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Chapter 7 · Integrals

Exercise 7.3 · Q2: ∫sin⁡3xcos⁡4x dx\int\sin 3x\cos 4x\,dx

Exercise 7.3Integration using trigonometric identities · Section 7.3.2
The question

Find the integral of the function

sin⁡3xcos⁡4x.\sin 3x\cos 4x.
Exam answer

Let I=∫sin⁡3xcos⁡4x dxI = \displaystyle\int \sin 3x\cos 4x\,dx⁠.

Using 2sin⁡Acos⁡B2\sin A\cos B =sin⁡(A+B)+sin⁡(A−B)= \sin(A + B) + \sin(A - B)⁠,

sin⁡3xcos⁡4x=12(sin⁡7x+sin⁡(−x))=12(sin⁡7x−sin⁡x)\begin{aligned} &\sin 3x\cos 4x \\ &= \frac12\big(\sin 7x + \sin(-x)\big) \\ &= \frac12(\sin 7x - \sin x) \end{aligned} ∴ I=12∫(sin⁡7x−sin⁡x) dx=12(−cos⁡7x7+cos⁡x)+C\begin{aligned} \therefore\ I &= \frac12\int(\sin 7x - \sin x)\,dx \\ &= \frac12\left(-\frac{\cos 7x}{7} + \cos x\right) + C \end{aligned}

∴ I=−114cos⁡7x+12cos⁡x\therefore\ I = -\dfrac{1}{14}\cos 7x + \dfrac12\cos x +C+ C⁠.

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

The one idea this problem is built on

A product of a sine and a cosine has no direct formula, but a sum of sines does. The product-to-sum identity sin⁡Acos⁡B=12[sin⁡(A+B)+sin⁡(A−B)]\sin A\cos B = \frac12[\sin(A+B) + \sin(A-B)] turns the product into a sum of two sines of linear angles, and each of those integrates at once.

Given, to find, and the plan
Given
The function sin⁡3xcos⁡4x\sin 3x\cos 4x⁠, defined for every real xx⁠.
To find
∫sin⁡3xcos⁡4x dx\displaystyle\int \sin 3x\cos 4x\,dx⁠.
Plan
(1) Recall where the identity comes from. (2) Apply it with A=3xA = 3x⁠, B=4xB = 4x⁠, and tidy the negative angle. (3) Integrate the two sines. (4) Differentiate back to check.

Basics you need

1Compound-angle formulas

sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A+B) = \sin A\cos B + \cos A\sin B and sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin(A-B) = \sin A\cos B - \cos A\sin B⁠. Basics: trigonometric identities for integrals

2Sine is odd

sin⁡(−θ)=−sin⁡θ\sin(-\theta) = -\sin\theta for every θ\theta (and cosine is even: cos⁡(−θ)=cos⁡θ\cos(-\theta) = \cos\theta⁠).

3Sine of a multiple of xx

∫sin⁡mx dx=−1mcos⁡mx+C\int\sin mx\,dx = -\frac1m\cos mx + C for m≠0m \ne 0 (Example 5(i), Section 7.3.1, by the substitution t=mxt = mx⁠). Basics: substitution

4Checking an integral

F(x)+CF(x) + C is right exactly when F′(x)F'(x) is the integrand (Property (I), Section 7.2.1). Basics: integration undoes differentiation

Solution

  1. Where the product-to-sum identity comes from

    Add the two compound-angle formulas. The cos⁡Asin⁡B\cos A\sin B terms cancel:

    sin⁡(A+B)+sin⁡(A−B)=2sin⁡Acos⁡B,\sin(A+B) + \sin(A-B) = 2\sin A\cos B,

    so

    sin⁡Acos⁡B=12[sin⁡(A+B)+sin⁡(A−B)].\sin A\cos B = \tfrac12\big[\sin(A+B) + \sin(A-B)\big].
    This is the identity the book recalls in Example 7(ii), Section 7.3.2 (its "Why?" is answered by this addition). Knowing how it is built means you never have to memorise which sign goes where: you can rebuild it in one line.
    Compound-angle formulas for sin⁡(A±B)\sin(A \pm B) (Class XI trigonometry). Basics: trigonometric identities

