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Chapter 7 · Integrals

Exercise 7.2 · Q4: ∫sin⁡x sin⁡(cos⁡x) dx\int \sin x\,\sin(\cos x)\,dx

Exercise 7.2Integration by substitution · Section 7.3.1
The question

Integrate the function:

sin⁡x sin⁡(cos⁡x)\sin x\,\sin(\cos x)
Exam answer

Let I=∫sin⁡x sin⁡(cos⁡x) dxI = \displaystyle\int\sin x\,\sin(\cos x)\,dx⁠.

Put cos⁡x=t\cos x = t⁠. Then −sin⁡x dx=dt-\sin x\,dx = dt⁠.

∴ I=−∫sin⁡t dt=cos⁡t+C\begin{aligned} \therefore\ I &= -\int\sin t\,dt \\ &= \cos t + C \end{aligned}

∴ I=cos⁡(cos⁡x)+C\therefore\ I = \cos(\cos x) + C⁠.

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

The one idea this problem is built on

The sine is applied to cos⁡x\cos x⁠, and the other factor sin⁡x\sin x is, apart from a minus sign, the derivative of cos⁡x\cos x⁠. So t=cos⁡xt = \cos x gives dt=−sin⁡x dxdt = -\sin x\,dx⁠, and the integral becomes −∫sin⁡t dt=cos⁡t-\int\sin t\,dt = \cos t⁠. The minus sign is the only trap.

Given, to find, and the plan
Given
f(x)=sin⁡x sin⁡(cos⁡x)f(x) = \sin x\,\sin(\cos x)⁠, defined and continuous for every real xx⁠.
To find
∫sin⁡x sin⁡(cos⁡x) dx\displaystyle\int \sin x\,\sin(\cos x)\,dx⁠.
Plan
(1) Identify the inner function cos⁡x\cos x and its derivative. (2) Substitute t=cos⁡xt = \cos x⁠, keeping track of the sign. (3) Integrate sin⁡t\sin t⁠. (4) Return to xx⁠. (5) Differentiate to check.

Basics you need

1Derivatives of sine and cosine

ddxsin⁡x=cos⁡x\frac{d}{dx}\sin x = \cos x and ddxcos⁡x=−sin⁡x\frac{d}{dx}\cos x = -\sin x (Chapter 5, Table 5.3). Basics: trigonometric facts

2Substitution

Put t=g(x)t = g(x)⁠, so dt=g′(x) dxdt = g'(x)\,dx (Section 7.3.1). Basics: substitution

3The integral of sin⁡t\sin t

∫sin⁡t dt=−cos⁡t+C\int\sin t\,dt = -\cos t + C (standard formula (iii), Section 7.2), because ddt(−cos⁡t)=sin⁡t\frac{d}{dt}(-\cos t) = \sin t⁠.

Solution

  1. Find the inner function and its derivative

    The integrand is a sine of cos⁡x\cos x⁠, multiplied by sin⁡x\sin x⁠. The inner function is cos⁡x\cos x⁠, and

    ddxcos⁡x=−sin⁡x.\frac{d}{dx}\cos x = -\sin x.
    The factor sin⁡x\sin x in front is the derivative of the inner function up to the constant factor −1-1⁠. A constant factor is no obstacle: it can be moved outside the integral (Property (IV), Section 7.2.1). So t=cos⁡xt = \cos x is the right substitution.
    (cos⁡x)′=−sin⁡x(\cos x)' = -\sin x⁠, Chapter 5, Table 5.3. Basics: trigonometric facts

  2. Substitute t=cos⁡xt = \cos x

    t=cos⁡x,dtdx=−sin⁡x,sin⁡x dx=− dt.\begin{aligned} t &= \cos x, \\ \frac{dt}{dx} &= -\sin x, \\ \sin x\,dx &= -\,dt. \end{aligned}

    Therefore

    ∫sin⁡(cos⁡x) (sin⁡x dx)=∫sin⁡t (− dt)=−∫sin⁡t dt.\begin{aligned} \int \sin(\cos x)\,(\sin x\,dx) &= \int \sin t\,(-\,dt) \\ &= -\int \sin t\,dt. \end{aligned}
    We solved dt=−sin⁡x dxdt = -\sin x\,dx for the piece that is actually in the integrand, sin⁡x dx\sin x\,dx⁠. Then sin⁡(cos⁡x)\sin(\cos x) becomes sin⁡t\sin t and sin⁡x dx\sin x\,dx becomes −dt-dt⁠. The constant −1-1 comes out in front.
    Substitution, Section 7.3.1; ∫kf=k∫f\int k f = k\int f⁠, Property (IV), Section 7.2.1. Lesson: integration by substitution

  3. Integrate sin⁡t\sin t

    −∫sin⁡t dt=−(−cos⁡t)+C=cos⁡t+C.-\int\sin t\,dt = -\left(-\cos t\right) + C = \cos t + C.
    Standard formula (iii) gives ∫sin⁡t dt=−cos⁡t+C\int\sin t\,dt = -\cos t + C⁠. The two minus signs cancel. (The constant CC is arbitrary, so −C-C is just another arbitrary constant; we keep writing CC⁠.)
    Standard formula (iii), Section 7.2.

