Chapter 7 · Integrals

Exercise 7.6 · Q24 (MCQ): $\int e^x\sec x(1+\tan x)\,dx$

Exercise 7.6Integral of the type eˣ[f + f′] · Section 7.6.1

The question

Choose the correct answer.

\int e^x\sec x\,(1 + \tan x)\,dx \text{ equals}

(A) $e^x\cos x + C$ (B) $e^x\sec x + C$ (C) $e^x\sin x + C$ (D) $e^x\tan x + C$

Exam answer

\begin{aligned} &\int e^x\sec x(1 + \tan x)\,dx \\ &= \int e^x(\sec x + \sec x\tan x)\,dx \end{aligned}

Here $f(x) = \sec x$ and ${f'(x) = \sec x\tan x}$ ⁠. Using $\displaystyle\int e^x\left[f(x) + f'(x)\right]dx$ $= e^xf(x) + C,$

Answer: (B) $e^x\sec x + C$

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

The one idea this problem is built on

Multiply out: $\sec x(1 + \tan x) = \sec x + \sec x\tan x$ ⁠, and $\sec x\tan x$ is the derivative of $\sec x$ ⁠. So the integrand is $e^x[f + f']$ with $f = \sec x$ ⁠, and Section 7.6.1 gives $e^x\sec x + C$ ⁠.

Given, to find, and the plan

Given: $e^x\sec x(1 + \tan x)$ ⁠, defined where $\cos x \ne 0$ ⁠, that is $x \ne (2n+1)\frac{\pi}{2}$ for integers $n$ ⁠; four candidate answers.
To find: The integral and the matching option.
Plan: (1) Expand and recognise $f + f'$ ⁠. (2) Apply the rule. (3) Differentiate back. (4) Examine every option.

Basics you need

1The rule $\int e^x[f + f']\,dx = e^x f + C$

Section 7.6.1 (the product rule read backwards). Lesson: integration by parts

2Derivative of $\sec x$

$\frac{d}{dx}\sec x = \sec x\tan x$ (standard formula (vi), Section 7.2). Basics: trigonometric identities

3Product rule

$\frac{d}{dx}(uv) = u'v + uv'$ ⁠, with $\frac{d}{dx}e^x = e^x$ ⁠. Basics: integration undoes differentiation

Solution

Expand and recognise $f + f'$
$e^x\sec x(1 + \tan x) = e^x\big(\sec x + \sec x\tan x\big).$
Take $f(x) = \sec x$ ⁠; then $f'(x) = \sec x\tan x$ ⁠, so the bracket is $f(x) + f'(x)$ ⁠.

Opening the bracket shows two terms, and one of them ( $\sec x\tan x$ ⁠) is the derivative of the other ( $\sec x$ ⁠). That is exactly the pattern of Section 7.6.1.

$\frac{d}{dx}\sec x = \sec x\tan x$ (formula (vi)). Basics: trigonometric identities
Apply the rule
$\int e^x\sec x(1 + \tan x)\,dx = e^x\sec x + C.$
$\int e^x[f + f']\,dx = e^x f + C$ with $f = \sec x$ ⁠. Since $\sec x$ is differentiable wherever $\cos x \ne 0$ ⁠, the answer holds on each interval between consecutive odd multiples of $\frac{\pi}{2}$ ⁠.

Section 7.6.1. Lesson: integration by parts
Check: differentiate the answer
$\begin{aligned} \frac{d}{dx}\big(e^x\sec x\big) &= e^x\sec x + e^x\sec x\tan x \\ &= e^x\sec x(1 + \tan x), \end{aligned}$
the integrand, wherever $\cos x \ne 0$ ⁠.

Product rule, then take out the common factor $e^x\sec x$ ⁠.

Product rule. Basics: integration undoes differentiation
Examine every option

Differentiate each option by the product rule and compare with the integrand. At the sample point $x = \frac{\pi}{4}$ the integrand is $e^{\pi/4}\cdot\sqrt2\cdot 2$ ⁠, that is
$2\sqrt2\,e^{\pi/4} = 6.2035327876\ldots$
(A) $\frac{d}{dx}(e^x\cos x) = e^x(\cos x - \sin x)$ ⁠: at $\frac{\pi}{4}$ this is $0$ ⁠. Wrong.