  2. Apply it with A=3xA = 3x⁠, B=4xB = 4x

    Here A+B=7xA + B = 7x and A−B=3x−4x=−xA - B = 3x - 4x = -x⁠:

    sin⁡3xcos⁡4x=12[sin⁡7x+sin⁡(−x)]=12[sin⁡7x−sin⁡x].\begin{aligned} \sin 3x\cos 4x &= \tfrac12\big[\sin 7x + \sin(-x)\big] \\ &= \tfrac12\big[\sin 7x - \sin x\big]. \end{aligned}
    The identity needs AA to be the angle of the sine and BB the angle of the cosine, so A=3xA = 3x⁠, B=4xB = 4x⁠, even though this makes A−BA - B negative. A negative angle is no problem: sine is odd, so sin⁡(−x)=−sin⁡x\sin(-x) = -\sin x⁠.
    sin⁡(−θ)=−sin⁡θ\sin(-\theta) = -\sin\theta⁠.

  3. Integrate the two sines

    ∫sin⁡3xcos⁡4x dx=12∫sin⁡7x dx−12∫sin⁡x dx=12(−17cos⁡7x)−12(−cos⁡x)+C=−114cos⁡7x+12cos⁡x+C.\begin{aligned} &\int \sin 3x\cos 4x\,dx \\ &= \frac12\int\sin 7x\,dx - \frac12\int\sin x\,dx \\ &= \frac12\Big(-\frac17\cos 7x\Big) - \frac12(-\cos x) + C \\ &= -\frac{1}{14}\cos 7x + \frac12\cos x + C. \end{aligned}
    Each sine is a standard integral: ∫sin⁡x dx=−cos⁡x\int\sin x\,dx = -\cos x⁠, and for sin⁡7x\sin 7x the substitution t=7xt = 7x⁠, dx=dt7dx = \frac{dt}{7} brings in the factor 17\frac17⁠. Watch the two minus signs in the middle term: −12⋅(−cos⁡x)=+12cos⁡x-\frac12\cdot(-\cos x) = +\frac12\cos x⁠.
    Properties (III), (IV), Section 7.2.1; ∫sin⁡x dx=−cos⁡x+C\int\sin x\,dx = -\cos x + C (Section 7.2); Example 5(i) for sin⁡mx\sin mx⁠. Lesson: integration by substitution

  4. Check: differentiate the answer back

    With F(x)=−114cos⁡7x+12cos⁡xF(x) = -\frac{1}{14}\cos 7x + \frac12\cos x and ddxcos⁡7x=−7sin⁡7x\frac{d}{dx}\cos 7x = -7\sin 7x (chain rule):

    F′(x)=−114(−7sin⁡7x)+12(−sin⁡x)=12[sin⁡7x−sin⁡x]=sin⁡3xcos⁡4x.\begin{aligned} F'(x) &= -\frac{1}{14}(-7\sin 7x) + \frac12(-\sin x) \\ &= \tfrac12\big[\sin 7x - \sin x\big] \\ &= \sin 3x\cos 4x. \end{aligned}

    The last line is Step 2 read backwards. The answer is correct for every real xx⁠.

    Differentiating returns us to the "sum" form; the identity then brings back the product. The factor 7 from the chain rule cancels the 17\frac17 from the integration.
    Chain rule, Theorem 4, Section 5.3.1; ddxcos⁡x=−sin⁡x\frac{d}{dx}\cos x = -\sin x⁠. Basics: integration undoes differentiation
Result ∫sin⁡3xcos⁡4x dx=−114cos⁡7x+12cos⁡x+C,\int \sin 3x\cos 4x\,dx = -\frac{1}{14}\cos 7x + \frac12\cos x + C,

valid for all real xx⁠. (Textbook answer: −114cos⁡7x+12cos⁡x+C-\dfrac{1}{14}\cos 7x + \dfrac12\cos x + C⁠.)

