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Chapter 7 · Integrals

Exercise 7.1 · Q20: ∫2−3sin⁡xcos⁡2x dx\int \frac{2 - 3\sin x}{\cos^2 x}\,dx

Exercise 7.1Integration as anti-differentiation · Section 7.2
The question

Find the following integral:

∫2−3sin⁡xcos⁡2x dx\int \frac{2 - 3\sin x}{\cos^2 x}\,dx
Exam answer

Let I=∫2−3sin⁡xcos⁡2x dxI = \displaystyle\int\frac{2 - 3\sin x}{\cos^2 x}\,dx⁠.

I=∫(2cos⁡2x−3sin⁡xcos⁡2x)dx=∫(2sec⁡2x−3sec⁡xtan⁡x)dx\begin{aligned} I &= \int\left(\frac{2}{\cos^2 x} - \frac{3\sin x}{\cos^2 x}\right)dx \\ &= \int\left(2\sec^2 x - 3\sec x\tan x\right)dx \end{aligned}

∴ I=2tan⁡x−3sec⁡x+C\therefore\ I = 2\tan x - 3\sec x + C⁠.

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

The one idea this problem is built on

The denominator cos⁡2x\cos^2 x is a single term, so the fraction splits: 2cos⁡2x−3sin⁡xcos⁡2x\frac{2}{\cos^2 x} - \frac{3\sin x}{\cos^2 x}⁠. Then write sin⁡xcos⁡2x\frac{\sin x}{\cos^2 x} as 1cos⁡x⋅sin⁡xcos⁡x=sec⁡xtan⁡x\frac{1}{\cos x}\cdot\frac{\sin x}{\cos x} = \sec x\tan x⁠. Both pieces, sec⁡2x\sec^2 x and sec⁡xtan⁡x\sec x\tan x⁠, are on the list of standard integrals.

Given, to find, and the plan
Given
The integrand f(x)=2−3sin⁡xcos⁡2xf(x) = \dfrac{2 - 3\sin x}{\cos^2 x}⁠.
To find
∫f(x) dx\int f(x)\,dx⁠.
Plan
(1) Find where ff is defined. (2) Split the fraction and rewrite each piece using sec⁡\sec and tan⁡\tan⁠. (3) Integrate with formulae (iv) and (vi). (4) Differentiate the answer to check it.

Basics you need

1Integration undoes differentiation

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

2Splitting over a single-term denominator

a−bd=ad−bd\dfrac{a - b}{d} = \dfrac{a}{d} - \dfrac{b}{d}⁠. Basics: splitting fractions

3Secant and tangent

sec⁡x=1cos⁡x\sec x = \frac{1}{\cos x}⁠, tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x}⁠, so 1cos⁡2x=sec⁡2x\frac{1}{\cos^2 x} = \sec^2 x and sin⁡xcos⁡2x=sec⁡xtan⁡x\frac{\sin x}{\cos^2 x} = \sec x\tan x⁠. Basics: trigonometric facts for integrals

4Two standard integrals

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

Solution

  1. Where is the integrand defined?

    The denominator cos⁡2x\cos^2 x must not be 0, so the integrand is defined when

    cos⁡x≠0,i.e.x≠(2n+1)π2, n∈Z.\cos x \ne 0, \quad\text{i.e.}\quad x \ne (2n+1)\tfrac{\pi}{2},\ n \in \mathbb{Z}.
    This is also exactly where sec⁡x\sec x and tan⁡x\tan x exist, so rewriting in terms of them in the next step loses nothing. The answer is an antiderivative on each interval between consecutive odd multiples of π2\frac{\pi}{2}⁠; like the textbook, we write one constant CC⁠.
    The Note in Section 7.2: always keep the interval of definition in mind.