  4. Return to xx

    cos⁡t+C=cos⁡(cos⁡x)+C.\cos t + C = \cos(\cos x) + C.
    Replace tt by cos⁡x\cos x⁠. No restriction is needed: cos⁡(cos⁡x)\cos(\cos x) is defined for every real xx⁠.

  5. Check: differentiate the answer

    Chain rule with inner u=cos⁡xu = \cos x and outer cos⁡u\cos u⁠:

    ddxcos⁡(cos⁡x)=−sin⁡(cos⁡x)⋅ddx(cos⁡x)=−sin⁡(cos⁡x)⋅(−sin⁡x)=sin⁡x sin⁡(cos⁡x).\begin{aligned} \frac{d}{dx}\cos(\cos x) &= -\sin(\cos x)\cdot\frac{d}{dx}(\cos x) \\ &= -\sin(\cos x)\cdot(-\sin x) \\ &= \sin x\,\sin(\cos x). \end{aligned}

    This is the integrand, so the answer is correct for all real xx⁠.

    The two minus signs (one from differentiating the outer cosine, one from the inner cosine) cancel, exactly mirroring the two that cancelled in Step 3.
    Chain rule, Chapter 5, Theorem 4; (cos⁡u)′=−sin⁡u(\cos u)' = -\sin u⁠, Table 5.3. Basics: integration undoes differentiation
Result

For every real xx⁠,

∫sin⁡x sin⁡(cos⁡x) dx=cos⁡(cos⁡x)+C.\int \sin x\,\sin(\cos x)\,dx = \cos(\cos x) + C.

(Textbook answer: cos⁡(cos⁡x)+C\cos(\cos x) + C⁠.)

Check it with numbers

Let F(x)=cos⁡(cos⁡x)F(x) = \cos(\cos x) (angles in radians). At x=1x = 1 the integrand is sin⁡1⋅sin⁡(cos⁡1)≈0.432849\sin 1\cdot\sin(\cos 1) \approx 0.432849⁠. With h=0.0001h = 0.0001⁠:

F(1.0001)≈0.8575964991,F(0.9999)≈0.8575099293,F(1.0001)−F(0.9999)0.0002≈0.432849.\begin{aligned} F(1.0001) &\approx 0.8575964991, \\ F(0.9999) &\approx 0.8575099293, \\ \frac{F(1.0001)-F(0.9999)}{0.0002} &\approx 0.432849. \end{aligned}

The slope of FF equals the integrand.

Common mistakes
  1. Losing the minus sign. dt=−sin⁡x dxdt = -\sin x\,dx⁠, not +sin⁡x dx+\sin x\,dx⁠. Forgetting it gives the wrong answer −cos⁡(cos⁡x)-\cos(\cos x)⁠.
  2. Treating sin⁡(cos⁡x)\sin(\cos x) as sin⁡xcos⁡x\sin x\cos x⁠. It is the sine of cos⁡x\cos x⁠, a composite function, not a product.
  3. Using degrees. Calculus formulas for trigonometric functions assume radians.
What to take away
  1. The derivative of the inner function only needs to be present up to a constant factor; the constant moves outside the integral.
  2. With t=cos⁡xt = \cos x⁠, always write sin⁡x dx=−dt\sin x\,dx = -dt before substituting.

Every fact used, and where it comes from

FactUsed inSource
(cos⁡x)′=−sin⁡x(\cos x)' = -\sin xSteps 1, 2, 5Chapter 5, Table 5.3
SubstitutionStep 2Section 7.3.1 (Lesson: substitution)
∫kf(x) dx=k∫f(x) dx\int k f(x)\,dx = k\int f(x)\,dxStep 2Property (IV), Section 7.2.1
∫sin⁡t dt=−cos⁡t+C\int\sin t\,dt = -\cos t + CStep 3Standard formula (iii), Section 7.2
Chain ruleStep 5Chapter 5, Theorem 4
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