Chapter 7 · Integrals

Exercise 7.4 · Q14: $\int\frac{dx}{\sqrt{8 + 3x - x^2}}$

Exercise 7.4Integrals of some particular functions · Section 7.4

The question

Integrate the function

\frac{1}{\sqrt{8 + 3x - x^2}}.

Exam answer

Let $I = \displaystyle\int\frac{dx}{\sqrt{8 + 3x - x^2}}$ ⁠.

\begin{aligned} &8 + 3x - x^2 \\ &= 8 -\left[\left(x - \frac32\right)^2 - \frac94\right] \\ &= \frac{41}{4} - \left(x - \frac32\right)^2 \\ &= \left(\frac{\sqrt{41}}{2}\right)^2 - \left(x - \frac32\right)^2 \end{aligned}

Put $x - \frac32 = t$ ⁠. Then $dx = dt$ ⁠.

I = \int\frac{dt}{\sqrt{\left(\frac{\sqrt{41}}{2}\right)^2 - t^2}}

[Using $\displaystyle\int\frac{dt}{\sqrt{a^2 - t^2}}$ $= \sin^{-1}\frac ta + C$ ]

I = \sin^{-1}\frac{2t}{\sqrt{41}} + C

$\therefore\ I = \sin^{-1}\left(\dfrac{2x - 3}{\sqrt{41}}\right) + C$ ⁠.

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

The one idea this problem is built on

Under the root is a quadratic with a negative $x^2$ term. Completing the square turns it into "a positive constant minus a square", $\left(\frac{\sqrt{41}}{2}\right)^2 - \left(x - \frac32\right)^2$ ⁠. That is exactly the shape $\frac{1}{\sqrt{a^2 - t^2}}$ of formula 7.4 (5), whose integral is an inverse sine.

Given, to find, and the plan

Given: The integrand $\frac{1}{\sqrt{8 + 3x - x^2}}$ ⁠.
To find: $\int\frac{dx}{\sqrt{8 + 3x - x^2}}$ ⁠, and where the answer is valid.
Plan: (1) Find the domain. (2) Complete the square. (3) Shift $t = x - \frac32$ and use 7.4 (5). (4) Return to $x$ ⁠. (5) Differentiate back.

Basics you need

1Completing the square

$x^2 - px = \left(x - \frac p2\right)^2 - \frac{p^2}{4}$ ⁠: take half the coefficient of $x$ ⁠, square it, and subtract what you added. Basics: completing the square

2A shift is a substitution

If $t = x - h$ with $h$ a constant, then $dt = dx$ ⁠, so the integral keeps its form with $t$ in place of $x$ ⁠. Basics: substitution

3Formula 7.4 (5)

$\int\frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac xa + C$ for $a > 0$ and $-a < x < a$ ⁠. Lesson: integrals of particular functions

4Derivative of $\sin^{-1}$

$\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^2}}$ for $-1 < x < 1$ ⁠, used with the chain rule in the check. Basics: integration undoes differentiation

Solution

Where the integrand exists

We need $8 + 3x - x^2 > 0$ ⁠, that is $x^2 - 3x - 8 < 0$ ⁠. The roots of $x^2 - 3x - 8 = 0$ are
$x = \frac{3 \pm \sqrt{9 + 32}}{2} = \frac{3 \pm \sqrt{41}}{2},$
so the domain is $\frac{3 - \sqrt{41}}{2} < x < \frac{3 + \sqrt{41}}{2}$ ⁠, roughly $-1.70 < x < 4.70$ ⁠.

The quadratic sits under a square root in a denominator, so it must be strictly positive. An upward parabola $x^2 - 3x - 8$ is negative only between its two roots.
Complete the square
$\begin{aligned} 8 + 3x - x^2 &= 8 - \left(x^2 - 3x\right) \\ &= 8 - \left[\left(x - \frac32\right)^2 - \frac94\right] \\ &= \frac{41}{4} - \left(x - \frac32\right)^2 \\ &= \left(\frac{\sqrt{41}}{2}\right)^2 - \left(x - \frac32\right)^2. \end{aligned}$
First take the minus sign outside the $x$ terms, so the $x^2$ inside has coefficient 1. Half of $-3$ is $-\frac32$ ⁠, and $\left(\frac32\right)^2 = \frac94$ ⁠. Then $8 + \frac94 = \frac{32 + 9}{4} = \frac{41}{4}$ ⁠, which is positive, so it can be written as the square $\left(\frac{\sqrt{41}}{2}\right)^2$ ⁠.

Type 7.4 (8): $\int\frac{dx}{\sqrt{ax^2 + bx + c}}$ by completing the square. Basics: completing the square
Shift $t = x - \frac32$ and use 7.4 (5)

Put $t = x - \frac32$ ⁠, so $dt = dx$ ⁠, and let $a = \frac{\sqrt{41}}{2}$ ⁠. On the domain, $-a < t < a$ ⁠. Then
$\begin{aligned} \int\frac{dx}{\sqrt{8 + 3x - x^2}} &= \int\frac{dt}{\sqrt{a^2 - t^2}} \\ &= \sin^{-1}\frac ta + C. \end{aligned}$
The domain found in Step 1 is exactly $\left(x - \frac32\right)^2 < \frac{41}{4}$ ⁠, that is $|t| < a$ ⁠, which is where formula 7.4 (5) holds. The shift does not change $dx$ ⁠, because the derivative of $x - \frac32$ is 1.

