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Chapter 7 · Integrals

Exercise 7.3 · Q4: ∫sin⁡3(2x+1) dx\int\sin^3(2x+1)\,dx

Exercise 7.3Integration using trigonometric identities · Section 7.3.2
The question

Find the integral of the function

sin⁡3(2x+1).\sin^3(2x+1).
Exam answer

Let I=∫sin⁡3(2x+1) dxI = \displaystyle\int \sin^3(2x + 1)\,dx⁠.

Using sin⁡2θ=1−cos⁡2θ\sin^2\theta = 1 - \cos^2\theta⁠,

I=∫(1−cos⁡2(2x+1))×sin⁡(2x+1) dx\begin{aligned} I &= \int\big(1 - \cos^2(2x + 1)\big) \\ &\qquad \times\sin(2x + 1)\,dx \end{aligned}

Put cos⁡(2x+1)=t\cos(2x + 1) = t⁠. Then −2sin⁡(2x+1) dx=dt-2\sin(2x + 1)\,dx = dt⁠.

∴ I=−12∫(1−t2) dt=−12(t−t33)+C\begin{aligned} \therefore\ I &= -\frac12\int(1 - t^2)\,dt \\ &= -\frac12\left(t - \frac{t^3}{3}\right) + C \end{aligned}

∴ I=−12cos⁡(2x+1)\therefore\ I = -\dfrac12\cos(2x + 1) +16cos⁡3(2x+1)+C+ \dfrac16\cos^3(2x + 1) + C⁠.

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

The one idea this problem is built on

An odd power of sine has a spare factor. Keep one sin⁡θ\sin\theta aside and turn the remaining sin⁡2θ\sin^2\theta into 1−cos⁡2θ1 - \cos^2\theta⁠. The spare sin⁡θ\sin\theta is (up to a constant) the derivative of cos⁡θ\cos\theta⁠, so the substitution t=cos⁡θt = \cos\theta turns the whole integral into a polynomial in tt⁠.

Given, to find, and the plan
Given
sin⁡3(2x+1)\sin^3(2x+1)⁠, defined for every real xx⁠.
To find
∫sin⁡3(2x+1) dx\displaystyle\int \sin^3(2x+1)\,dx⁠.
Plan
(1) Write sin⁡3θ=(1−cos⁡2θ)sin⁡θ\sin^3\theta = (1 - \cos^2\theta)\sin\theta with θ=2x+1\theta = 2x+1⁠. (2) Substitute t=cos⁡(2x+1)t = \cos(2x+1)⁠. (3) Integrate the polynomial and return to xx⁠. (4) Differentiate back to check. (5) Compare with the triple-angle route used in Example 7(iii).

Basics you need

1The Pythagorean identity

sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1⁠, so sin⁡2θ=1−cos⁡2θ\sin^2\theta = 1 - \cos^2\theta for every θ\theta⁠. Basics: trigonometric identities for integrals

2Substitution

If t=g(x)t = g(x) then dt=g′(x) dxdt = g'(x)\,dx⁠, and ∫f(g(x)) g′(x) dx=∫f(t) dt\int f(g(x))\,g'(x)\,dx = \int f(t)\,dt (Section 7.3.1). Here t=cos⁡(2x+1)t = \cos(2x+1) and dt=−2sin⁡(2x+1) dxdt = -2\sin(2x+1)\,dx⁠. Basics: substitution

3The power formula

∫tn dt=tn+1n+1+C\int t^n\,dt = \frac{t^{n+1}}{n+1} + C⁠, n≠−1n \ne -1 (standard formulae, Section 7.2). Basics: powers and the power rule

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. Keep one sine aside

    Write θ=2x+1\theta = 2x+1⁠. Then

    sin⁡3θ=sin⁡2θ⋅sin⁡θ=(1−cos⁡2θ)sin⁡θ.\sin^3\theta = \sin^2\theta\cdot\sin\theta = (1 - \cos^2\theta)\sin\theta.
    With an odd power we can split off one factor and still have an even power left, which the Pythagorean identity converts entirely into cosines. Everything is then a function of cos⁡θ\cos\theta times sin⁡θ\sin\theta⁠, and sin⁡θ\sin\theta is what the derivative of cos⁡θ\cos\theta supplies. This is the method of Example 6(i) and of the "Alternatively" part of Example 7(iii).
    sin⁡2θ=1−cos⁡2θ\sin^2\theta = 1 - \cos^2\theta⁠. Basics: trigonometric identities