(B) $\frac{d}{dx}(e^x\sec x) = e^x\sec x(1 + \tan x)$ ⁠: the integrand exactly (Step 3). Correct.

(C) $\frac{d}{dx}(e^x\sin x) = e^x(\sin x + \cos x)$ ⁠: at $\frac{\pi}{4}$ this is $\sqrt2\,e^{\pi/4} = 3.1017663938\ldots$ Wrong.

(D) $\frac{d}{dx}(e^x\tan x) = e^x(\tan x + \sec^2 x)$ ⁠: at $\frac{\pi}{4}$ this is $3e^{\pi/4} = 6.5798401522\ldots$ Wrong.

One sample point where the values differ is enough to rule an option out. (At $x = 0$ option (A)'s derivative happens to equal the integrand, both being 1, so a single point can only disprove, never prove; option (B) is proved for all $x$ in Step 3.) Each wrong option is the $e^x[f + f']$ rule applied with a different $f$ ⁠: (A) $f = \cos x$ ⁠, (C) $f = \sin x$ (this is Question 16), (D) $f = \tan x$ ⁠.

Product rule; derivatives of $\cos x$ ⁠, $\sin x$ ⁠, $\tan x$ (formulae (ii)–(iv)). Basics: integration undoes differentiation

Result

\int e^x\sec x(1 + \tan x)\,dx = e^x\sec x + C,

valid on any interval on which $\cos x \ne 0$ ⁠. The correct option is (B). (Textbook answer: B.)

Check it with numbers

On $\left[0, \frac{\pi}{4}\right]$ ⁠, where $\cos x \ne 0$ ⁠:

\begin{aligned} F\left(\tfrac{\pi}{4}\right) - F(0) &= e^{\pi/4}\sec\frac{\pi}{4} - e^0\sec 0 \\ &= \sqrt2\,e^{\pi/4} - 1 = 2.1017663938\ldots \end{aligned}

Numerical integration of $e^x\sec x(1 + \tan x)$ over $\left[0, \frac{\pi}{4}\right]$ gives $2.1017663938\ldots$ ⁠, the same.

Common mistakes

Choosing $f = \tan x$ because $\tan x$ appears: its derivative is $\sec^2 x$ ⁠, not $\sec x$ ⁠.
Choosing (A) after testing only $x = 0$ ⁠: there (A)'s derivative also equals 1. Test at a point like $\frac{\pi}{4}$ ⁠, or better, differentiate symbolically.
Ignoring where $\sec x$ is undefined: the answer is valid only away from odd multiples of $\frac{\pi}{2}$ ⁠.

What to take away

Open the bracket and look for a function together with its derivative; here $\sec x$ and $\sec x\tan x$ ⁠.
$\int e^x\sec x(1 + \tan x)\,dx = e^x\sec x + C$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
$\int e^x[f + f']\,dx = e^x f + C$	Step 2	Section 7.6.1 (Lesson: integration by parts)
$\frac{d}{dx}\sec x = \sec x\tan x$	Steps 1, 3	Standard formula (vi), Section 7.2
Derivatives of $\cos x$ ⁠, $\sin x$ ⁠, $\tan x$	Step 4	Standard formulae (ii)–(iv), Section 7.2
Product rule	Steps 3, 4	Chapter 5, Section 5.3

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Exercise 7.6 · Q24 (MCQ): ∫exsec⁡x(1+tan⁡x) dx\int e^x\sec x(1+\tan x)\,dx∫exsecx(1+tanx)dx

Basics you need

Solution

Expand and recognise f+f′f + f'f+f′

Apply the rule

Check: differentiate the answer

Examine every option

Every fact used, and where it comes from

Exercise 7.6 · Q24 (MCQ): $\int e^x\sec x(1+\tan x)\,dx$

Expand and recognise $f + f'$