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Chapter 7 · Integrals

Exercise 7.1 · Q5: An anti derivative of sin⁡2x−4e3x\sin 2x - 4e^{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.

sin⁡2x−4e3x\sin 2x - 4e^{3x}
Exam answer

We know that ddx(cos⁡2x)=−2sin⁡2x\frac{d}{dx}(\cos 2x) = -2\sin 2x and ddx(e3x)=3e3x\frac{d}{dx}\left(e^{3x}\right) = 3e^{3x}⁠.

∴ ddx(−12cos⁡2x−43e3x)=sin⁡2x−4e3x\begin{aligned} &\therefore\ \frac{d}{dx}\left(-\frac12\cos 2x - \frac43e^{3x}\right) \\ &= \sin 2x - 4e^{3x} \end{aligned}

∴\therefore An anti derivative is −12cos⁡2x−43e3x-\dfrac12\cos 2x - \dfrac43e^{3x}⁠, and ∫(sin⁡2x−4e3x)dx\int\left(\sin 2x - 4e^{3x}\right)dx =−12cos⁡2x−43e3x= -\dfrac12\cos 2x - \dfrac43e^{3x} +C+ C⁠.

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

The one idea this problem is built on

The derivative of a difference is the difference of the derivatives, and constants come along unchanged. So we may find an anti derivative of each term separately, sin⁡2x\sin 2x and e3xe^{3x}⁠, and combine them with the same sign and coefficient: −12cos⁡2x-\frac12\cos 2x and −4⋅13e3x-4\cdot\frac13 e^{3x}⁠.

Given, to find, and the plan
Given
f(x)=sin⁡2x−4e3xf(x) = \sin 2x - 4e^{3x}⁠, defined for every real xx⁠.
To find
One function FF with F′(x)=sin⁡2x−4e3xF'(x) = \sin 2x - 4e^{3x}⁠.
Plan
(1) Find an anti derivative of sin⁡2x\sin 2x⁠. (2) Find one of e3xe^{3x}⁠. (3) Combine them as the function combines its terms (Property (V)). (4) Differentiate back to check. (5) Add CC⁠.

Basics you need

1Term by term

∫[k1f1+k2f2] dx=k1∫f1 dx+k2∫f2 dx\int [k_1f_1 + k_2f_2]\,dx = k_1\int f_1\,dx + k_2\int f_2\,dx (Property (V), Section 7.2.1). It holds because (k1F1+k2F2)′(k_1F_1 + k_2F_2)' equals k1F1′+k2F2′k_1F_1' + k_2F_2'⁠. Lesson: integration as anti differentiation

2Sine and exponential

ddx(−cos⁡x)=sin⁡x\frac{d}{dx}(-\cos x) = \sin x and ddxex=ex\frac{d}{dx}e^x = e^x (items (iii) and (xi) of the standard formulae, Section 7.2). Basics: trigonometric facts for integrals

3Linear inner function

By the chain rule, an inner axax multiplies a derivative by aa⁠; so when integrating we divide by aa⁠. Basics: the chain rule backwards

Solution

  1. Anti derivative of sin⁡2x\sin 2x

    Guess −cos⁡2x-\cos 2x⁠: its derivative is sin⁡2x⋅2\sin 2x\cdot 2 (chain rule, inner 2x2x⁠). Halving it fixes the factor:

    ddx(−12cos⁡2x)=12⋅2sin⁡2x=sin⁡2x.\begin{aligned} \frac{d}{dx}\left(-\tfrac12\cos 2x\right) &= \tfrac12\cdot 2\sin 2x \\ &= \sin 2x. \end{aligned}
    This is Question 1 of this exercise: −cos⁡-\cos differentiates to sin⁡\sin⁠, and the inner 2x2x contributes a factor 22 that we cancel with 12\frac12⁠.
    Item (iii) of the standard formulae; chain rule, Theorem 4, Section 5.3.1. Exercise 7.1 Q1

  2. Anti derivative of e3xe^{3x}

    Guess e3xe^{3x}⁠: its derivative is e3x⋅3e^{3x}\cdot 3 (chain rule, inner 3x3x⁠). Divide by 33⁠:

    ddx(13e3x)=13⋅3e3x=e3x.\begin{aligned} \frac{d}{dx}\left(\tfrac13 e^{3x}\right) &= \tfrac13\cdot 3e^{3x} \\ &= e^{3x}. \end{aligned}
    The exponential is its own derivative; the only adjustment is the factor 33 from the exponent 3x3x⁠. Compare Question 3, where the exponent 2x2x gave the factor 12\frac12⁠.
    Item (xi) of the standard formulae; Theorem 5, Section 5.4 (Chapter 5). Exercise 7.1 Q3

  3. Combine the pieces with the same sign and coefficient

    The function is 1⋅sin⁡2x+(−4)⋅e3x1\cdot\sin 2x + (-4)\cdot e^{3x}⁠, so we take 11 times the first anti derivative and −4-4 times the second:

