Example 24: $\int\sqrt{3-2x-x^2}\,dx$

Basics you need

Solution

1. Complete the square and find where the root exists

   Group the $x$ terms and take out the minus sign:

   $$\begin{aligned}3 - 2x - x^2 &= 3 - (x^2 + 2x) \\ &= 3 - \left[(x + 1)^2 - 1\right] \\ &= 4 - (x + 1)^2 = 2^2 - (x+1)^2.\end{aligned}$$

   The root needs $4 - (x+1)^2 \ge 0$⁠, that is $(x+1)^2 \le 4$⁠, that is $|x + 1| \le 2$⁠, which means $-2 \le x + 1 \le 2$⁠, or $-3 \le x \le 1$. Outside this interval the integrand does not exist, so the question only makes sense on $[-3, 1]$⁠.

   The form "constant minus a square" is the pattern $a^2 - y^2$ of formula (iii). It also shows at once that the quantity is largest (equal to 4) at $x = -1$ and falls to 0 at $x = -3$ and $x = 1$⁠. Indeed, a point of the graph at height $h = \sqrt{4 - (x+1)^2}$ satisfies $(x+1)^2 + h^2 = 4$ with $h \ge 0$⁠: the graph is the upper half of the circle with centre $(-1, 0)$ and radius 2 (figure). (We call the height $h$ here because from Step 2 on the letter $y$ stands for $x + 1$⁠.)
   Expand to check: $4 - (x^2 + 2x + 1) = 3 - 2x - x^2$⁠. Basics: completing the square

2. Shift: put $y = x + 1$

   With $y = x + 1$ we have $dy = dx$⁠, and $y$ runs over $[-2, 2]$ as $x$ runs over $[-3, 1]$⁠:

   $$\int\sqrt{3 - 2x - x^2}\,dx = \int\sqrt{4 - y^2}\,dy.$$
   This makes the integrand exactly $\sqrt{a^2 - y^2}$ with $a = 2$⁠. As in any shift, the derivative of $x + 1$ is 1, so $dx$ becomes $dy$ with no extra factor.
   Integration by substitution, Section 7.3.1. Lesson: integration by substitution

3. Apply formula 7.6.2 (iii) with $a = 2$

   Here $a^2 = 4$⁠, $\frac{a^2}{2} = 2$ and $\frac{y}{a} = \frac{y}{2}$⁠:

   $$\int\sqrt{4 - y^2}\,dy = \frac{1}{2}y\sqrt{4 - y^2} + 2\sin^{-1}\frac{y}{2} + C.$$

   The book writes the middle step as $\frac{4}{2}\sin^{-1}\frac{y}{2}$⁠; that is the same 2.

   Because $-2 \le y \le 2$⁠, the number $\frac{y}{2}$ lies in $[-1, 1]$⁠, the domain of $\sin^{-1}$⁠, so the formula's right-hand side is defined wherever the integrand is.
   Formula (iii) is the one with $a^2 - x^2$ under the root: its answer has $\sin^{-1}$⁠, not a log (compare Example 23). Lesson: integration by parts

4. Return to $x$

   Put back $y = x + 1$ and $4 - y^2 = 3 - 2x - x^2$ (Step 1):

   $$\begin{aligned}\int\sqrt{3 - 2x - x^2}\,dx = {}&\tfrac{1}{2}(x + 1)\sqrt{3 - 2x - x^2} \\ &+ 2\sin^{-1}\!\left(\frac{x + 1}{2}\right) + C.\end{aligned}$$

   Where it is valid. Both sides are defined for $-3 \le x \le 1$⁠. The derivative check in Step 5 is done on the open interval $-3 < x < 1$⁠, where the root is positive. At the two end points the answer is still continuous, and its one-sided derivatives there are 0, which is the value of the integrand at those points.

   The answer must be in the variable of the question. Writing $4 - y^2$ back as $3 - 2x - x^2$ gives the book's printed form.

5. Check: differentiate the answer

   Let $y = x + 1$ and $R = \sqrt{4 - y^2}$⁠, for $-3 < x < 1$ (so $R > 0$⁠). By the chain rule $\frac{dR}{dx} = \frac{-2y}{2R} = -\frac{y}{R}$⁠.

   First term, product rule:

   $$\frac{d}{dx}\left(\tfrac{1}{2}yR\right) = \tfrac{1}{2}R - \tfrac{1}{2}y\cdot\frac{y}{R} = \frac{R^2 - y^2}{2R}.$$

   Second term, chain rule with $u = \frac{y}{2}$⁠, $\frac{du}{dx} = \frac{1}{2}$⁠:

   $$\frac{d}{dx}\,2\sin^{-1}\frac{y}{2} = \frac{2\cdot\frac{1}{2}}{\sqrt{1 - \frac{y^2}{4}}} = \frac{1}{\frac{R}{2}} = \frac{2}{R}.$$

   (We used $\sqrt{1 - \frac{y^2}{4}} = \frac{\sqrt{4 - y^2}}{2} = \frac{R}{2}$⁠.) Add, using $R^2 = 4 - y^2$⁠:

   $$\frac{R^2 - y^2 + 4}{2R} = \frac{8 - 2y^2}{2R} = \frac{4 - y^2}{R} = R.$$

   So the derivative is $R = \sqrt{3 - 2x - x^2}$⁠, the integrand. The answer is confirmed.

   By definition, $F$ is an antiderivative of $f$ exactly when $F' = f$⁠. The same computation proves formula 7.6.2 (iii) for $a = 2$⁠.
   $\frac{d}{dx}\sin^{-1}u = \frac{u'}{\sqrt{1-u^2}}$ is formula (viii) of the list in Section 7.2 read backwards. Basics: integration undoes differentiation

6. A check from the picture

   Call the answer without C $F(x)$⁠. The definite integral $F(1) - F(-3)$ is the area of the whole half disc, which is $\frac{1}{2}\pi\cdot 2^2 = 2\pi$⁠. From the formula:

   $$\begin{aligned}F(1) &= \tfrac{1}{2}\cdot 2\cdot 0 + 2\sin^{-1}1 = 2\cdot\tfrac{\pi}{2} = \pi, \\ F(-3) &= \tfrac{1}{2}\cdot(-2)\cdot 0 + 2\sin^{-1}(-1) = -\pi.\end{aligned}$$

   So $F(1) - F(-3) = 2\pi$⁠, exactly the area of the half disc.

   This is an independent check: the area of a half disc comes from geometry, not from the formula. The $\sin^{-1}$ term produces the $\pi$⁠, which is why a circle's integral has an inverse sine in it.
   A definite integral of a positive function is the area under its graph (Sections 7.7 and 7.8). Basics: area as a sum of thin strips

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