Chapter 7 · Integrals

Exercise 7.7 · Q1: $\int\sqrt{4 - x^2}\,dx$

Exercise 7.7Integrals of some more types · Section 7.6.2

The question

Integrate the function

\sqrt{4 - x^2}.

Exam answer

Let $I = \displaystyle\int\sqrt{4 - x^2}\,dx$ $= \displaystyle\int\sqrt{2^2 - x^2}\,dx$ ⁠.

Using $\displaystyle\int\sqrt{a^2 - x^2}\,dx = \frac{x}{2}\sqrt{a^2 - x^2}$ $+ \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C$ with ${a = 2,}$

\begin{aligned} I &= \frac{x}{2}\sqrt{4 - x^2} \\ &\quad + \frac42\sin^{-1}\frac{x}{2} + C \end{aligned}

$\therefore\ I = \dfrac{x}{2}\sqrt{4 - x^2} + 2\sin^{-1}\dfrac{x}{2}$ $+ C$ ⁠.

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

The one idea this problem is built on

$4 - x^2 = 2^2 - x^2$ is exactly the shape $a^2 - x^2$ with $a = 2$ ⁠, so the textbook's formula 7.6.2 (iii) applies at once:

\begin{aligned} \int\sqrt{a^2 - x^2}\,dx &= \frac{x}{2}\sqrt{a^2 - x^2} \\ &\quad + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C. \end{aligned}

We first see where that formula comes from (integration by parts, taking the constant function 1 as the second function), then use it. The two terms have a picture: under the semicircle $y = \sqrt{4 - x^2}$ ⁠, the area from 0 to $x$ is a right triangle plus a circular sector.

Given, to find, and the plan

Given: The integrand $\sqrt{4 - x^2}$ ⁠.
To find: $\int\sqrt{4 - x^2}\,dx$ ⁠, and where the answer is valid.
Plan: (1) Find where the square root is real. (2) Derive formula 7.6.2 (iii) by parts. (3) Apply it with $a = 2$ ⁠. (4) Read the answer as triangle plus sector. (5) Differentiate back.

Basics you need

1Integration undoes differentiation

$\int f(x)\,dx = F(x) + C$ means $F'(x) = f(x)$ ⁠. So any answer can be tested by differentiating it. Basics: integration undoes differentiation

2Integration by parts

$\int f_1 f_2\,dx = f_1\int f_2\,dx$ $- \int\left(f_1'\int f_2\,dx\right)dx$ ⁠. Here we take $f_2 = 1$ ⁠, so $\int f_2\,dx = x$ ⁠. Lesson: integration by parts

3Formula 7.4 (5)

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

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

$\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^2}}$ for $|x| < 1$ (standard formula (viii), Section 7.2), and by the chain rule $\frac{d}{dx}\sin^{-1}\frac{x}{2} = \frac{1}{\sqrt{4 - x^2}}$ ⁠. Basics: substitution and the chain rule

Solution

Where the square root is real

We need $4 - x^2 \ge 0$ ⁠, that is $x^2 \le 4$ ⁠, that is
$-2 \le x \le 2.$
So the integrand lives on the closed interval $[-2, 2]$ ⁠. Its graph $y = \sqrt{4 - x^2}$ is the upper half of the circle $x^2 + y^2 = 4$ (radius 2, centre the origin).

A square root of a negative number is not a real number, so outside $[-2, 2]$ there is nothing to integrate. Squaring $y = \sqrt{4 - x^2}$ gives $x^2 + y^2 = 4$ with $y \ge 0$ ⁠: the upper semicircle.
Where formula 7.6.2 (iii) comes from

