Chapter 7 · Integrals

Exercise 7.6 · Q12: $\int x\sec^2x\,dx$

Exercise 7.6Integration by parts · Section 7.6

The question

Integrate the functions in Exercises 1 to 22.

12. $x\sec^2x$

Exam answer

Let $I = \displaystyle\int x\sec^2x\,dx$ ⁠.

Taking $x$ as the first function, by parts:

\begin{aligned} I &= x\tan x - \int 1\cdot\tan x\,dx \\ &= x\tan x - \log|\sec x| + C \end{aligned}

$\therefore\ I = x\tan x + \log|\cos x| + C$ ⁠.

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

The one idea this problem is built on

$\sec^2x$ is the derivative of $\tan x$ ⁠, so it is easy to integrate. Taking $u = x$ and $\frac{dv}{dx} = \sec^2x$ ⁠, one use of integration by parts leaves $\int\tan x\,dx$ ⁠, which the book has already found by a substitution: $\log|\sec x|$ ⁠.

Given, to find, and the plan

Given: The function $x\sec^2x = \frac{x}{\cos^2x}$ ⁠, defined wherever $\cos x \ne 0$ ⁠, that is for $x \ne (2n+1)\frac\pi2$ ⁠, $n$ an integer.
To find: $\int x\sec^2x\,dx$ on any interval where $\cos x \ne 0$ ⁠.
Plan: (1) Name $u$ and $\frac{dv}{dx}$ ⁠. (2) Find $\frac{du}{dx}$ and $v$ ⁠. (3) Apply the formula. (4) Use $\int\tan x\,dx = \log|\sec x|$ ⁠. (5) Differentiate back.

Basics you need

1Integration by parts

Section 7.6, equation (1):

\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx,

with $u$ the first function (differentiated) and $\frac{dv}{dx}$ the second (integrated to give $v$ ⁠). Lesson: integration by parts

2Which factor should be $u$ ⁠?

Remark (iii) after Example 17: a power of $x$ is usually the first function. Memory aid ILATE (Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential; not in the book): A before T.

3Two trigonometric integrals

$\int\sec^2x\,dx = \tan x + C$ (standard formula (iv)) and $\int\tan x\,dx = \log|\sec x| + C$ (Section 7.3.1, found there with the substitution $t = \cos x$ ⁠). Basics: trigonometric identities

4Logarithm of an absolute value

$\frac{d}{dx}\log|x| = \frac1x$ for $x \ne 0$ (standard formula (xii)), and $\log|\sec x| = -\log|\cos x|$ because $|\sec x| = \frac{1}{|\cos x|}$ ⁠. Basics: logarithms and 1/x

Solution

Name $u$ and $\frac{dv}{dx}$
$\begin{aligned} u &= x \quad (\text{first}), \\ \frac{dv}{dx} &= \sec^2x \quad (\text{second}). \end{aligned}$
ILATE: Algebraic before Trigonometric. Differentiating $x$ gives 1, and $\sec^2x$ integrates at once to $\tan x$ ⁠. The other choice would need $\frac{d}{dx}\sec^2x = 2\sec^2x\tan x$ and $\int x\,dx = \frac{x^2}{2}$ ⁠, making the new integral harder.

Remark (iii) after Example 17, Section 7.6. Lesson: integration by parts
Find $\frac{du}{dx}$ and $v$
$\frac{du}{dx} = 1, \qquad v = \int\sec^2x\,dx = \tan x.$
$\frac{d}{dx}\tan x = \sec^2x$ ⁠, so $\tan x$ is an integral of $\sec^2x$ on every interval where $\cos x \ne 0$ ⁠. No constant in $v$ (Remark (ii) after Example 17).

Standard formula (iv), Section 7.2. Basics: trigonometric identities
Apply the formula
$\begin{aligned} \int x\sec^2x\,dx &= uv - \int v\,\frac{du}{dx}\,dx \\ &= x\tan x - \int\tan x\,dx. \end{aligned}$
First function times integral of second, minus the integral of (derivative of first, which is 1) times (integral of second).

Section 7.6, equation (1). Lesson: integration by parts
Use $\int\tan x\,dx = \log|\sec x|$
$\begin{aligned} \int x\sec^2x\,dx &= x\tan x - \log|\sec x| + C \\ &= x\tan x + \log|\cos x| + C. \end{aligned}$
The book shows $\int\tan x\,dx = -\log|\cos x| + C$ $= \log|\sec x| + C$ in Section 7.3.1. Since $|\sec x| = \frac{1}{|\cos x|}$ ⁠, $-\log|\sec x| = \log|\cos x|$ ⁠. The absolute value matters: $\cos x$ is negative on some intervals, and $\log$ needs a positive argument.