Check it with numbers

First the identity at x=1x = 1⁠: sin⁡3cos⁡4\sin 3\cos 4 ≈0.1411200×(−0.6536436)\approx 0.1411200\times(-0.6536436) ≈−0.0922422\approx -0.0922422⁠, and 12(sin⁡7−sin⁡1)\frac12(\sin 7 - \sin 1) ≈12(0.6569866−0.8414710)\approx \frac12(0.6569866 - 0.8414710) ≈−0.0922422\approx -0.0922422⁠. They agree.

Then the derivative at x=1x = 1 with h=0.001h = 0.001⁠: F(1.001)≈0.2162099314F(1.001) \approx 0.2162099314⁠, F(0.999)≈0.2163944210F(0.999) \approx 0.2163944210⁠, so the slope is about −0.00018448960.002\dfrac{-0.0001844896}{0.002} ≈−0.0922448\approx -0.0922448⁠, matching the integrand −0.0922422-0.0922422 to four significant figures (the gap is the error of the difference quotient).

Common mistakes
  1. Writing sin⁡(−x)=sin⁡x\sin(-x) = \sin x⁠. Sine is odd: sin⁡(−x)=−sin⁡x\sin(-x) = -\sin x⁠. (Cosine is even, which is why the cosine identities never cause this trouble.)
  2. Swapping the roles of AA and BB⁠. If you take A=4xA = 4x in sin⁡Acos⁡B\sin A\cos B you are expanding sin⁡4xcos⁡3x\sin 4x\cos 3x⁠, a different function.
  3. Forgetting the 17\frac17⁠. ∫sin⁡7x dx=−17cos⁡7x\int\sin 7x\,dx = -\frac17\cos 7x⁠.
  4. Losing the factor 12\frac12 from the identity.
What to take away
  1. Products of sines and cosines of multiples of xx become sums with the product-to-sum identities; Questions 3, 6 and 7 use the same tool.
  2. The three identities: 2sin⁡Acos⁡B=sin⁡(A+B)+sin⁡(A−B)2\sin A\cos B = \sin(A+B) + \sin(A-B)⁠, 2cos⁡Acos⁡B=cos⁡(A+B)+cos⁡(A−B)2\cos A\cos B = \cos(A+B) + \cos(A-B)⁠, 2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)2\sin A\sin B = \cos(A-B) - \cos(A+B)⁠.
  3. A negative angle is fine: tidy it with sin⁡(−θ)=−sin⁡θ\sin(-\theta) = -\sin\theta⁠, cos⁡(−θ)=cos⁡θ\cos(-\theta) = \cos\theta⁠.

Every fact used, and where it comes from

FactUsed inSource
sin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡B\sin(A\pm B) = \sin A\cos B \pm \cos A\sin BStep 1Compound-angle formulas (Class XI trigonometry)
sin⁡Acos⁡B=12[sin⁡(A+B)+sin⁡(A−B)]\sin A\cos B = \frac12[\sin(A+B) + \sin(A-B)]Steps 1, 2, 4Example 7(ii), Section 7.3.2
sin⁡(−θ)=−sin⁡θ\sin(-\theta) = -\sin\thetaStep 2Sine is an odd function (Class XI)
∫sin⁡mx dx=−1mcos⁡mx+C\int\sin mx\,dx = -\frac1m\cos mx + CStep 3Example 5(i), Section 7.3.1
Sum and constant-multiple rulesStep 3Properties (III), (IV), Section 7.2.1
Chain rule; differentiation undoes integrationStep 4Theorem 4, Section 5.3.1; Property (I), Section 7.2.1
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