1}⁠, not a logarithm.
  3. Writing sin⁡−1x+316\sin^{-1}\frac{x + 3}{16}⁠. The formula divides by a=4a = 4⁠, not by a2a^2⁠.
What to take away
  1. With a negative x2x^2 term, write the quadratic as k2−(x+h)2k^2 - (x + h)^2⁠.
  2. ∫dxk2−(x+h)2=sin⁡−1x+hk+C\int\frac{dx}{\sqrt{k^2 - (x + h)^2}} = \sin^{-1}\frac{x + h}{k} + C⁠, on ∣x+h∣<k|x + h| < k⁠.
  3. The graph of the quadratic shows the domain at once: where it is above the axis.

Every fact used, and where it comes from

FactUsed inSource
Completing the squareSteps 1, 2Type (8), Section 7.4 (Basics: completing the square)
Shift t=x+3t = x + 3⁠, dt=dxdt = dxStep 3Section 7.3.1 (Lesson: substitution)
∫dxa2−x2=sin⁡−1xa+C\int\frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac xa + CStep 3Formula 7.4 (5), Section 7.4
Principal range of sin⁡−1\sin^{-1} is [−π2,π2]\left[-\frac\pi2, \frac\pi2\right]Step 3Chapter 2, Section 2.2
ddxsin⁡−1x=11−x2\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^2}}⁠; chain ruleStep 5Standard formula (viii), Section 7.2; Theorem 4, Section 5.3.1
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