Let $a > 0$ and $I = \int\sqrt{a^2 - x^2}\cdot 1\,dx$ for $|x| < a$ ⁠. By parts with first function $\sqrt{a^2 - x^2}$ and second function 1:
$\begin{aligned} I &= x\sqrt{a^2 - x^2} - \int x\cdot\frac{-x}{\sqrt{a^2 - x^2}}\,dx \\ &= x\sqrt{a^2 - x^2} + \int\frac{x^2}{\sqrt{a^2 - x^2}}\,dx. \end{aligned}$
Now write $x^2 = a^2 - (a^2 - x^2)$ in the last numerator:
$\begin{aligned} \int\frac{x^2\,dx}{\sqrt{a^2 - x^2}} &= a^2\int\frac{dx}{\sqrt{a^2 - x^2}} \\ &\quad - \int\sqrt{a^2 - x^2}\,dx \\ &= a^2\sin^{-1}\frac{x}{a} - I. \end{aligned}$
So $I = x\sqrt{a^2 - x^2} + a^2\sin^{-1}\frac{x}{a} - I$ ⁠, and solving for $I$ (with the constant added at the end):
$\begin{aligned} I &= \frac{x}{2}\sqrt{a^2 - x^2} \\ &\quad + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C. \end{aligned}$
The derivative of $\sqrt{a^2 - x^2}$ is $\frac{-2x}{2\sqrt{a^2 - x^2}} = \frac{-x}{\sqrt{a^2 - x^2}}$ (chain rule), which exists for $|x| < a$ ⁠. The trick $x^2 = a^2 - (a^2 - x^2)$ splits $\frac{x^2}{\sqrt{a^2 - x^2}}$ into $\frac{a^2}{\sqrt{a^2 - x^2}}$ minus $\frac{a^2 - x^2}{\sqrt{a^2 - x^2}} = \sqrt{a^2 - x^2}$ ⁠, so the unknown $I$ reappears and we can solve for it, just as the textbook does for formula (i) in Section 7.6.2. The textbook also mentions the alternative substitution $x = a\sin\theta$ ⁠.

Integration by parts, Section 7.6; formula 7.4 (5). Lesson: integration by parts
Apply it with $a = 2$

Here $a^2 = 4$ ⁠, $a = 2$ ⁠, and $\frac{a^2}{2} = 2$ ⁠:
$\begin{aligned} \int\sqrt{4 - x^2}\,dx &= \frac{x}{2}\sqrt{4 - x^2} \\ &\quad + 2\sin^{-1}\frac{x}{2} + C. \end{aligned}$
Always take $a$ positive ( $a = 2$ ⁠, not $-2$ ⁠), because the formula was derived for $a > 0$ ⁠: then $\frac{x}{a}$ lies in $[-1, 1]$ exactly when $|x| \le a$ ⁠, so $\sin^{-1}\frac{x}{a}$ is defined on the whole domain of the integrand.

Formula 7.6.2 (iii). Lesson: integration by parts
What the two terms mean: triangle plus sector

Take $0 < x \le 2$ and the point $P = \left(x, \sqrt{4 - x^2}\right)$ on the semicircle. The area under the semicircle from 0 to $x$ is the answer $F(x) = \frac{x}{2}\sqrt{4 - x^2} + 2\sin^{-1}\frac{x}{2}$ minus $F(0) = 0$ ⁠. The segment $OP$ cuts that region into two pieces:
- a right triangle with base $x$ and height $\sqrt{4 - x^2}$ ⁠: area $\frac{x}{2}\sqrt{4 - x^2}$ ⁠;
- a circular sector of radius 2 between $OA$ (the $y$ -axis) and $OP$ ⁠. Its angle $\theta$ satisfies $\sin\theta = \frac{x}{2}$ ⁠, so $\theta = \sin^{-1}\frac{x}{2}$ ⁠, and its area is $\frac12\cdot 2^2\cdot\theta = 2\sin^{-1}\frac{x}{2}$ ⁠.
At $x = 1.2$ ⁠: triangle $\frac{1.2}{2}\cdot 1.6 = 0.96$ plus sector $2\sin^{-1}0.6 \approx 1.2870022$ gives $F(1.2) - F(0) \approx 2.2470022$ ⁠, the area under the semicircle from 0 to 1.2.

By Section 7.8, $F(x) - F(0)$ is the area under the curve from 0 to $x$ ⁠. The picture confirms the formula with plain geometry: the "algebraic" term is a triangle, the $\sin^{-1}$ term is a sector. At $x = 2$ the triangle vanishes and the sector is a quarter disc of area $\pi$ ⁠, and indeed $F(2) = 2\cdot\frac{\pi}{2} = \pi$ ⁠.