  2. Split the fraction and rewrite each piece

    2−3sin⁡xcos⁡2x=2cos⁡2x−3sin⁡xcos⁡2x=2⋅1cos⁡2x−3⋅1cos⁡x⋅sin⁡xcos⁡x=2sec⁡2x−3sec⁡xtan⁡x.\begin{aligned} \frac{2 - 3\sin x}{\cos^2 x} &= \frac{2}{\cos^2 x} - \frac{3\sin x}{\cos^2 x} \\ &= 2\cdot\frac{1}{\cos^2 x} - 3\cdot\frac{1}{\cos x}\cdot\frac{\sin x}{\cos x} \\ &= 2\sec^2 x - 3\sec x\tan x. \end{aligned}
    Splitting is allowed because cos⁡2x\cos^2 x is a single term. The key move is in the second piece: cos⁡2x=cos⁡x⋅cos⁡x\cos^2 x = \cos x\cdot\cos x⁠, and sharing one cos⁡x\cos x with sin⁡x\sin x produces tan⁡x\tan x while the other produces sec⁡x\sec x⁠. Now both pieces match formulae on the list.
    1cos⁡x=sec⁡x\frac{1}{\cos x} = \sec x⁠, sin⁡xcos⁡x=tan⁡x\frac{\sin x}{\cos x} = \tan x⁠. Basics: trigonometric facts Basics: splitting fractions

  3. Integrate term by term

    ∫(2sec⁡2x−3sec⁡xtan⁡x) dx=2 ⁣∫sec⁡2x dx−3 ⁣∫sec⁡xtan⁡x dx=2tan⁡x−3sec⁡x+C.\begin{aligned} &\int (2\sec^2 x - 3\sec x\tan x)\,dx \\ &\quad = 2\!\int \sec^2 x\,dx - 3\!\int \sec x\tan x\,dx \\ &\quad = 2\tan x - 3\sec x + C. \end{aligned}
    Property V takes out the constants 2 and −3-3⁠. Formula (iv) gives tan⁡x\tan x since (tan⁡x)′=sec⁡2x(\tan x)' = \sec^2 x⁠; formula (vi) gives sec⁡x\sec x since (sec⁡x)′=sec⁡xtan⁡x(\sec x)' = \sec x\tan x⁠. One constant CC collects both arbitrary constants.
    Formulae (iv) and (vi), Section 7.2; Property V, Section 7.2.1. The textbook's Example 3(iii) is the same problem with 1 and 1 in place of 2 and 3: ∫1−sin⁡xcos⁡2x dx=tan⁡x−sec⁡x+C\int\frac{1 - \sin x}{\cos^2 x}\,dx = \tan x - \sec x + C⁠.

  4. Check: differentiate the answer back

    ddx(2tan⁡x−3sec⁡x+C)=2sec⁡2x−3sec⁡xtan⁡x=2cos⁡2x−3sin⁡xcos⁡2x=2−3sin⁡xcos⁡2x.\begin{aligned} &\frac{d}{dx}(2\tan x - 3\sec x + C) \\ &\quad = 2\sec^2 x - 3\sec x\tan x \\ &\quad = \frac{2}{\cos^2 x} - \frac{3\sin x}{\cos^2 x} \\ &\quad = \frac{2 - 3\sin x}{\cos^2 x}. \end{aligned}

    This is the integrand wherever cos⁡x≠0\cos x \ne 0⁠, so the answer is correct.

    The second and third lines are Step 2 read backwards: write sec⁡\sec and tan⁡\tan in terms of sin⁡\sin and cos⁡\cos⁠, then put both pieces over the common denominator cos⁡2x\cos^2 x⁠.
    Derivatives (iv) and (vi), list in Section 7.2; Property I, Section 7.2.1. Basics: integration undoes differentiation
Result

Wherever cos⁡x≠0\cos x \ne 0⁠,

∫2−3sin⁡xcos⁡2x dx=2tan⁡x−3sec⁡x+C.\int \frac{2 - 3\sin x}{\cos^2 x}\,dx = 2\tan x - 3\sec x + C.

(Textbook answer: 2tan⁡x−3sec⁡x+C2\tan x - 3\sec x + C⁠.)

