Chapter 7 · Integrals

Exercise 7.1 · Q19: $\int \frac{\sec^2 x}{\operatorname{cosec}^2 x}\,dx$

Exercise 7.1Integration as anti-differentiation · Section 7.2

The question

Find the following integral:

\int \frac{\sec^2 x}{\operatorname{cosec}^2 x}\,dx

Exam answer

Let $I = \displaystyle\int\frac{\sec^2 x}{\operatorname{cosec}^2 x}\,dx$ ⁠.

\begin{aligned} \frac{\sec^2 x}{\operatorname{cosec}^2 x} &= \frac{\sin^2 x}{\cos^2 x} = \tan^2 x \\ &= \sec^2 x - 1 \end{aligned}

\therefore\ I = \int\left(\sec^2 x - 1\right)dx

$\therefore\ I = \tan x - x + C$ ⁠.

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

The one idea this problem is built on

Rewrite in sines and cosines: $\frac{\sec^2 x}{\operatorname{cosec}^2 x} = \frac{\sin^2 x}{\cos^2 x} = \tan^2 x$ ⁠. There is no standard formula for $\int \tan^2 x\,dx$ ⁠, but the identity $\tan^2 x = \sec^2 x - 1$ turns it into two integrals that are standard: $\int\sec^2 x\,dx = \tan x$ and $\int 1\,dx = x$ ⁠.

Given, to find, and the plan

Given: The integrand $f(x) = \dfrac{\sec^2 x}{\operatorname{cosec}^2 x}$ ⁠.
To find: $\int f(x)\,dx$ ⁠.
Plan: (1) Find where $f$ is defined. (2) Simplify $f$ to $\tan^2 x$ ⁠. (3) Use $\tan^2 x = \sec^2 x - 1$ ⁠. (4) Integrate with formulae (iv) and (i). (5) 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

2Reciprocal functions

$\sec x = \frac{1}{\cos x}$ (needs $\cos x \ne 0$ ⁠) and $\operatorname{cosec} x = \frac{1}{\sin x}$ (needs $\sin x \ne 0$ ⁠). Basics: trigonometric facts for integrals

3The Pythagorean identity in tangent form

$1 + \tan^2 x = \sec^2 x$ ⁠, so $\tan^2 x = \sec^2 x - 1$ ⁠. (Divide $\sin^2 x + \cos^2 x = 1$ by $\cos^2 x$ ⁠.) Basics: trigonometric identities for integrals

4Two standard integrals

(iv) $\int \sec^2 x\,dx = \tan x + C$ and $\int dx = x + C$ (list in Section 7.2). Lesson: integration as anti-differentiation

Solution

Where is the integrand defined?

$\sec x$ needs $\cos x \ne 0$ ⁠, and $\operatorname{cosec} x$ needs $\sin x \ne 0$ (it is also never 0, so dividing by $\operatorname{cosec}^2 x$ is safe). So the integrand is defined when
$\begin{aligned} &\sin x \ne 0 \text{ and } \cos x \ne 0, \\ &\text{i.e. } x \ne \tfrac{n\pi}{2},\ n \in \mathbb{Z}. \end{aligned}$
$\sin x = 0$ at multiples of $\pi$ and $\cos x = 0$ at odd multiples of $\frac{\pi}{2}$ ⁠; together these are all multiples of $\frac{\pi}{2}$ ⁠. The answer will be an antiderivative on each interval between consecutive multiples of $\frac{\pi}{2}$ ⁠. (The simplified form $\tan^2 x$ also makes sense at $x = n\pi$ ⁠, but the function as given does not.)

The Note in Section 7.2: always keep the interval of definition in mind.
Simplify the integrand to $\tan^2 x$
$\begin{aligned} \frac{\sec^2 x}{\operatorname{cosec}^2 x} &= \frac{1/\cos^2 x}{1/\sin^2 x} \\ &= \frac{1}{\cos^2 x}\cdot\frac{\sin^2 x}{1} \\ &= \frac{\sin^2 x}{\cos^2 x} = \tan^2 x. \end{aligned}$
Writing everything in terms of $\sin x$ and $\cos x$ is the safe first move with any trigonometric quotient. Dividing by a fraction means multiplying by its reciprocal, and $\frac{\sin x}{\cos x} = \tan x$ ⁠.

$\sec x = \frac{1}{\cos x}$ ⁠, $\operatorname{cosec} x = \frac{1}{\sin x}$ ⁠, $\tan x = \frac{\sin x}{\cos x}$ ⁠. Basics: trigonometric facts
Turn $\tan^2 x$ into something we can integrate
$\tan^2 x = \sec^2 x - 1.$
The list of Section 7.2 contains $\int\sec^2 x\,dx$ but not $\int\tan^2 x\,dx$ ⁠. The identity $1 + \tan^2 x = \sec^2 x$ (true wherever $\cos x \ne 0$ ⁠) swaps the unknown integral for a known one plus a constant.