Area of a sector of radius $r$ and angle $\theta$ (radians) is $\frac12 r^2\theta$ ⁠. Basics: area as a sum of thin strips
Check: differentiate the answer

For $-2 < x < 2$ ⁠, by the product rule and the chain rule,
$\begin{aligned} \frac{d}{dx}\left[\frac{x}{2}\sqrt{4 - x^2}\right] &= \frac12\sqrt{4 - x^2} - \frac{x^2}{2\sqrt{4 - x^2}}, \\ \frac{d}{dx}\left[2\sin^{-1}\frac{x}{2}\right] &= 2\cdot\frac{1}{\sqrt{1 - \frac{x^2}{4}}}\cdot\frac12 \\ &= \frac{2}{\sqrt{4 - x^2}}. \end{aligned}$
Put everything over $2\sqrt{4 - x^2}$ ⁠:
$\begin{aligned} F'(x) &= \frac{(4 - x^2) - x^2 + 4}{2\sqrt{4 - x^2}} \\ &= \frac{2(4 - x^2)}{2\sqrt{4 - x^2}} = \sqrt{4 - x^2}.\ \checkmark \end{aligned}$
$\frac12\sqrt{4 - x^2} = \frac{4 - x^2}{2\sqrt{4 - x^2}}$ and $\frac{2}{\sqrt{4 - x^2}} = \frac{4}{2\sqrt{4 - x^2}}$ ⁠. Also $\sqrt{1 - \frac{x^2}{4}} = \frac{\sqrt{4 - x^2}}{2}$ ⁠. Finally $\frac{4 - x^2}{\sqrt{4 - x^2}} = \sqrt{4 - x^2}$ because $4 - x^2 > 0$ inside the interval. At $x = \pm 2$ the integrand is 0, and the answer's one-sided slope there is also 0, so the answer works on all of $[-2, 2]$ ⁠.

Product rule, chain rule, standard formula (viii), Section 7.2. Basics: integration undoes differentiation

Result

\begin{aligned} \int\sqrt{4 - x^2}\,dx &= \frac{x}{2}\sqrt{4 - x^2} \\ &\quad + 2\sin^{-1}\frac{x}{2} + C, \end{aligned}

valid for $-2 \le x \le 2$ ⁠. (Textbook answer: $\frac12 x\sqrt{4 - x^2} + 2\sin^{-1}\frac{x}{2} + C$ ⁠, the same.)

Check it with numbers

By Section 7.8, $\int_a^b f(x)\,dx = F(b) - F(a)$ for our answer $F$ ⁠, 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.

Interval	Sum of thin strips	$F(b) - F(a)$
$[0, 1]$	$1.9132230$	$1.9132230$
$[-1, 1.8]$	$4.9373638$	$4.9373638$

Common mistakes

Using $\frac{a^2}{2}\sin^{-1}\frac{x}{a}$ with $a^2$ in place of $a$ ⁠. Inside $\sin^{-1}$ it is $\frac{x}{a} = \frac{x}{2}$ ⁠, not $\frac{x}{4}$ ⁠.
Mixing up the formulas. $\sqrt{a^2 - x^2}$ gives $\sin^{-1}$ ⁠; $\sqrt{x^2 \pm a^2}$ gives a logarithm. Look at which term carries the minus sign.
Forgetting the $\frac12$ in front of $x\sqrt{4 - x^2}$ ⁠. It comes from solving $2I = \dots$ in Step 2; the derivative check catches it.

What to take away

$\int\sqrt{a^2 - x^2}\,dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C$ ⁠, valid for $|x| \le a$ ⁠.
By parts with second function 1, the original integral reappears; solve for it.
Geometrically: triangle plus sector under a semicircle.

Every fact used, and where it comes from

Fact	Used in	Source
Integration by parts	Step 2	Section 7.6 (Lesson: integration by parts)
$\int\frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac{x}{a} + C$	Step 2	Formula 7.4 (5)
$\int\sqrt{a^2 - x^2}\,dx$ formula	Steps 2, 3	Formula 7.6.2 (iii), Section 7.6.2
Area of a sector $\frac12 r^2\theta$	Step 4	Geometry (Basics: area)
$\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^2}}$	Step 5	Standard formula (viii), Section 7.2
$\int_a^b f(x)\,dx = F(b) - F(a)$	Step 4, numerical check	Section 7.8 (Lesson: definite integrals)

Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →

Exercise 7.7 · Q1: ∫4−x2 dx\int\sqrt{4 - x^2}\,dx∫4−x2​dx

Basics you need

Solution

Where the square root is real

Where formula 7.6.2 (iii) comes from

Apply it with a=2a = 2a=2

What the two terms mean: triangle plus sector

Check: differentiate the answer

Every fact used, and where it comes from

Exercise 7.7 · Q1: $\int\sqrt{4 - x^2}\,dx$

Apply it with $a = 2$