Chapter 7 · Integrals

Exercise 7.1 · Q18: $\int \sec x\,(\sec x + \tan x)\,dx$

Exercise 7.1Integration as anti-differentiation · Section 7.2

The question

Find the following integral:

\int \sec x\,(\sec x + \tan x)\,dx

Exam answer

Let $I = \displaystyle\int\sec x\left(\sec x + \tan x\right)dx$ ⁠.

I = \int\left(\sec^2 x + \sec x\tan x\right)dx

$\therefore\ I = \tan x + \sec x + C$ ⁠.

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

The one idea this problem is built on

Multiplying out gives $\sec^2 x + \sec x\tan x$ ⁠, and both terms are derivatives we already know: $\frac{d}{dx}\tan x = \sec^2 x$ and $\frac{d}{dx}\sec x = \sec x\tan x$ ⁠. So the integral is read straight off the list of standard integrals: $\tan x + \sec x + C$ ⁠.

Given, to find, and the plan

Given: The integrand $f(x) = \sec x\,(\sec x + \tan x)$ ⁠.
To find: $\int f(x)\,dx$ ⁠.
Plan: (1) Note where $f$ is defined. (2) Expand the product. (3) Use formulae (iv) and (vi) of Section 7.2. (4) Differentiate the answer to check it.

Basics you need

1Integration undoes differentiation

$\int f(x)\,dx = F(x) + C$ means $F'(x) = f(x)$ ⁠. Basics: integration undoes differentiation

2Secant and tangent

$\sec x = \frac{1}{\cos x}$ and $\tan x = \frac{\sin x}{\cos x}$ ⁠; both exist exactly when $\cos x \ne 0$ ⁠, that is $x \ne (2n+1)\frac{\pi}{2}$ ⁠, $n \in \mathbb{Z}$ ⁠. Basics: trigonometric facts for integrals

3Two standard integrals

(iv) $\int \sec^2 x\,dx = \tan x + C$ and (vi) $\int \sec x\tan x\,dx = \sec x + C$ (list in Section 7.2). Basics: trigonometric facts for integrals

4Sums

$\int (f + g)\,dx = \int f\,dx + \int g\,dx$ (Property III, Section 7.2.1). Lesson: integration as anti-differentiation

Solution

Where is the integrand defined?

$\sec x$ and $\tan x$ both have $\cos x$ in the denominator, so the integrand is defined exactly when
$\cos x \ne 0, \quad\text{i.e.}\quad x \ne (2n+1)\tfrac{\pi}{2},\ n \in \mathbb{Z}.$
The domain is a string of open intervals such as $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ ⁠, $\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)$ ⁠, and so on. The answer is an antiderivative on each of them. (Strictly the constant may differ from one interval to the next; like the textbook we write one $C$ ⁠.)

The Note in Section 7.2: always keep the interval of definition in mind.
Expand the product
$\sec x\,(\sec x + \tan x) = \sec^2 x + \sec x\tan x.$
There is no product rule for integrals, but after multiplying out, each term is one of the derivatives in the list of Section 7.2. Spotting that is the whole problem.

Distributive law. Basics: trigonometric facts for integrals
Integrate each term
$\begin{aligned} &\int (\sec^2 x + \sec x\tan x)\,dx \\ &\quad = \int \sec^2 x\,dx + \int \sec x\tan x\,dx \\ &\quad = \tan x + \sec x + C. \end{aligned}$
Property III splits the sum. Formula (iv) holds because $\frac{d}{dx}\tan x = \sec^2 x$ ⁠; formula (vi) holds because $\frac{d}{dx}\sec x = \sec x\tan x$ ⁠. The two constants combine into one $C$ ⁠.