$\sin^2 x + \cos^2 x = 1$ ⁠; dividing by $\cos^2 x$ gives $\tan^2 x + 1 = \sec^2 x$ ⁠. Basics: trigonometric identities for integrals
Integrate
$\begin{aligned} \int \frac{\sec^2 x}{\operatorname{cosec}^2 x}\,dx &= \int (\sec^2 x - 1)\,dx \\ &= \int \sec^2 x\,dx - \int dx \\ &= \tan x - x + C. \end{aligned}$
Property V splits the difference. Formula (iv) gives $\tan x$ because $\frac{d}{dx}\tan x = \sec^2 x$ ⁠; $\int dx = x$ because $\frac{d}{dx}x = 1$ ⁠. One constant $C$ collects both.

Formula (iv) and formula (i) with $n = 0$ ⁠, Section 7.2; Property V, Section 7.2.1. Lesson: integration as anti-differentiation
Check: differentiate the answer back
$\begin{aligned} \frac{d}{dx}(\tan x - x + C) &= \sec^2 x - 1 \\ &= \tan^2 x = \frac{\sec^2 x}{\operatorname{cosec}^2 x}. \end{aligned}$
This is the integrand at every $x$ with $x \ne \frac{n\pi}{2}$ ⁠, so the answer is correct.

The last line is Steps 3 and 2 read backwards. The equalities hold wherever the integrand is defined, which is all that is required.

Property I, Section 7.2.1. Basics: integration undoes differentiation

Result

For $x \ne \frac{n\pi}{2}$ ( $n$ an integer),

\int \frac{\sec^2 x}{\operatorname{cosec}^2 x}\,dx = \tan x - x + C.

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

Check it with numbers

At $x = \frac{\pi}{3}$ ⁠: $\sec^2\frac{\pi}{3} = 4$ and $\operatorname{cosec}^2\frac{\pi}{3} = \frac43$ ⁠, so the integrand is $4 \div \frac43 = 3$ ⁠, which equals $\tan^2\frac{\pi}{3} = (\sqrt3)^2 = 3$ ⁠. With $F(x) = \tan x - x$ ⁠: $F\!\left(\frac{\pi}{3} + 0.001\right) \approx 0.68786020$ and $F\!\left(\frac{\pi}{3} - 0.001\right) \approx 0.68186017$ ⁠, so the slope of $F$ there is about

\frac{0.68786020 - 0.68186017}{0.002} \approx 3.0000.

It matches the integrand value 3.

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

Common mistakes

$\int \tan^2 x\,dx = \frac{\tan^3 x}{3}$ ⁠. The power formula is for powers of $x$ ⁠, not powers of $\tan x$ ⁠. (Differentiating $\frac{\tan^3 x}{3}$ gives $\tan^2 x\sec^2 x$ ⁠, not $\tan^2 x$ ⁠.)
Wrong identity. $\tan^2 x = \sec^2 x - 1$ ⁠, not $1 - \sec^2 x$ and not $\sec^2 x + 1$ ⁠.
Dropping the $-x$ ⁠. The constant $-1$ in the integrand integrates to $-x$ ⁠, not to 0.

What to take away

Rewrite trigonometric quotients in $\sin$ and $\cos$ first.
$\int \tan^2 x\,dx = \tan x - x + C$ and, the same way, $\int\cot^2 x\,dx = -\cot x - x + C$ (from $\cot^2 x = \operatorname{cosec}^2 x - 1$ ⁠).
When a function is not on the list, look for an identity that turns it into functions that are.

Every fact used, and where it comes from

Fact	Used in	Source
$\sec x = \frac{1}{\cos x}$ ⁠, $\operatorname{cosec} x = \frac{1}{\sin x}$ ⁠, $\tan x = \frac{\sin x}{\cos x}$	Steps 1, 2	Trigonometric functions (Class XI)
$1 + \tan^2 x = \sec^2 x$	Steps 3, 5	Trigonometric identities (Class XI)
$\int \sec^2 x\,dx = \tan x + C$	Step 4	Formula (iv), Section 7.2
$\int dx = x + C$	Step 4	Formula (i) with $n = 0$ ⁠, Section 7.2
$\int (f - g)\,dx = \int f\,dx - \int g\,dx$	Step 4	Property V, Section 7.2.1 (Lesson: integration as anti-differentiation)
$\frac{d}{dx}\int f\,dx = f$	Step 5	Property I, Section 7.2.1

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Exercise 7.1 · Q19: ∫sec⁡2xcosec⁡2x dx\int \frac{\sec^2 x}{\operatorname{cosec}^2 x}\,dx∫cosec2xsec2x​dx

Basics you need

Solution

Where is the integrand defined?

Simplify the integrand to tan⁡2x\tan^2 xtan2x

Turn tan⁡2x\tan^2 xtan2x into something we can integrate

Integrate

Check: differentiate the answer back

Every fact used, and where it comes from

Exercise 7.1 · Q19: $\int \frac{\sec^2 x}{\operatorname{cosec}^2 x}\,dx$

Simplify the integrand to $\tan^2 x$

Turn $\tan^2 x$ into something we can integrate