  2. Substitute t=cos⁡(2x+1)t = \cos(2x+1)

    By the chain rule, dtdx=−sin⁡(2x+1)⋅2\frac{dt}{dx} = -\sin(2x+1)\cdot 2⁠, so

    dt=−2sin⁡(2x+1) dx,i.e.sin⁡(2x+1) dx=−12 dt.dt = -2\sin(2x+1)\,dx, \quad\text{i.e.}\quad \sin(2x+1)\,dx = -\tfrac12\,dt.

    Therefore

    ∫sin⁡3(2x+1) dx=∫(1−cos⁡2(2x+1))sin⁡(2x+1) dx=−12∫(1−t2) dt.\begin{aligned} &\int \sin^3(2x+1)\,dx \\ &= \int \big(1 - \cos^2(2x+1)\big)\sin(2x+1)\,dx \\ &= -\frac12\int (1 - t^2)\,dt. \end{aligned}
    Two things to watch: the minus sign (the derivative of cosine is minus sine) and the factor 2 (the derivative of the inner angle 2x+12x+1⁠). Both end up in the constant −12-\frac12⁠. No xx is left inside the integral, so the substitution is complete. It is valid for all xx⁠, since cos⁡(2x+1)\cos(2x+1) is differentiable everywhere.
    Change of variable, Section 7.3.1; chain rule, Theorem 4, Section 5.3.1. Lesson: integration by substitution

  3. Integrate and return to xx

    −12∫(1−t2) dt=−12(t−t33)+C=−12t+16t3+C.\begin{aligned} -\frac12\int (1 - t^2)\,dt &= -\frac12\Big(t - \frac{t^3}{3}\Big) + C \\ &= -\frac12 t + \frac16 t^3 + C. \end{aligned}

    Put back t=cos⁡(2x+1)t = \cos(2x+1)⁠:

    ∫sin⁡3(2x+1) dx=−12cos⁡(2x+1)+16cos⁡3(2x+1)+C.\begin{aligned} &\int \sin^3(2x+1)\,dx \\ &= -\frac12\cos(2x+1) + \frac16\cos^3(2x+1) + C. \end{aligned}
    The integrand in tt is a polynomial, so the power formula does everything. Multiplying −12-\frac12 into the bracket: −12⋅(−t33)=+t36-\frac12\cdot\big(-\frac{t^3}{3}\big) = +\frac{t^3}{6}⁠.
    Power formula with n=0n = 0 and n=2n = 2⁠. Basics: the power rule

  4. Check: differentiate the answer back

    Let c=cos⁡(2x+1)c = \cos(2x+1)⁠, so dcdx=−2sin⁡(2x+1)\frac{dc}{dx} = -2\sin(2x+1)⁠. Then

    F′(x)=−12⋅(−2sin⁡(2x+1))+16⋅3c2⋅(−2sin⁡(2x+1))=sin⁡(2x+1)−c2sin⁡(2x+1)=(1−c2)sin⁡(2x+1)=sin⁡3(2x+1).\begin{aligned} F'(x) &= -\frac12\cdot\big(-2\sin(2x+1)\big) \\ &\qquad + \frac16\cdot 3c^2\cdot\big(-2\sin(2x+1)\big) \\ &= \sin(2x+1) - c^2\sin(2x+1) \\ &= (1 - c^2)\sin(2x+1) = \sin^3(2x+1). \end{aligned}

    The answer is correct for every real xx⁠.

    In the last line 1−cos⁡2(2x+1)=sin⁡2(2x+1)1 - \cos^2(2x+1) = \sin^2(2x+1)⁠, which is Step 1 read backwards.
    Chain rule, Theorem 4, Section 5.3.1. Basics: integration undoes differentiation

  5. The triple-angle route gives the same function

    Example 7(iii) also uses sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin 3\theta = 3\sin\theta - 4\sin^3\theta⁠, that is sin⁡3θ=3sin⁡θ−sin⁡3θ4\sin^3\theta = \frac{3\sin\theta - \sin 3\theta}{4}⁠. With θ=2x+1\theta = 2x+1 (so 3θ=6x+33\theta = 6x+3⁠):

