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Chapter 7 · Integrals

Exercise 7.1 · Q2: An anti derivative of cos⁡3x\cos 3x

Exercise 7.1Integration by inspection · Section 7.2
The question

Find an anti derivative (or integral) of the following function by the method of inspection.

cos⁡3x\cos 3x
Exam answer

We know that ddx(sin⁡3x)=3cos⁡3x\frac{d}{dx}(\sin 3x) = 3\cos 3x⁠.

∴ cos⁡3x=ddx(13sin⁡3x)\therefore\ \cos 3x = \frac{d}{dx}\left(\frac13\sin 3x\right)

∴\therefore An anti derivative of cos⁡3x\cos 3x is 13sin⁡3x\dfrac13\sin 3x⁠, and ∫cos⁡3x dx\int\cos 3x\,dx =13sin⁡3x+C= \dfrac13\sin 3x + C⁠.

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

The one idea this problem is built on

We want a function whose derivative is cos⁡3x\cos 3x⁠. Sine differentiates to cosine, so try sin⁡3x\sin 3x⁠. The inner 3x3x makes its derivative three times too big (chain rule), so we divide by 33⁠. The answer 13sin⁡3x\frac13\sin 3x is then confirmed by differentiating it back.

Given, to find, and the plan
Given
f(x)=cos⁡3xf(x) = \cos 3x⁠, defined for every real xx⁠.
To find
One function FF with F′(x)=cos⁡3xF'(x) = \cos 3x for every real xx⁠.
Plan
(1) Start from ddx(sin⁡x)=cos⁡x\frac{d}{dx}(\sin x) = \cos x and guess sin⁡3x\sin 3x⁠. (2) Differentiate the guess with the chain rule. (3) Remove the extra factor. (4) Differentiate the answer back to check it. (5) Write the family with + C+\,C⁠.

Basics you need

1Anti derivative

FF is an anti derivative of ff if F′(x)=f(x)F'(x) = f(x) at every point of the interval. To find one by inspection we look for a derivative we already know that produces ff⁠. Basics: integration undoes differentiation

2Derivative of sine

ddx(sin⁡x)=cos⁡x\frac{d}{dx}(\sin x) = \cos x⁠, which is item (ii) of the standard formulae in Section 7.2: ∫cos⁡x dx=sin⁡x+C\int \cos x\,dx = \sin x + C⁠. Basics: trigonometric facts for integrals

3The chain rule

ddxsin⁡(3x)=cos⁡(3x)⋅3\frac{d}{dx}\sin(3x) = \cos(3x)\cdot 3⁠: differentiate the outer sine at the inner 3x3x⁠, then multiply by the derivative of 3x3x⁠. Basics: the chain rule backwards

4The constant of integration

Adding a constant changes no derivative, so one anti derivative FF gives the whole family F+CF + C⁠. Lesson: integration as anti differentiation

Solution

  1. Make a first guess from a known derivative

    The standard formulae tell us which function differentiates to a cosine:

    ddx(sin⁡x)=cos⁡x.\frac{d}{dx}(\sin x) = \cos x.

    Our cosine has the angle 3x3x⁠, so the first guess is sin⁡3x\sin 3x⁠.

    Sine is the only basic function whose derivative is a plain cosine, so the answer must be built from sin⁡3x\sin 3x⁠. We keep the angle 3x3x unchanged and only adjust a constant factor afterwards.
    Item (ii) of the standard formulae, Section 7.2: ∫cos⁡x dx=sin⁡x+C\int \cos x\,dx = \sin x + C⁠. Basics: trigonometric facts for integrals

  2. Differentiate the guess (chain rule)

    Inner function u=3xu = 3x (so dudx=3\frac{du}{dx} = 3⁠), outer function sin⁡u\sin u⁠:

    ddx(sin⁡3x)=cos⁡u⋅dudx=3cos⁡3x.\begin{aligned} \frac{d}{dx}(\sin 3x) &= \cos u \cdot \frac{du}{dx} \\ &= 3\cos 3x. \end{aligned}

    The guess gives 3cos⁡3x3\cos 3x⁠: three times what we want.

    The factor 33 is the derivative of the inner function 3x3x⁠. Since sin⁡3x\sin 3x runs through its values three times as fast as sin⁡x\sin x⁠, its slopes are three times as steep.
    Chain rule, Theorem 4, Section 5.3.1 (Chapter 5). Basics: the chain rule backwards

  3. Remove the extra factor 3

    Multiply the guess by 13\frac13⁠; its derivative is multiplied by 13\frac13 too:

    ddx(13sin⁡3x)=13⋅3cos⁡3x=cos⁡3x.\begin{aligned} \frac{d}{dx}\left(\tfrac13\sin 3x\right) &= \tfrac13\cdot 3\cos 3x \\ &= \cos 3x. \end{aligned}

    So 13sin⁡3x\frac13\sin 3x is an anti derivative of cos⁡3x\cos 3x.