Section 7.3.1; $\log\frac1a = -\log a$ for $a > 0$ ⁠. Basics: logarithms and 1/x
Check: differentiate the answer

Product rule on $x\tan x$ ⁠, chain rule on $\log|\cos x|$ (outer $\log|\cdot|$ ⁠, inner $\cos x$ ⁠):
$\begin{aligned} &\frac{d}{dx}\big(x\tan x + \log|\cos x|\big) \\ &= \tan x + x\sec^2x + \frac{-\sin x}{\cos x} \\ &= \tan x + x\sec^2x - \tan x \\ &= x\sec^2x. \end{aligned}$
The two $\tan x$ terms cancel and the integrand is left, at every $x$ with $\cos x \ne 0$ ⁠. The derivative of $\log|u|$ is $\frac{u'}{u}$ whether $u$ is positive or negative.

Product rule; chain rule; standard formula (xii). Basics: integration undoes differentiation

Result

\int x\sec^2x\,dx = x\tan x + \log|\cos x| + C,

valid on every interval where $\cos x \ne 0$ (that is, $x \ne (2n+1)\frac\pi2$ ⁠). (Textbook answer: $x\tan x + \log|\cos x| + C$ ⁠, the same.)

Check it with numbers

With $F(x) = x\tan x + \log|\cos x|$ ⁠, on $\left[0, \frac\pi4\right]$ ⁠, where $\tan\frac\pi4 = 1$ and $\cos\frac\pi4 = \frac{1}{\sqrt2}$ ⁠:

\begin{aligned} F\left(\tfrac\pi4\right) - F(0) &= \frac\pi4 + \log\frac{1}{\sqrt2} - 0 \\ &= \frac\pi4 - \frac12\log 2 = 0.4388245731\ldots \end{aligned}

Numerical integration of $x\sec^2x$ over $\left[0, \frac\pi4\right]$ gives $0.4388245731\ldots$ ⁠, the same.

Common mistakes

$\int\tan x\,dx = \sec^2x$ ⁠. That is the derivative of $\tan x$ ⁠, not its integral; $\int\tan x\,dx = \log|\sec x|$ ⁠.
Sign at the end. $-\log|\sec x| = +\log|\cos x|$ ⁠.
Dropping the absolute value. $\log\cos x$ is undefined where $\cos x < 0$ ⁠.

What to take away

$x$ times the derivative of a known function (here $\sec^2x = (\tan x)'$ ⁠): take $u = x$ ⁠.
Remember $\int\tan x\,dx = \log|\sec x| + C$ $= -\log|\cos x| + C$ ⁠.
State the intervals of validity when the integrand has points where it is undefined.

Every fact used, and where it comes from

Fact	Used in	Source
Integration by parts	Step 3	Section 7.6, equation (1) (Lesson: integration by parts)
Choice of the first function (ILATE)	Step 1	Remark (iii) after Example 17, Section 7.6
No constant needed in $v$	Step 2	Remark (ii) after Example 17, Section 7.6
$\int\sec^2x\,dx = \tan x + C$	Step 2	Standard formula (iv), Section 7.2
$\int\tan x\,dx = \log\|\sec x\| + C$	Step 4	Section 7.3.1 (Lesson: substitution)
$\frac{d}{dx}\log\|x\| = \frac1x$	Step 5	Standard formula (xii), Section 7.2
Product and chain rules	Step 5	Chapter 5, Sections 5.3 and 5.3.1

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Exercise 7.6 · Q12: ∫xsec⁡2x dx\int x\sec^2x\,dx∫xsec2xdx

Basics you need

Solution

Name uuu and dvdx\frac{dv}{dx}dxdv​

Find dudx\frac{du}{dx}dxdu​ and vvv

Apply the formula

Use ∫tan⁡x dx=log⁡∣sec⁡x∣\int\tan x\,dx = \log|\sec x|∫tanxdx=log∣secx∣

Check: differentiate the answer

Every fact used, and where it comes from

Exercise 7.6 · Q12: $\int x\sec^2x\,dx$

Name $u$ and $\frac{dv}{dx}$

Find $\frac{du}{dx}$ and $v$

Use $\int\tan x\,dx = \log|\sec x|$