Formula 7.4 (5); substitution, Section 7.3.1. Lesson: integrals of particular functions
Return to $x$
$\frac ta = \frac{x - \frac32}{\frac{\sqrt{41}}{2}} = \frac{2x - 3}{\sqrt{41}},$
so
$\int\frac{dx}{\sqrt{8 + 3x - x^2}} = \sin^{-1}\frac{2x - 3}{\sqrt{41}} + C.$
Multiply the top and bottom of the fraction by 2 to clear the halves. This is exactly the textbook's form.
Check: differentiate the answer

By the chain rule, with inner function $u = \frac{2x - 3}{\sqrt{41}}$ and $u' = \frac{2}{\sqrt{41}}$ ⁠:
$\begin{aligned} \frac{d}{dx}\sin^{-1}u &= \frac{u'}{\sqrt{1 - u^2}} = \frac{\frac{2}{\sqrt{41}}}{\sqrt{\frac{41 - (2x - 3)^2}{41}}} \\ &= \frac{2}{\sqrt{41 - (2x - 3)^2}} = \frac{2}{\sqrt{32 + 12x - 4x^2}} \\ &= \frac{2}{2\sqrt{8 + 3x - x^2}} = \frac{1}{\sqrt{8 + 3x - x^2}}. \ \checkmark \end{aligned}$
$(2x - 3)^2 = 4x^2 - 12x + 9$ ⁠, so $41 - (2x - 3)^2 = 32 + 12x - 4x^2$ $= 4(8 + 3x - x^2)$ ⁠, and $\sqrt4 = 2$ ⁠. On the domain $|u| < 1$ ⁠, so the derivative of $\sin^{-1}$ is defined.

Chain rule (Theorem 4, Section 5.3.1); derivative (viii), Section 7.2. Basics: integration undoes differentiation

Result

\int\frac{dx}{\sqrt{8 + 3x - x^2}} = \sin^{-1}\frac{2x - 3}{\sqrt{41}} + C,

valid for $\frac{3 - \sqrt{41}}{2} < x < \frac{3 + \sqrt{41}}{2}$ ⁠. (Textbook answer: $\sin^{-1}\left(\frac{2x - 3}{\sqrt{41}}\right) + C$ ⁠.)

Check it with numbers

By Section 7.8, $\int_a^b f(x)\,dx = F(b) - F(a)$ for our answer $F$ ⁠. The middle column is a computer's sum of very many thin strips, found without our answer:

Interval	Sum of thin strips	$F(b) - F(a)$
$[0, 1]$	$0.3308006$	$0.3308006$
$[-1, 4]$	$1.7921108$	$1.7921108$

On $[0, 1]$ ⁠, $F(1) - F(0) = \sin^{-1}\frac{3}{\sqrt{41}} - \sin^{-1}\frac{1}{\sqrt{41}}$ ⁠; on $[-1, 4]$ ⁠, it is $2\sin^{-1}\frac{5}{\sqrt{41}}$ ⁠.

Common mistakes

Sign slip when taking out the minus. $8 + 3x - x^2 = 8 - (x^2 - 3x)$ ⁠, not $8 - (x^2 + 3x)$ ⁠.
Forgetting to add the $\frac94$ ⁠. Inside the bracket we subtract $\frac94$ ⁠; the minus sign outside turns it into $+\frac94$ ⁠, giving $\frac{41}{4}$ ⁠, not $\frac{23}{4}$ ⁠.
Using $a = \frac{41}{4}$ instead of $a = \frac{\sqrt{41}}{2}$ ⁠. In 7.4 (5) the constant is $a^2$ ⁠; $a$ is its square root.
Choosing a log formula. "Constant minus square" under a root gives $\sin^{-1}$ ⁠; "square minus or plus constant" gives a log.

What to take away

For $\int\frac{dx}{\sqrt{c + bx - x^2}}$ ⁠: complete the square to $a^2 - (x - h)^2$ ⁠, and the answer is $\sin^{-1}\frac{x - h}{a} + C$ ⁠.
The domain of the integrand is exactly $|x - h| < a$ ⁠, the range where the inverse sine formula works.

Every fact used, and where it comes from

Fact	Used in	Source
Completing the square	Step 2	Type 7.4 (8) (Basics: completing the square)
$\int\frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac xa + C$	Step 3	Formula 7.4 (5)
Substitution $t = x - \frac32$ ⁠, $dt = dx$	Step 3	Section 7.3.1 (Basics: substitution)
$\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^2}}$	Step 5	Derivative (viii), Section 7.2
Chain rule	Step 5	Theorem 4, Section 5.3.1
$\int_a^b f(x)\,dx = F(b) - F(a)$	Numerical check	Section 7.8 (Lesson: definite integrals)

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Exercise 7.4 · Q14: ∫dx8+3x−x2\int\frac{dx}{\sqrt{8 + 3x - x^2}}∫8+3x−x2​dx​

Basics you need

Solution

Where the integrand exists

Complete the square

Shift t=x−32t = x - \frac32t=x−23​ and use 7.4 (5)

Return to xxx

Check: differentiate the answer

Every fact used, and where it comes from

Exercise 7.4 · Q14: $\int\frac{dx}{\sqrt{8 + 3x - x^2}}$

Shift $t = x - \frac32$ and use 7.4 (5)

Return to $x$