Chapter 7 · Integrals

Exercise 7.4 · Q9: $\int\frac{\sec^2 x}{\sqrt{\tan^2 x+4}}\,dx$

Exercise 7.4Integrals of some particular functions · Section 7.4

The question

Integrate the function (Exercise 7.4, "Integrate the functions in Exercises 1 to 23"):

\frac{\sec^2 x}{\sqrt{\tan^2 x + 4}}

Exam answer

Let $I = \displaystyle\int\frac{\sec^2 x}{\sqrt{\tan^2 x + 4}}\,dx$ ⁠.

Put $\tan x = t$ ⁠. Then $\sec^2 x\,dx = dt$ ⁠.

I = \int\frac{dt}{\sqrt{t^2 + 2^2}}

[Using $\displaystyle\int\frac{dt}{\sqrt{t^2 + a^2}}$ $= \log\left|t + \sqrt{t^2 + a^2}\right| + C$ ]

I = \log\left|t + \sqrt{t^2 + 4}\right| + C

$\therefore\ I = \log\left|\tan x + \sqrt{\tan^2 x + 4}\right|$ $+ C$ ⁠.

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

The one idea this problem is built on

The numerator $\sec^2 x\,dx$ is exactly $d(\tan x)$ ⁠, and the root contains $(\tan x)^2 + 2^2$ ⁠. So $t = \tan x$ turns a trigonometric integral into the algebraic one $\int\frac{dt}{\sqrt{t^2 + 2^2}}$ ⁠, which is formula 7.4 (6).

Given, to find, and the plan

Given: $f(x) = \dfrac{\sec^2 x}{\sqrt{\tan^2 x + 4}}$ ⁠.
To find: $\int f(x)\,dx$ ⁠, and where the answer is valid.
Plan: (1) Find where $f$ is defined. (2) Substitute $t = \tan x$ ⁠. (3) Use formula 7.4 (6) with $a = 2$ ⁠. (4) Return to $x$ ⁠. (5) Differentiate back.

Basics you need

1Derivative of $\tan x$

$\frac{d}{dx}\tan x = \sec^2 x$ ⁠, so with $t = \tan x$ ⁠, $dt = \sec^2 x\,dx$ ⁠. Basics: trigonometric facts

2Substitution

A factor $g'(x)\,dx$ becomes $dt$ when $t = g(x)$ ⁠. Basics: substitution (the chain rule backwards)

3Formula 7.4 (6)

$\int\frac{dx}{\sqrt{x^2 + a^2}} = \log\left|x + \sqrt{x^2 + a^2}\right| + C$ ⁠. Lesson: integrals of particular functions

Solution

Where is the integrand defined?

$\sec x$ and $\tan x$ need $\cos x \ne 0$ ⁠, that is, $x \ne (2n + 1)\frac\pi2$ for integers $n$ ⁠. Where they exist, $\tan^2 x + 4 \ge 4 > 0$ ⁠, so the root is fine. The integrand is defined on each interval
$(2n - 1)\frac\pi2 < x < (2n + 1)\frac\pi2.$
The only restriction comes from $\sec$ and $\tan$ themselves; the root never causes trouble. The answer holds on each such interval (with its own constant on each).
Substitute $t = \tan x$

Put $t = \tan x$ ⁠. Then $dt = \sec^2 x\,dx$ ⁠, and
$\int\frac{\sec^2 x\,dx}{\sqrt{\tan^2 x + 4}} = \int\frac{dt}{\sqrt{t^2 + 4}}.$
The whole numerator $\sec^2 x\,dx$ is used up as $dt$ ⁠, and the rest is a function of $\tan x$ alone. On each interval of Step 1, $t = \tan x$ runs over all real numbers.

Integration by substitution, Section 7.3.1. Lesson: integration by substitution
Use formula 7.4 (6) with $a = 2$
$\int\frac{dt}{\sqrt{t^2 + 2^2}} = \log\left|t + \sqrt{t^2 + 4}\right| + C.$
"Square plus a positive constant, under a root", with $4 = 2^2$ ⁠: formula 7.4 (6), valid for every real $t$ ⁠.

Formula 7.4 (6). Lesson: integrals of particular functions
Return to $x$
$\int\frac{\sec^2 x}{\sqrt{\tan^2 x + 4}}\,dx = \log\left|\tan x + \sqrt{\tan^2 x + 4}\right| + C.$
The quantity inside the modulus is always positive, because $\sqrt{\tan^2 x + 4} > |\tan x|$ ⁠.