    F(x)=−12cos⁡2x−4⋅13e3x=−12cos⁡2x−43e3x.\begin{aligned} F(x) &= -\tfrac12\cos 2x - 4\cdot\tfrac13 e^{3x} \\ &= -\tfrac12\cos 2x - \tfrac43 e^{3x}. \end{aligned}
    Differentiation respects sums and constant multiples, so the anti derivative of a combination is the same combination of anti derivatives. The coefficient −4-4 is carried along unchanged: it is not differentiated or integrated, only multiplied.
    Property (V), Section 7.2.1 (with k1=1k_1 = 1⁠, k2=−4k_2 = -4⁠). Lesson: integration as anti differentiation

  4. Check: differentiate the answer back

    F′(x)=−12(−sin⁡2x)(2)−43e3x(3)=sin⁡2x−4e3x.  ✓\begin{aligned} F'(x) &= -\tfrac12(-\sin 2x)(2) - \tfrac43 e^{3x}(3) \\ &= \sin 2x - 4e^{3x}. \;\checkmark \end{aligned}
    Each term is differentiated with the chain rule: ddxcos⁡2x=−2sin⁡2x\frac{d}{dx}\cos 2x = -2\sin 2x and ddxe3x=3e3x\frac{d}{dx}e^{3x} = 3e^{3x}⁠. The result is the given function for every real xx⁠.
    Property (I), Section 7.2.1. Basics: integration undoes differentiation

  5. All anti derivatives

    ∫(sin⁡2x−4e3x)dx=−12cos⁡2x−43e3x+C.\begin{aligned} &\int\left(\sin 2x - 4e^{3x}\right)dx \\ &= -\tfrac12\cos 2x - \tfrac43 e^{3x} + C. \end{aligned}
    Each piece could carry its own constant, but a sum of constants is just one constant, so a single CC is enough (the book's Note after Example 2).
    Remark in Section 7.2; Note after Example 2 (one constant of integration). Example 2
Result

An anti derivative of sin⁡2x−4e3x\sin 2x - 4e^{3x} is −12cos⁡2x−43e3x-\frac12\cos 2x - \frac43 e^{3x}⁠, valid for every real xx⁠; in general add + C+\,C⁠. (Textbook answer: −12cos⁡2x−43e3x-\frac12\cos 2x - \frac43 e^{3x}⁠.)

Check it with numbers

At x=0.2x = 0.2 with h=0.001h = 0.001⁠, for F(x)=−12cos⁡2x−43e3xF(x) = -\frac12\cos 2x - \frac43 e^{3x}⁠:

F(0.201)−F(0.199)0.002=−2.896931311−(−2.883133174)0.002≈−6.899068.\begin{aligned} &\frac{F(0.201) - F(0.199)}{0.002} \\ &= \frac{-2.896931311 - (-2.883133174)}{0.002} \\ &\approx -6.899068. \end{aligned}

The integrand is sin⁡0.4−4e0.6\sin 0.4 - 4e^{0.6}⁠, about 0.389418−7.2884750.389418 - 7.288475⁠, which is −6.899057-6.899057⁠. They agree to four decimal places (the gap of about 0.000010.00001 comes from the step hh⁠).

Common mistakes
  1. Losing the coefficient 4. The anti derivative of 4e3x4e^{3x} is 43e3x\frac43 e^{3x}⁠, not 13e3x\frac13 e^{3x}⁠.
  2. Sign of the first term. sin⁡\sin integrates to −cos⁡-\cos⁠, so the first term is −12cos⁡2x-\frac12\cos 2x⁠.
  3. Multiplying the factors instead of dividing. −2cos⁡2x−12e3x-2\cos 2x - 12e^{3x} is what you get by differentiating the inner functions into the answer; differentiate it back and you get 4sin⁡2x−36e3x4\sin 2x - 36e^{3x}⁠.
What to take away
  1. Split a sum or difference into terms, integrate each, and keep every coefficient and sign.
  2. ∫sin⁡ax dx=−1acos⁡ax+C\int \sin ax\,dx = -\frac1a\cos ax + C and ∫eax dx=1aeax+C\int e^{ax}\,dx = \frac1a e^{ax} + C for a≠0a \ne 0⁠.

Every fact used, and where it comes from

FactUsed inSource
∫sin⁡x dx=−cos⁡x+C\int \sin x\,dx = -\cos x + CStep 1Standard formulae, item (iii), Section 7.2
∫ex dx=ex+C\int e^x\,dx = e^x + CStep 2Standard formulae, item (xi), Section 7.2; Theorem 5, Section 5.4 (Chapter 5)
Chain ruleSteps 1, 2, 4Theorem 4, Section 5.3.1 (Chapter 5)
Integrate term by term with constant multiplesStep 3Property (V), Section 7.2.1
Differentiation and integration undo each otherStep 4Property (I), Section 7.2.1
One constant of integration is enoughStep 5Note after Example 2, Section 7.2.1
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