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Chapter 7 · Integrals

Exercise 7.3 · Q20: ∫cos⁡2x(cos⁡x+sin⁡x)2 dx\int\frac{\cos 2x}{(\cos x+\sin x)^2}\,dx

Exercise 7.3Integration using trigonometric identities · Section 7.3.2
The question

Find the integral of the function

cos⁡2x(cos⁡x+sin⁡x)2.\frac{\cos 2x}{(\cos x + \sin x)^2}.
Exam answer

Let I=∫cos⁡2x(cos⁡x+sin⁡x)2 dxI = \displaystyle\int\frac{\cos 2x}{(\cos x + \sin x)^2}\,dx⁠.

Using cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = \cos^2 x - \sin^2 x⁠,

cos⁡2x=(cos⁡x−sin⁡x)×(cos⁡x+sin⁡x)\begin{aligned} \cos 2x &= (\cos x - \sin x) \\ &\qquad \times(\cos x + \sin x) \end{aligned} ∴ I=∫cos⁡x−sin⁡xcos⁡x+sin⁡x dx\therefore\ I = \int\frac{\cos x - \sin x}{\cos x + \sin x}\,dx

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

∴ I=∫dtt=log⁡∣t∣+C\therefore\ I = \int\frac{dt}{t} = \log|t| + C

∴ I=log⁡∣cos⁡x+sin⁡x∣+C\therefore\ I = \log|\cos x + \sin x| + C⁠.

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

The one idea this problem is built on

cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = \cos^2 x - \sin^2 x is a difference of squares: (cos⁡x−sin⁡x)(cos⁡x+sin⁡x)(\cos x - \sin x)(\cos x + \sin x)⁠. One factor cos⁡x+sin⁡x\cos x + \sin x cancels with the denominator, leaving cos⁡x−sin⁡xcos⁡x+sin⁡x\frac{\cos x - \sin x}{\cos x + \sin x}⁠. Its numerator is the derivative of its denominator, so the integral is a logarithm: log⁡∣cos⁡x+sin⁡x∣+C\log|\cos x + \sin x| + C⁠.

Given, to find, and the plan
Given
f(x)=cos⁡2x(cos⁡x+sin⁡x)2f(x) = \dfrac{\cos 2x}{(\cos x + \sin x)^2}⁠, defined wherever cos⁡x+sin⁡x≠0\cos x + \sin x \ne 0⁠.
To find
∫cos⁡2x(cos⁡x+sin⁡x)2 dx\displaystyle\int \frac{\cos 2x}{(\cos x + \sin x)^2}\,dx⁠.
Plan
(1) Find the domain. (2) Factorise cos⁡2x\cos 2x and cancel. (3) Substitute t=cos⁡x+sin⁡xt = \cos x + \sin x⁠. (4) Integrate 1t\frac1t and return to xx⁠. (5) Differentiate back.

Basics you need

1cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = \cos^2 x - \sin^2 x

One of the three forms of the double-angle formula; with A2−B2=(A−B)(A+B)A^2 - B^2 = (A - B)(A + B) it factorises. Basics: trigonometric identities for integrals

2Substitution

If t=g(x)t = g(x)⁠, then dt=g′(x) dxdt = g'(x)\,dx⁠. Basics: substitution

3∫dtt=log⁡∣t∣+C\int\frac{dt}{t} = \log|t| + C

Standard formula (xii), Section 7.2, valid for t≠0t \ne 0⁠. Basics: logarithms and 1/x

Solution

  1. Where is the integrand defined?

    We need cos⁡x+sin⁡x≠0\cos x + \sin x \ne 0⁠. If cos⁡x=0\cos x = 0 then sin⁡x=±1\sin x = \pm 1⁠, so the sum is not zero there; otherwise divide by cos⁡x\cos x⁠: the condition is tan⁡x≠−1\tan x \ne -1⁠. So

    x≠nπ−π4,n an integer.x \ne n\pi - \frac{\pi}{4}, \quad n \text{ an integer}.
    The answer will be valid on each interval between consecutive excluded points, for example (−π4,3π4)\left(-\frac{\pi}{4}, \frac{3\pi}{4}\right)⁠. On such an interval cos⁡x+sin⁡x\cos x + \sin x keeps one sign, and the cancellation in Step 2 is legitimate.

