Chapter 7 · Integrals

Exercise 7.8 · Q10: $\int_0^1\frac{dx}{1 + x^2}$

Exercise 7.8Fundamental theorem of calculus · Section 7.8

The question

Evaluate the definite integral

\int_0^1\frac{dx}{1 + x^2}.

Exam answer

Let $I = \displaystyle\int_0^1\frac{dx}{1 + x^2}$ ⁠.

Using $\displaystyle\int\frac{dx}{1 + x^2} = \tan^{-1}x + C,$

\begin{aligned} I &= \Big[\tan^{-1}x\Big]_0^1 \\ &= \tan^{-1}1 - \tan^{-1}0 \\ &= \frac{\pi}{4} - 0 \end{aligned}

$\therefore\ I = \dfrac{\pi}{4}$ ⁠.

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

The one idea this problem is built on

The second fundamental theorem (Theorem 2, Section 7.8.3): $\int_a^b f(x)\,dx = F(b) - F(a)$ ⁠. Here $F(x) = \tan^{-1}x$ (standard formula (x), Section 7.2), and the answer comes from two principal values: $\tan^{-1}1 = \frac{\pi}{4}$ and $\tan^{-1}0 = 0$ ⁠. So a simple fraction has an area involving $\pi$ ⁠.

Given, to find, and the plan

Given: $f(x) = \frac{1}{1 + x^2}$ on $[0, 1]$ ⁠.
To find: The exact value of $\int_0^1\frac{dx}{1 + x^2}$ ⁠.
Plan: (1) Continuity. (2) Antiderivative $\tan^{-1}x$ ⁠, checked by differentiating. (3) $F(1) - F(0)$ with principal values. (4) Size check against the graph.

Basics you need

1Second fundamental theorem

$\int_a^b f(x)\,dx = \big[F(x)\big]_a^b = F(b) - F(a)$ for $f$ continuous on $[a, b]$ and $F' = f$ ⁠. Lesson: definite integrals and the fundamental theorem

2The $\tan^{-1}$ formula

$\frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^2}$ for every real $x$ ⁠, so $\int\frac{dx}{1 + x^2} = \tan^{-1}x + C$ (standard formula (x), Section 7.2). Lesson: integrals of particular functions

3Principal values

$\tan^{-1}x$ takes values in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ ⁠: $\tan^{-1}0 = 0$ ⁠, $\tan^{-1}1 = \frac{\pi}{4}$ (Chapter 2). Basics: trigonometric values

Solution

Check continuity on $[0, 1]$

The denominator satisfies $1 + x^2 \ge 1 > 0$ for every real $x$ ⁠, so it is never 0. The quotient $\frac{1}{1 + x^2}$ of continuous functions is therefore continuous everywhere, in particular on $[0, 1]$ ⁠.

Theorem 2 needs a continuous integrand on the closed interval (Remark (iv) after Theorem 2). A square is never negative, so $1 + x^2$ can never vanish.

Quotient of continuous functions: Chapter 5, Theorem 1, Section 5.2.1. Lesson: the fundamental theorem
Antiderivative, and check it

By standard formula (x),
$\int\frac{dx}{1 + x^2} = \tan^{-1}x + C, \qquad F(x) = \tan^{-1}x.$
Check: $F'(x) = \frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^2} = f(x)$ for every real $x$ ⁠.

The formula is the Chapter 5 derivative of $\tan^{-1}x$ read backwards. It is valid for all real $x$ ⁠, so certainly on $[0, 1]$ ⁠.

Standard formula (x), Section 7.2; the derivative is in Chapter 5, Section 5.3.3. Basics: integration undoes differentiation
Evaluate $F(1) - F(0)$
$\begin{aligned}\int_0^1\frac{dx}{1 + x^2} &= \tan^{-1}1 - \tan^{-1}0 \\ &= \frac{\pi}{4} - 0 = \frac{\pi}{4} \approx 0.7853982.\end{aligned}$
$\tan^{-1}1$ is the angle in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ whose tangent is 1, namely $\frac{\pi}{4}$ (not $\frac{5\pi}{4}$ ⁠, which is outside the principal range). Likewise $\tan^{-1}0 = 0$ ⁠.