Check it with numbers

At x=π6x = \frac{\pi}{6}⁠: sin⁡x=12\sin x = \frac12⁠, cos⁡2x=34\cos^2 x = \frac34⁠, so the integrand is 2−1.50.75=23\frac{2 - 1.5}{0.75} = \frac23⁠. The derivative of our answer there is exactly

2sec⁡2π6−3sec⁡π6tan⁡π6=2⋅43−3⋅23⋅13=83−2=23.\begin{aligned} &2\sec^2\tfrac{\pi}{6} - 3\sec\tfrac{\pi}{6}\tan\tfrac{\pi}{6} \\ &\quad = 2\cdot\tfrac43 - 3\cdot\tfrac{2}{\sqrt3}\cdot\tfrac{1}{\sqrt3} \\ &\quad = \tfrac83 - 2 = \tfrac23. \end{aligned}

Numerically, with F(x)=2tan⁡x−3sec⁡xF(x) = 2\tan x - 3\sec x⁠: F ⁣(π6+0.001)≈−2.30873576F\!\left(\frac{\pi}{6} + 0.001\right) \approx -2.30873576 and F ⁣(π6−0.001)≈−2.31006909F\!\left(\frac{\pi}{6} - 0.001\right) \approx -2.31006909⁠; their difference divided by 0.002 is about 0.6667, matching 23≈0.6667\frac23 \approx 0.6667⁠.

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

Common mistakes
  1. Integrating top and bottom separately. ∫uv dx≠∫u dx∫v dx\int\frac{u}{v}\,dx \ne \frac{\int u\,dx}{\int v\,dx}⁠. Split and rewrite first.
  2. Writing sin⁡xcos⁡2x=tan⁡2x\frac{\sin x}{\cos^2 x} = \tan^2 x⁠. tan⁡2x=sin⁡2xcos⁡2x\tan^2 x = \frac{\sin^2 x}{\cos^2 x}⁠; with only one sin⁡x\sin x on top, it is sec⁡xtan⁡x\sec x\tan x⁠.
  3. Sign of the second term. −3∫sec⁡xtan⁡x dx=−3sec⁡x-3\int\sec x\tan x\,dx = -3\sec x⁠, so the answer has −3sec⁡x-3\sec x⁠.
What to take away
  1. Over a cos⁡2x\cos^2 x denominator: a constant gives sec⁡2x\sec^2 x⁠, and sin⁡x\sin x gives sec⁡xtan⁡x\sec x\tan x⁠. Both are standard.
  2. The same idea over a sin⁡2x\sin^2 x denominator: a constant gives cosec⁡2x\operatorname{cosec}^2 x⁠, and cos⁡x\cos x gives cosec⁡xcot⁡x\operatorname{cosec} x\cot x⁠.
  3. Differentiate back and recombine over the common denominator to confirm.

Every fact used, and where it comes from

FactUsed inSource
Keep the interval of definition in mindStep 1Note after the list of formulae, Section 7.2
a−bd=ad−bd\frac{a - b}{d} = \frac ad - \frac bdStep 2Arithmetic of fractions
sec⁡x=1cos⁡x\sec x = \frac{1}{\cos x}⁠, tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x}Steps 2, 4Trigonometric functions (Class XI)
∫sec⁡2x dx=tan⁡x+C\int \sec^2 x\,dx = \tan x + C⁠; ∫sec⁡xtan⁡x dx=sec⁡x+C\int \sec x\tan x\,dx = \sec x + CStep 3Formulae (iv) and (vi), Section 7.2
Integrate term by term, constants in frontStep 3Property V, Section 7.2.1 (Lesson: integration as anti-differentiation)
Same pattern: ∫1−sin⁡xcos⁡2xdx=tan⁡x−sec⁡x+C\int\frac{1-\sin x}{\cos^2 x}dx = \tan x - \sec x + CStep 3Example 3(iii), Section 7.2.1
ddx∫f dx=f\frac{d}{dx}\int f\,dx = fStep 4Property I, Section 7.2.1
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