  2. Factorise cos⁡2x\cos 2x and cancel

    cos⁡2x=cos⁡2x−sin⁡2x=(cos⁡x−sin⁡x)(cos⁡x+sin⁡x).\begin{aligned} \cos 2x &= \cos^2 x - \sin^2 x \\ &= (\cos x - \sin x)(\cos x + \sin x). \end{aligned}

    So, for cos⁡x+sin⁡x≠0\cos x + \sin x \ne 0⁠,

    f(x)=(cos⁡x−sin⁡x)(cos⁡x+sin⁡x)(cos⁡x+sin⁡x)2=cos⁡x−sin⁡xcos⁡x+sin⁡x.\begin{aligned} f(x) &= \frac{(\cos x - \sin x)(\cos x + \sin x)}{(\cos x + \sin x)^2} \\ &= \frac{\cos x - \sin x}{\cos x + \sin x}. \end{aligned}
    The denominator is a power of cos⁡x+sin⁡x\cos x + \sin x⁠, so we want the numerator in terms of the same expression. The form cos⁡2x−sin⁡2x\cos^2 x - \sin^2 x of cos⁡2x\cos 2x is a difference of squares that contains exactly that factor.
    cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = \cos^2 x - \sin^2 x⁠; A2−B2=(A−B)(A+B)A^2 - B^2 = (A - B)(A + B)⁠. Basics: trigonometric identities for integrals

  3. Substitute t=cos⁡x+sin⁡xt = \cos x + \sin x

    Let t=cos⁡x+sin⁡xt = \cos x + \sin x⁠. Then

    dt=(−sin⁡x+cos⁡x) dx=(cos⁡x−sin⁡x) dx.dt = (-\sin x + \cos x)\,dx = (\cos x - \sin x)\,dx.

    This is the numerator times dxdx⁠, so

    ∫cos⁡x−sin⁡xcos⁡x+sin⁡x dx=∫dtt.\int\frac{\cos x - \sin x}{\cos x + \sin x}\,dx = \int\frac{dt}{t}.
    The numerator is exactly the derivative of the denominator, the pattern g′(x)g(x)\frac{g'(x)}{g(x)}⁠. The substitution t=g(x)t = g(x) turns it into dtt\frac{dt}{t}⁠.
    ddxcos⁡x=−sin⁡x\frac{d}{dx}\cos x = -\sin x⁠, ddxsin⁡x=cos⁡x\frac{d}{dx}\sin x = \cos x⁠; change of variable, Section 7.3.1. Lesson: integration by substitution

  4. Integrate and return to xx

    ∫dtt=log⁡∣t∣+C=log⁡∣cos⁡x+sin⁡x∣+C.\begin{aligned} \int\frac{dt}{t} &= \log|t| + C \\ &= \log|\cos x + \sin x| + C. \end{aligned}
    Formula (xii) applies since t≠0t \ne 0 on our domain. Keep the modulus: cos⁡x+sin⁡x\cos x + \sin x is negative on some intervals, for example near x=πx = \pi⁠.
    Standard formula (xii), Section 7.2. Basics: logarithms and 1/x

  5. Check: differentiate the answer back

    With u=cos⁡x+sin⁡xu = \cos x + \sin x and ddxlog⁡∣u∣=u′u\frac{d}{dx}\log|u| = \frac{u'}{u}⁠:

    ddxlog⁡∣cos⁡x+sin⁡x∣=cos⁡x−sin⁡xcos⁡x+sin⁡x=(cos⁡x−sin⁡x)(cos⁡x+sin⁡x)(cos⁡x+sin⁡x)2=cos⁡2x(cos⁡x+sin⁡x)2.\begin{aligned} \frac{d}{dx}\log|\cos x + \sin x| &= \frac{\cos x - \sin x}{\cos x + \sin x} \\ &= \frac{(\cos x - \sin x)(\cos x + \sin x)}{(\cos x + \sin x)^2} \\ &= \frac{\cos 2x}{(\cos x + \sin x)^2}. \end{aligned}

    The second line multiplies top and bottom by cos⁡x+sin⁡x≠0\cos x + \sin x \ne 0⁠. This is the integrand, so the answer is correct on every interval where cos⁡x+sin⁡x≠0\cos x + \sin x \ne 0⁠.

