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Chapter 7 · Integrals

Exercise 7.3 · Q9: ∫cos⁡x1+cos⁡x dx\int\frac{\cos x}{1+\cos x}\,dx

Exercise 7.3Integration using trigonometric identities · Section 7.3.2
The question

Find the integral of the function

cos⁡x1+cos⁡x.\frac{\cos x}{1 + \cos x}.
Exam answer

Let I=∫cos⁡x1+cos⁡x dxI = \displaystyle\int\frac{\cos x}{1 + \cos x}\,dx⁠.

I=∫(1+cos⁡x)−11+cos⁡x dx=∫(1−11+cos⁡x)dx\begin{aligned} I &= \int\frac{(1 + \cos x) - 1}{1 + \cos x}\,dx \\ &= \int\left(1 - \frac{1}{1 + \cos x}\right)dx \end{aligned}

Using 1+cos⁡x=2cos⁡2x21 + \cos x = 2\cos^2\frac{x}{2}⁠,

I=∫(1−12sec⁡2x2)dx=x−12⋅2tan⁡x2+C\begin{aligned} I &= \int\left(1 - \frac12\sec^2\frac{x}{2}\right)dx \\ &= x - \frac12\cdot 2\tan\frac{x}{2} + C \end{aligned}

∴ I=x−tan⁡x2+C\therefore\ I = x - \tan\dfrac{x}{2} + C⁠.

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

The one idea this problem is built on

Make the numerator look like the denominator: cos⁡x=(1+cos⁡x)−1\cos x = (1 + \cos x) - 1⁠. Then the fraction splits into 1−11+cos⁡x1 - \frac{1}{1+\cos x}⁠, and the half-angle identity 1+cos⁡x=2cos⁡2x21 + \cos x = 2\cos^2\frac{x}{2} turns 11+cos⁡x\frac{1}{1+\cos x} into 12sec⁡2x2\frac12\sec^2\frac{x}{2}⁠, which integrates to a tangent.

Given, to find, and the plan
Given
cos⁡x1+cos⁡x\dfrac{\cos x}{1+\cos x}⁠, defined where cos⁡x≠−1\cos x \ne -1⁠, that is, x≠(2k+1)πx \ne (2k+1)\pi for integers kk⁠.
To find
∫cos⁡x1+cos⁡x dx\displaystyle\int \frac{\cos x}{1+\cos x}\,dx⁠.
Plan
(1) Split off 1 from the fraction. (2) Rewrite 11+cos⁡x\frac{1}{1+\cos x} with the half-angle identity. (3) Integrate. (4) Differentiate back to check and state where the answer is valid.

Basics you need

1Adding and subtracting to split a fraction

ND=D−(D−N)D=1−D−ND\frac{N}{D} = \frac{D - (D - N)}{D} = 1 - \frac{D - N}{D}⁠. Here D−N=1D - N = 1⁠. Basics: splitting fractions

2Half-angle form

1+cos⁡x=2cos⁡2x21 + \cos x = 2\cos^2\frac{x}{2} (the double-angle formula cos⁡2θ=2cos⁡2θ−1\cos 2\theta = 2\cos^2\theta - 1 with θ=x2\theta = \frac{x}{2}⁠). Basics: trigonometric identities for integrals

3The integral of sec⁡2\sec^2

∫sec⁡2x dx=tan⁡x+C\int\sec^2 x\,dx = \tan x + C (Section 7.2); with t=x2t = \frac{x}{2}⁠, ∫sec⁡2x2 dx=2tan⁡x2+C\int\sec^2\frac{x}{2}\,dx = 2\tan\frac{x}{2} + C⁠. 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. Split off 1

    For cos⁡x≠−1\cos x \ne -1⁠:

    cos⁡x1+cos⁡x=(1+cos⁡x)−11+cos⁡x=1−11+cos⁡x.\frac{\cos x}{1 + \cos x} = \frac{(1 + \cos x) - 1}{1 + \cos x} = 1 - \frac{1}{1 + \cos x}.
    Writing the numerator as "denominator minus something" produces a whole part (here 1) and a simpler fraction. This is the same move as dividing polynomials, and it often removes the variable from the numerator.
    Splitting a fraction. Basics: splitting fractions

  2. Rewrite the remaining fraction with half-angles

    11+cos⁡x=12cos⁡2x2=12sec⁡2x2.\frac{1}{1 + \cos x} = \frac{1}{2\cos^2\frac{x}{2}} = \frac12\sec^2\frac{x}{2}.

    So the integrand is 1−12sec⁡2x21 - \frac12\sec^2\frac{x}{2}⁠.

    11+cos⁡x\frac{1}{1+\cos x} has no standard integral as it stands, but sec⁡2\sec^2 does. The condition cos⁡x2≠0\cos\frac{x}{2} \ne 0 is the same as 1+cos⁡x≠01 + \cos x \ne 0⁠, so the rewrite is valid at every point where the integrand is defined.
    1+cos⁡x=2cos⁡2x21 + \cos x = 2\cos^2\frac{x}{2}⁠; sec⁡θ=1cos⁡θ\sec\theta = \frac{1}{\cos\theta}⁠. Basics: trigonometric identities

  3. Integrate

    With t=x2t = \frac{x}{2}⁠, dx=2 dtdx = 2\,dt⁠: ∫sec⁡2x2 dx=2tan⁡x2\int\sec^2\frac{x}{2}\,dx = 2\tan\frac{x}{2}⁠. Hence