    ∫sin⁡3(2x+1) dx=34∫sin⁡(2x+1) dx−14∫sin⁡(6x+3) dx=−38cos⁡(2x+1)+124cos⁡(6x+3)+C.\begin{aligned} &\int \sin^3(2x+1)\,dx \\ &= \frac34\int\sin(2x+1)\,dx - \frac14\int\sin(6x+3)\,dx \\ &= -\frac38\cos(2x+1) + \frac{1}{24}\cos(6x+3) + C. \end{aligned}

    Now use cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4\cos^3\theta - 3\cos\theta with c=cos⁡(2x+1)c = \cos(2x+1)⁠:

    −38c+124(4c3−3c)=−38c−18c+16c3=−12c+16c3.\begin{aligned} -\frac38 c + \frac{1}{24}(4c^3 - 3c) &= -\frac38 c - \frac18 c + \frac16 c^3 \\ &= -\frac12 c + \frac16 c^3. \end{aligned}

    This is exactly our answer: here the two routes agree with no difference in the constant at all.

    The book's Remark after Example 7 says the two answers there are equivalent "using trigonometric identities"; this step shows how, for our function. In general two correct answers may differ by a constant; here they happen to be identical.
Result ∫sin⁡3(2x+1) dx=−12cos⁡(2x+1)+16cos⁡3(2x+1)+C,\begin{aligned} &\int \sin^3(2x+1)\,dx \\ &= -\frac12\cos(2x+1) + \frac16\cos^3(2x+1) + C, \end{aligned}

valid for all real xx⁠. This is exactly the textbook answer, −12cos⁡(2x+1)-\dfrac12\cos(2x+1) + 16cos⁡3(2x+1)+C+\, \dfrac16\cos^3(2x+1) + C⁠.

Check it with numbers

At x=1x = 1 with h=0.001h = 0.001⁠: F(1.001)≈0.333286177097F(1.001) \approx 0.333286177097 and F(0.999)≈0.333280555232F(0.999) \approx 0.333280555232⁠, so the slope is about 0.0000056218650.002≈0.0028109\dfrac{0.000005621865}{0.002} \approx 0.0028109⁠. The integrand is sin⁡33≈0.0028104\sin^3 3 \approx 0.0028104⁠. They agree to three significant figures (the gap is the error of the difference quotient).

Common mistakes
  1. Treating sin⁡3\sin^3 like sin⁡2\sin^2⁠. The half-angle identity alone does not lower a cube; split off one factor first (or use the triple-angle identity).
  2. Losing the minus sign in dt=−2sin⁡(2x+1) dxdt = -2\sin(2x+1)\,dx⁠.
  3. Losing the factor 2 from the inner angle 2x+12x+1⁠.
  4. Thinking two different-looking answers must be wrong. Step 5 shows that very different-looking answers can be the same function.
What to take away
  1. Odd power of sine: keep one sin⁡θ\sin\theta⁠, convert the rest with sin⁡2θ=1−cos⁡2θ\sin^2\theta = 1 - \cos^2\theta⁠, put t=cos⁡θt = \cos\theta⁠. (Odd power of cosine: the mirror image, t=sin⁡θt = \sin\theta⁠.)
  2. For an inner angle ax+bax+b⁠, dtdt carries the factor aa⁠.
  3. Compare answers by rewriting one into the other or by checking that their difference is constant.

Every fact used, and where it comes from

FactUsed inSource
sin⁡2θ=1−cos⁡2θ\sin^2\theta = 1 - \cos^2\thetaSteps 1, 4Pythagorean identity (Class XI trigonometry)
Substitution t=cos⁡(2x+1)t = \cos(2x+1)Step 2Section 7.3.1; same method as Example 6(i) and Example 7(iii)
∫tn dt=tn+1n+1+C\int t^n\,dt = \frac{t^{n+1}}{n+1} + CStep 3Standard formulae, Section 7.2
Chain ruleSteps 2, 4Theorem 4, Section 5.3.1
sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin 3\theta = 3\sin\theta - 4\sin^3\theta⁠, cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4\cos^3\theta - 3\cos\thetaStep 5Triple-angle formulas (Class XI); Example 7(iii), Section 7.3.2
Differentiation undoes integrationStep 4Property (I), Section 7.2.1
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