    Chain rule (Theorem 4, Section 5.3.1); Property (I), Section 7.2.1. Basics: integration undoes differentiation
Result ∫cos⁡2x(cos⁡x+sin⁡x)2 dx=log⁡∣cos⁡x+sin⁡x∣+C,\int\frac{\cos 2x}{(\cos x + \sin x)^2}\,dx = \log|\cos x + \sin x| + C,

valid on any interval on which cos⁡x+sin⁡x≠0\cos x + \sin x \ne 0⁠. (Textbook answer: log⁡∣cos⁡x+sin⁡x∣+C\log|\cos x + \sin x| + C⁠.)

Check it with numbers

Let F(x)=log⁡∣cos⁡x+sin⁡x∣F(x) = \log|\cos x + \sin x| and look near x=0.5x = 0.5 with h=0.0001h = 0.0001⁠:

  • F(0.5001)≈0.305311685618F(0.5001) \approx 0.305311685618 and F(0.4999)≈0.305253004019F(0.4999) \approx 0.305253004019⁠.
  • Slope estimate: F(0.5001)−F(0.4999)0.0002≈0.293408\dfrac{F(0.5001) - F(0.4999)}{0.0002} \approx 0.293408⁠.
  • Integrand at x=0.5x = 0.5⁠: cos⁡1(cos⁡0.5+sin⁡0.5)2≈0.293408\dfrac{\cos 1}{(\cos 0.5 + \sin 0.5)^2} \approx 0.293408⁠.

They agree.

Common mistakes
  1. Expanding the denominator instead. (cos⁡x+sin⁡x)2=1+sin⁡2x(\cos x + \sin x)^2 = 1 + \sin 2x is true (it is the key to Question 14), but here it hides the cancellation.
  2. Writing ∫dtt2\int\frac{dt}{t^2}⁠. After cancelling one factor only the first power of tt remains in the denominator; the answer is a logarithm, not −1t-\frac1t⁠.
  3. Dropping the modulus in log⁡∣cos⁡x+sin⁡x∣\log|\cos x + \sin x|⁠.
What to take away
  1. cos⁡2x=(cos⁡x−sin⁡x)(cos⁡x+sin⁡x)\cos 2x = (\cos x - \sin x)(\cos x + \sin x)⁠: use it whenever cos⁡x±sin⁡x\cos x \pm \sin x appears in a denominator.
  2. ∫g′(x)g(x) dx=log⁡∣g(x)∣+C\int\frac{g'(x)}{g(x)}\,dx = \log|g(x)| + C⁠.
  3. Compare Question 14: the same square in the denominator, but there the numerator was only cos⁡x−sin⁡x\cos x - \sin x⁠, so a power of tt remained.

Every fact used, and where it comes from

FactUsed inSource
cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = \cos^2 x - \sin^2 xSteps 2, 5Class XI trigonometry (Basics: trigonometric identities)
A2−B2=(A−B)(A+B)A^2 - B^2 = (A - B)(A + B)Step 2Algebra
ddxsin⁡x=cos⁡x\frac{d}{dx}\sin x = \cos x⁠, ddxcos⁡x=−sin⁡x\frac{d}{dx}\cos x = -\sin xSteps 3, 5List in Section 7.2, (ii) and (iii)
Change of variable (substitution)Step 3Section 7.3.1 (Lesson: integration by substitution)
∫dtt=log⁡∣t∣+C\int\frac{dt}{t} = \log|t| + CStep 4Standard formula (xii), Section 7.2
Chain ruleStep 5Theorem 4, Section 5.3.1
Differentiation undoes integrationStep 5Property (I), Section 7.2.1
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