Exercise 7.3 · Q8: $\int\frac{1-\cos x}{1+\cos x}\,dx$

Basics you need

Solution

1. Where the integrand is defined

   The denominator vanishes when $\cos x = -1$⁠, that is, at $x = \pm\pi, \pm 3\pi, \dots$⁠, the odd multiples of $\pi$⁠. We work on any interval between two of them, such as $-\pi < x < \pi$⁠.

   An antiderivative is only promised on an interval where the integrand is defined. Here the integrand also blows up near those points (the top tends to 2, the bottom to 0), so the answer will have the same break points.

2. Rewrite with half-angles

   $$\begin{aligned} \frac{1 - \cos x}{1 + \cos x} &= \frac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}} \\ &= \tan^2\frac{x}{2}. \end{aligned}$$
   Both identities come from the double-angle formula with $\theta = \frac{x}{2}$ (Basic 1). The 2s cancel and $\frac{\sin}{\cos} = \tan$⁠. The condition $\cos\frac{x}{2} \ne 0$ is the same as $1 + \cos x \ne 0$⁠, because $1 + \cos x = 2\cos^2\frac{x}{2}$⁠; so this step loses no points of the domain and adds none.
   $\cos 2\theta = 1 - 2\sin^2\theta = 2\cos^2\theta - 1$ (Class XI; Example 7(i), Section 7.3.2). Basics: trigonometric identities

3. Turn $\tan^2$ into $\sec^2 - 1$

   $$\tan^2\frac{x}{2} = \sec^2\frac{x}{2} - 1.$$
   There is no standard formula for $\tan^2$⁠, but $\sec^2$ is exactly the derivative of $\tan$⁠. This is the same trick as in Exercise 7.2 Q21 ($\tan^2(2x-3)$⁠).
   $1 + \tan^2\theta = \sec^2\theta$ (Pythagorean identity divided by $\cos^2\theta$⁠).

4. Integrate

   For $\int\sec^2\frac{x}{2}\,dx$ put $t = \frac{x}{2}$⁠, $dx = 2\,dt$⁠: $\int\sec^2 t\cdot 2\,dt = 2\tan t = 2\tan\frac{x}{2}$⁠. Hence

   $$\begin{aligned} \int \frac{1 - \cos x}{1 + \cos x}\,dx &= \int \Big(\sec^2\frac{x}{2} - 1\Big)dx \\ &= 2\tan\frac{x}{2} - x + C. \end{aligned}$$
   Since $t = \frac{x}{2}$ changes half as fast as $x$⁠, we multiply by 2 (dividing by the coefficient $\frac12$⁠). A common slip is to write $\frac12\tan\frac{x}{2}$⁠.
   $\int\sec^2 x\,dx = \tan x + C$ (Section 7.2); substitution, Section 7.3.1. Lesson: integration by substitution

5. Check: differentiate the answer back

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

   $$\begin{aligned} F'(x) &= 2\cdot\sec^2\frac{x}{2}\cdot\frac12 - 1 = \sec^2\frac{x}{2} - 1 \\ &= \tan^2\frac{x}{2} = \frac{1 - \cos x}{1 + \cos x}. \end{aligned}$$

   The answer is valid on every interval $\big((2k-1)\pi,\ (2k+1)\pi\big)$⁠, that is, wherever $\cos x \ne -1$⁠. (Strictly, the constant may be chosen separately on each such interval.)

   $\tan\frac{x}{2}$ is undefined exactly where $\frac{x}{2}$ is an odd multiple of $\frac{\pi}{2}$⁠, that is, where $x$ is an odd multiple of $\pi$⁠: the same points the integrand excludes.
   Chain rule, Theorem 4, Section 5.3.1; $\frac{d}{dx}\tan x = \sec^2 x$⁠. Basics: integration undoes differentiation

6. A second route gives the same answer

   Multiply top and bottom by $1 - \cos x$ (allowed where also $\cos x \ne 1$⁠, i.e. $\sin x \ne 0$⁠):

   $$\begin{aligned} \frac{(1-\cos x)^2}{1 - \cos^2 x} &= \frac{1 - 2\cos x + \cos^2 x}{\sin^2 x} \\ &= \operatorname{cosec}^2 x - 2\operatorname{cosec} x\cot x + \cot^2 x \\ &= 2\operatorname{cosec}^2 x - 2\operatorname{cosec} x\cot x - 1, \end{aligned}$$

   using $\cot^2 x = \operatorname{cosec}^2 x - 1$⁠. The standard formulae give $\int = -2\cot x + 2\operatorname{cosec} x - x + C$⁠. And

   $$\begin{aligned} \operatorname{cosec} x - \cot x &= \frac{1 - \cos x}{\sin x} \\ &= \frac{2\sin^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}} = \tan\frac{x}{2}, \end{aligned}$$

   so this is $2\tan\frac{x}{2} - x + C$ again.

   Getting the same result two ways is strong evidence that it is right. The half-angle route is shorter and does not need the extra condition $\sin x \ne 0$⁠.
   $\int\operatorname{cosec}^2 x\,dx = -\cot x + C$⁠, $\int\operatorname{cosec} x\cot x\,dx = -\operatorname{cosec} x + C$ (standard formulae, Section 7.2).

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