    ∫cos⁡x1+cos⁡x dx=∫dx−12∫sec⁡2x2 dx=x−12⋅2tan⁡x2+C=x−tan⁡x2+C.\begin{aligned} \int \frac{\cos x}{1 + \cos x}\,dx &= \int dx - \frac12\int\sec^2\frac{x}{2}\,dx \\ &= x - \frac12\cdot 2\tan\frac{x}{2} + C \\ &= x - \tan\frac{x}{2} + C. \end{aligned}
    The 12\frac12 from the identity and the 2 from the substitution cancel, leaving a clean tan⁡x2\tan\frac{x}{2}⁠.
    Properties (III), (IV), Section 7.2.1; ∫sec⁡2x dx=tan⁡x+C\int\sec^2 x\,dx = \tan x + C (Section 7.2); substitution, Section 7.3.1. Lesson: integration by substitution

  4. Check: differentiate the answer back

    With F(x)=x−tan⁡x2F(x) = x - \tan\frac{x}{2} and the chain rule:

    F′(x)=1−12sec⁡2x2=1−12cos⁡2x2=1−11+cos⁡x=cos⁡x1+cos⁡x.\begin{aligned} F'(x) &= 1 - \frac12\sec^2\frac{x}{2} = 1 - \frac{1}{2\cos^2\frac{x}{2}} \\ &= 1 - \frac{1}{1 + \cos x} = \frac{\cos x}{1 + \cos x}. \end{aligned}

    The answer is valid on every interval ((2k−1)π, (2k+1)π)\big((2k-1)\pi,\ (2k+1)\pi\big)⁠, that is, wherever cos⁡x≠−1\cos x \ne -1⁠.

    tan⁡x2\tan\frac{x}{2} is undefined exactly at the odd multiples of π\pi⁠, the same points the integrand excludes.
    Chain rule, Theorem 4, Section 5.3.1. Basics: integration undoes differentiation
Result ∫cos⁡x1+cos⁡x dx=x−tan⁡x2+C,\int \frac{\cos x}{1 + \cos x}\,dx = x - \tan\frac{x}{2} + C,

valid wherever cos⁡x≠−1\cos x \ne -1⁠, i.e. x≠(2k+1)πx \ne (2k+1)\pi⁠. (Textbook answer: x−tan⁡x2+Cx - \tan\dfrac{x}{2} + C⁠.)

Check it with numbers

At x=1x = 1 with h=0.001h = 0.001⁠: F(1.001)≈0.4540481095F(1.001) \approx 0.4540481095 and F(0.999)≈0.4533465561F(0.999) \approx 0.4533465561⁠, so the slope is about 0.00070155340.002≈0.3507767\dfrac{0.0007015534}{0.002} \approx 0.3507767⁠. The integrand is cos⁡11+cos⁡1\dfrac{\cos 1}{1 + \cos 1} ≈0.54030231.5403023\approx \dfrac{0.5403023}{1.5403023} ≈0.3507768\approx 0.3507768⁠. They agree to six significant figures.

A cross-check with Question 8: adding twice this integrand to Question 8's integrand gives

1−cos⁡x1+cos⁡x+2cos⁡x1+cos⁡x=1+cos⁡x1+cos⁡x=1,\frac{1 - \cos x}{1 + \cos x} + \frac{2\cos x}{1 + \cos x} = \frac{1 + \cos x}{1 + \cos x} = 1,

so the two answers must add up to xx plus a constant. They do: (2tan⁡x2−x)\big(2\tan\frac{x}{2} - x\big) + 2(x−tan⁡x2)=x+\, 2\big(x - \tan\frac{x}{2}\big) = x⁠.

Common mistakes
  1. Splitting the fraction the wrong way. cos⁡x1+cos⁡x≠cos⁡x1+cos⁡xcos⁡x\frac{\cos x}{1 + \cos x} \ne \frac{\cos x}{1} + \frac{\cos x}{\cos x}⁠: only a numerator that is a sum can be split.
  2. Using 1+cos⁡x=2sin⁡2x21 + \cos x = 2\sin^2\frac{x}{2}⁠. That is 1−cos⁡x1 - \cos x⁠; with the plus sign it is 2cos⁡2x22\cos^2\frac{x}{2}⁠.
  3. Losing the factor bookkeeping: 12\frac12 from the identity times 2 from t=x2t = \frac{x}{2} gives 1.
What to take away
  1. Numerator close to the denominator: subtract and add to split off a whole part.
  2. 11+cos⁡x=12sec⁡2x2\frac{1}{1 + \cos x} = \frac12\sec^2\frac{x}{2} and 11−cos⁡x=12cosec⁡2x2\frac{1}{1 - \cos x} = \frac12\operatorname{cosec}^2\frac{x}{2} are worth remembering.
  3. Questions 8, 9 and 12 all hinge on half-angle forms of 1±cos⁡x1 \pm \cos x⁠.

Every fact used, and where it comes from

FactUsed inSource
ND=1−D−ND\frac{N}{D} = 1 - \frac{D - N}{D}Step 1Algebra of fractions
1+cos⁡x=2cos⁡2x21 + \cos x = 2\cos^2\frac{x}{2}Steps 2, 4Double-angle formula with θ=x2\theta = \frac{x}{2} (Class XI; Example 7(i), Section 7.3.2)
∫sec⁡2x dx=tan⁡x+C\int\sec^2 x\,dx = \tan x + CStep 3Standard formulae, Section 7.2
Substitution t=x2t = \frac{x}{2}Step 3Section 7.3.1
Chain rule; differentiation undoes integrationStep 4Theorem 4, Section 5.3.1; Property (I), Section 7.2.1
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