Formulae (iv) and (vi), Section 7.2. The textbook's Example 3(ii) is the same problem with cosec and cot: $\int \operatorname{cosec} x(\operatorname{cosec} x + \cot x)\,dx$ $= -\cot x - \operatorname{cosec} x + C$ ⁠.
Check: differentiate the answer back
$\begin{aligned} \frac{d}{dx}(\tan x + \sec x + C) &= \sec^2 x + \sec x\tan x \\ &= \sec x\,(\sec x + \tan x). \end{aligned}$
This is the integrand at every $x$ with $\cos x \ne 0$ ⁠, so the answer is correct.

The last line takes out the common factor $\sec x$ ⁠, which reverses Step 2. Both derivatives exist exactly where $\cos x \ne 0$ ⁠, the same set on which the integrand is defined.

Derivatives (iv) and (vi) in the list of Section 7.2; Property I, Section 7.2.1. Basics: integration undoes differentiation

Result

Wherever $\cos x \ne 0$ ⁠,

\int \sec x\,(\sec x + \tan x)\,dx = \tan x + \sec x + C.

(Textbook answer: $\tan x + \sec x + C$ ⁠.)

Check it with numbers

At $x = \frac{\pi}{4}$ ⁠, $\sec x = \sqrt2$ and $\tan x = 1$ ⁠, so the integrand is $\sqrt2(\sqrt2 + 1) = 2 + \sqrt2 \approx 3.4142136$ ⁠. With $F(x) = \tan x + \sec x$ ⁠: $F\!\left(\frac{\pi}{4} + 0.001\right) \approx 2.41763190$ and $F\!\left(\frac{\pi}{4} - 0.001\right) \approx 2.41080346$ ⁠, so the slope of $F$ there is about

\frac{2.41763190 - 2.41080346}{0.002} \approx 3.4142,

which matches $2 + \sqrt2$ to four decimal places.

(Each value of $F$ is rounded to 8 decimal places, so the slope is reliable to about 4 decimal places.)

Common mistakes

Mixing up the pairs. $\sec^2 x$ integrates to $\tan x$ and $\sec x\tan x$ integrates to $\sec x$ ⁠, not the other way round.
Trying $\int \sec x\,dx\cdot\int(\sec x + \tan x)\,dx$ ⁠. There is no product rule for integrals; expand instead.
Forgetting where $\cos x = 0$ ⁠. The answer is not valid at odd multiples of $\frac{\pi}{2}$ ⁠.

What to take away

Learn the six trigonometric derivatives of Section 7.2 well enough to recognise them inside an integrand.
$\sec x(\sec x + \tan x)$ is the derivative of $\sec x + \tan x$ ⁠: a neat identity worth remembering.
State the domain: $\cos x \ne 0$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
$\sec x = \frac{1}{\cos x}$ ⁠, $\tan x = \frac{\sin x}{\cos x}$ ⁠; defined when $\cos x \ne 0$	Step 1	Trigonometric functions (Class XI)
$\int (f+g)\,dx = \int f\,dx + \int g\,dx$	Step 3	Property III, Section 7.2.1 (Lesson: integration as anti-differentiation)
$\int \sec^2 x\,dx = \tan x + C$	Step 3	Formula (iv), Section 7.2
$\int \sec x\tan x\,dx = \sec x + C$	Step 3	Formula (vi), Section 7.2
Same pattern with cosec and cot	Step 3	Example 3(ii), Section 7.2.1
$(\tan x)' = \sec^2 x$ ⁠, $(\sec x)' = \sec x\tan x$	Step 4	Derivatives column, list in Section 7.2; Property I, Section 7.2.1

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Exercise 7.1 · Q18: ∫sec⁡x (sec⁡x+tan⁡x) dx\int \sec x\,(\sec x + \tan x)\,dx∫secx(secx+tanx)dx

Basics you need

Solution

Where is the integrand defined?

Expand the product

Integrate each term

Check: differentiate the answer back

Every fact used, and where it comes from

Exercise 7.1 · Q18: $\int \sec x\,(\sec x + \tan x)\,dx$