    A constant multiplier passes straight through differentiation, (kF)′=kF′(kF)' = kF'⁠. Choosing k=13k = \frac13 cancels the factor 33 exactly. The book does the same for cos⁡2x\cos 2x in Example 1 (i), where the factor is 12\frac12⁠.
    Constant factor rule for derivatives (Section 5.3). Example 1

  4. Check: differentiate the answer back

    Let F(x)=13sin⁡3xF(x) = \frac13\sin 3x⁠. With inner u=3xu = 3x⁠, u′=3u' = 3⁠:

    F′(x)=13⋅cos⁡3x⋅3=cos⁡3x.  ✓\begin{aligned} F'(x) &= \tfrac13 \cdot \cos 3x \cdot 3 \\ &= \cos 3x. \;\checkmark \end{aligned}
    This is the definition of "anti derivative" checked directly: F′F' equals the given function at every real xx⁠, so FF qualifies.
    Property (I), Section 7.2.1: differentiation undoes integration. Basics: integration undoes differentiation

  5. All anti derivatives: add the constant

    ∫cos⁡3x dx=13sin⁡3x+C,C∈R.\int \cos 3x\,dx = \tfrac13\sin 3x + C, \quad C \in \R.
    The question asks for one anti derivative, and 13sin⁡3x\frac13\sin 3x is the simplest (the one with C=0C = 0⁠). Every other anti derivative differs from it by a constant, because two functions with the same derivative on an interval differ by a constant.
    Remark in Section 7.2 (functions with the same derivative differ by a constant). Lesson: integration as anti differentiation
Result

An anti derivative of cos⁡3x\cos 3x is 13sin⁡3x\frac13\sin 3x⁠, valid for every real xx⁠; in general

∫cos⁡3x dx=13sin⁡3x+C.\int \cos 3x\,dx = \tfrac13\sin 3x + C.

(Textbook answer: 13sin⁡3x\frac13\sin 3x⁠.)

Check it with numbers

At x=0.5x = 0.5 with h=0.001h = 0.001⁠, for F(x)=13sin⁡3xF(x) = \frac13\sin 3x⁠:

F(0.501)−F(0.499)0.002=0.332567569722−0.3324260955310.002≈0.0707371.\begin{aligned} &\frac{F(0.501) - F(0.499)}{0.002} \\ &= \frac{0.332567569722 - 0.332426095531}{0.002} \\ &\approx 0.0707371. \end{aligned}

The integrand there is cos⁡1.5≈0.0707372\cos 1.5 \approx 0.0707372⁠. The slope of FF matches the given function.

Common mistakes
  1. Adding a minus sign. −13sin⁡3x-\frac13\sin 3x differentiates to −cos⁡3x-\cos 3x⁠. The minus sign belongs to the integral of sine, not of cosine.
  2. Multiplying by 3. 3sin⁡3x3\sin 3x differentiates to 9cos⁡3x9\cos 3x⁠. Divide by the inner factor, never multiply.
  3. Changing the angle. 13sin⁡x\frac13\sin x differentiates to 13cos⁡x\frac13\cos x⁠, a different function. The factor 13\frac13 goes in front; the angle 3x3x stays inside.
What to take away
  1. For a constant a≠0a \ne 0⁠: ∫cos⁡ax dx=1asin⁡ax+C\int \cos ax\,dx = \frac{1}{a}\sin ax + C⁠.
  2. The inner linear function axax always costs a factor 1a\frac1a when integrating, because it gave a factor aa when differentiating.
  3. Differentiate the answer back to catch a wrong sign or factor.

Every fact used, and where it comes from

FactUsed inSource
ddx(sin⁡x)=cos⁡x\frac{d}{dx}(\sin x) = \cos x⁠; ∫cos⁡x dx=sin⁡x+C\int \cos x\,dx = \sin x + CSteps 1, 4Standard formulae, item (ii), Section 7.2
Chain ruleSteps 2, 4Theorem 4, Section 5.3.1 (Chapter 5)
ddx(kF)=kF′\frac{d}{dx}(kF) = kF'Step 3Algebra of derivatives, Section 5.3 (product rule with a constant factor)
Inspection applied to cos⁡2x\cos 2xStep 3Example 1 (i), Section 7.2.1
Differentiation and integration undo each otherStep 4Property (I), Section 7.2.1
Anti derivatives differ by a constantStep 5Remark, Section 7.2 (Lesson: anti differentiation)
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