Replace $t$ by $\tan x$ ⁠. The modulus is kept as in the book, though it changes nothing here.
Check: differentiate the answer back

Let $s = \sqrt{\tan^2 x + 4}$ ⁠. Chain rule: $s' = \frac{2\tan x\sec^2 x}{2s} = \frac{\tan x\sec^2 x}{s}$ ⁠. With $F(x) = \log(\tan x + s)$ ⁠,
$\begin{aligned} F'(x) &= \frac{\sec^2 x + \frac{\tan x\sec^2 x}{s}}{\tan x + s} \\ &= \frac{\sec^2 x\,(s + \tan x)}{s\,(\tan x + s)} \\ &= \frac{\sec^2 x}{\sqrt{\tan^2 x + 4}}. \ \checkmark \end{aligned}$
Factor $\sec^2 x$ out of the numerator; the bracket $s + \tan x$ then cancels with the denominator. This holds at every $x$ with $\cos x \ne 0$ ⁠.

Basics: integration undoes differentiation

Result

$\begin{aligned} &\int\frac{\sec^2 x}{\sqrt{\tan^2 x + 4}}\,dx \\ &= \log\left|\tan x + \sqrt{\tan^2 x + 4}\right| + C, \end{aligned}$ valid wherever $\cos x \ne 0$ ⁠. (Textbook answer: $\log\left|\tan x + \sqrt{\tan^2 x + 4}\right| + C$ ⁠, the same.)

Check it with numbers

Take $x = 0.5$ and $h = 0.001$ ⁠:

$F(0.501) = 0.96363751$ and $F(0.499) = 0.96238495$ ⁠,
slope $\approx \frac{0.96363751 - 0.96238495}{0.002} = 0.6263$ ⁠,
integrand: $\sec^2 0.5 = 1.29845$ and $\sqrt{\tan^2 0.5 + 4} = 2.07327$ ⁠, so the integrand is $\frac{1.29845}{2.07327} = 0.6263$ ⁠. ✓

Common mistakes

Trying to simplify $\sec^2 x$ with $1 + \tan^2 x$ ⁠. That is true but leads nowhere here; $\sec^2 x\,dx$ is wanted as $dt$ ⁠.
Using $a = 4$ instead of $a = 2$ ⁠. The formula is written with $a^2$ under the root.
Forgetting to return to $x$ ⁠. The answer must be in terms of $\tan x$ ⁠, not $t$ ⁠.

What to take away

$\sec^2 x\,dx$ next to a function of $\tan x$ signals $t = \tan x$ ⁠.
After a substitution, a trigonometric integral can become one of the algebraic forms of Section 7.4.
The domain is set by $\sec x$ and $\tan x$ ⁠: $\cos x \ne 0$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
$\tan x$ ⁠, $\sec x$ defined where $\cos x \ne 0$	Step 1	Trigonometric functions (Class XI, Chapter 3)
$\frac{d}{dx}\tan x = \sec^2 x$	Steps 2, 5	Standard formula (iv), Section 7.2
Substitution $t = \tan x$	Step 2	Section 7.3.1 (Lesson: substitution)
$\int\frac{dx}{\sqrt{x^2 + a^2}} = \log\left\|x + \sqrt{x^2 + a^2}\right\| + C$	Step 3	Formula 7.4 (6), Section 7.4
Chain rule; $F' = f$ checks the answer	Step 5	Theorem 4, Section 5.3.1; Property (I), Section 7.2.1

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 7.4 · Q9: ∫sec⁡2xtan⁡2x+4 dx\int\frac{\sec^2 x}{\sqrt{\tan^2 x+4}}\,dx∫tan2x+4​sec2x​dx

Basics you need

Solution

Where is the integrand defined?

Substitute t=tan⁡xt = \tan xt=tanx

Use formula 7.4 (6) with a=2a = 2a=2

Return to xxx

Check: differentiate the answer back

Every fact used, and where it comes from

Exercise 7.4 · Q9: $\int\frac{\sec^2 x}{\sqrt{\tan^2 x+4}}\,dx$

Substitute $t = \tan x$

Use formula 7.4 (6) with $a = 2$

Return to $x$