$\tan\frac{\pi}{4} = 1$ ⁠, $\tan 0 = 0$ ⁠. Basics: trigonometric values
Does the size make sense?

On $[0, 1]$ the curve $y = \frac{1}{1 + x^2}$ falls from $f(0) = 1$ to $f(1) = \frac12$ ⁠. On a base of width 1, the area is between the two rectangles:
$\frac12 < \frac{\pi}{4} \approx 0.785 < 1.$

The shaded region under $y = \frac{1}{1 + x^2}$ from 0 to 1 has area $\frac{\pi}{4} \approx 0.785$ ⁠. The curve falls from 1 to $\frac12$ ⁠, so the area lies between $\frac12$ and 1.

Since $f > 0$ ⁠, the integral is the shaded area (Section 7.8.1). Read backwards, this result says $\pi = 4\int_0^1\frac{dx}{1 + x^2}$ ⁠: an area under a simple curve that pins down $\pi$ ⁠.

Definite integral as area, Section 7.8.1. Basics: area as a sum of thin strips

Result

\int_0^1\frac{dx}{1 + x^2} = \frac{\pi}{4} \approx 0.7853982.

(Textbook answer: $\frac{\pi}{4}$ ⁠, the same.)

Check it with numbers

Numerical integration of $\frac{1}{1 + x^2}$ from 0 to 1 (thin strips, no antiderivative used) gives $0.7853982$ ⁠, and $\frac{\pi}{4} \approx \frac{3.1415927}{4} \approx 0.7853982$ ⁠. They agree.

Common mistakes

Using the log rule. $\int\frac{dx}{1 + x^2}$ is not $\log(1 + x^2)$ ⁠: the numerator is not the derivative $2x$ of the denominator.
Degrees instead of radians. $\tan^{-1}1 = \frac{\pi}{4}$ ⁠, not 45.
Mixing up with $\sin^{-1}$ ⁠. $\frac{1}{\sqrt{1 - x^2}}$ gives $\sin^{-1}x$ ⁠; $\frac{1}{1 + x^2}$ gives $\tan^{-1}x$ ⁠.

What to take away

$\int\frac{dx}{1 + x^2} = \tan^{-1}x + C$ ⁠, valid for all real $x$ ⁠.
Inverse-trigonometric antiderivatives are evaluated with principal values.
$\int_0^1\frac{dx}{1 + x^2} = \frac{\pi}{4}$ is worth remembering.

Every fact used, and where it comes from

Fact	Used in	Source
$\int_a^b f = F(b) - F(a)$ for continuous $f$	Steps 1, 3	Theorem 2, Section 7.8.3 (Lesson)
Quotient of continuous functions is continuous	Step 1	Chapter 5, Theorem 1, Section 5.2.1
$\int\frac{dx}{1 + x^2} = \tan^{-1}x + C$	Step 2	Standard formula (x), Section 7.2 (Lesson)
$\frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^2}$	Step 2	Chapter 5, Section 5.3.3
Principal values of $\tan^{-1}$	Step 3	Chapter 2 (Basics)
Definite integral as area	Step 4	Section 7.8.1 (Basics)

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Exercise 7.8 · Q10: ∫01dx1+x2\int_0^1\frac{dx}{1 + x^2}∫01​1+x2dx​

Basics you need

Solution

Check continuity on [0,1][0, 1][0,1]

Antiderivative, and check it

Evaluate F(1)−F(0)F(1) - F(0)F(1)−F(0)

Does the size make sense?

Every fact used, and where it comes from

Exercise 7.8 · Q10: $\int_0^1\frac{dx}{1 + x^2}$

Check continuity on $[0, 1]$

Evaluate $F(1) - F(0)$