Exercise 7.2 · Q32: $\int\frac{1}{1+\cot x}\,dx$

Basics you need

Solution

1. Where is the integrand defined?

   Two conditions:

   • $\cot x$ must exist: $\sin x \ne 0$⁠, that is, $x \ne n\pi$⁠.
   • The denominator $1 + \cot x$ must not be zero. Multiplying by $\sin x \ne 0$⁠, $1 + \cot x = 0$ means $\sin x + \cos x = 0$⁠, that is, $\tan x = -1$⁠, that is, $x = -\frac{\pi}{4} + n\pi$⁠.

   So $f$ is defined for all $x$ except $x = n\pi$ and $x = -\frac{\pi}{4} + n\pi$ ($n$ an integer).

   These excluded points tell us where the answer cannot be claimed (Note, Section 7.2). In particular $\sin x + \cos x \ne 0$ on the domain, which we need for the logarithm in Step 4.
   $\tan x = -1$ exactly when $x = -\frac{\pi}{4} + n\pi$⁠. Basics: trigonometric facts

2. Rewrite without the cotangent

   Multiply top and bottom by $\sin x$ (non-zero on the domain):

   $$\begin{aligned} \frac{1}{1 + \cot x} &= \frac{\sin x}{\sin x + \sin x\cot x} \\ &= \frac{\sin x}{\sin x + \cos x}. \end{aligned}$$
   Working with $\sin$ and $\cos$ only makes the structure visible: the denominator $\sin x + \cos x$ has a simple derivative, $\cos x - \sin x$⁠. Multiplying by $\sin x$ is legitimate only because $\sin x \ne 0$ on the domain.
   $\sin x\cdot\cot x = \cos x$⁠. Basics: trigonometric facts

3. Split the numerator into "denominator" and "derivative of denominator"

   Let $D = \sin x + \cos x$⁠, so $D' = \cos x - \sin x$⁠. Then

   $$\begin{aligned} \tfrac12 D - \tfrac12 D' &= \tfrac12(\sin x + \cos x) \\ &\quad - \tfrac12(\cos x - \sin x) \\ &= \sin x. \end{aligned}$$

   Hence

   $$\begin{aligned} \frac{\sin x}{\sin x + \cos x} &= \frac{\frac12 D - \frac12 D'}{D} \\ &= \tfrac12 - \tfrac12\cdot\frac{\cos x - \sin x}{\sin x + \cos x}. \end{aligned}$$
   Why this split? A fraction $\frac{D}{D}$ is just 1, and $\frac{D'}{D}$ integrates to $\log|D|$⁠. Writing the numerator as a combination $aD + bD'$ turns a hard fraction into these two easy ones. To find $a, b$⁠: $a(\sin x + \cos x) + b(\cos x - \sin x) = \sin x$ needs $a - b = 1$ (coefficients of $\sin x$⁠) and $a + b = 0$ (coefficients of $\cos x$⁠), so $a = \frac12$⁠, $b = -\frac12$⁠.
   Same trick as Example 6(iii), Section 7.3.1. Basics: splitting fractions

4. Integrate each piece

   First piece: $\int\frac12\,dx = \frac{x}{2}$⁠.

   Second piece: put $t = \sin x + \cos x$⁠, so $dt = (\cos x - \sin x)\,dx$⁠:

   $$\int\frac{\cos x - \sin x}{\sin x + \cos x}\,dx = \int\frac{dt}{t} = \log|t|.$$

   Together (Property (III) of Section 7.2.1, integral of a sum):

   $$\int\frac{dx}{1 + \cot x} = \frac{x}{2} - \tfrac12\log|\sin x + \cos x| + C.$$
   The substitution works because the numerator $\cos x - \sin x$ is exactly the derivative of $t = \sin x + \cos x$ (Section 7.3.1). The modulus is needed: $\sin x + \cos x$ is negative on parts of the domain (for example, at $x = 2.5$ it is $-0.2027\ldots$⁠).
   Standard formula (xii), Section 7.2: $\int\frac{dt}{t} = \log|t| + C$⁠. Lesson: integration by substitution

5. Check: differentiate the answer back

   Let $F(x) = \frac{x}{2} - \frac12\log|\sin x + \cos x|$⁠. Using $\frac{d}{dx}\log|g| = \frac{g'}{g}$⁠:

   $$\begin{aligned} F'(x) &= \tfrac12 - \tfrac12\cdot\frac{\cos x - \sin x}{\sin x + \cos x} \\ &= \frac{(\sin x + \cos x) - (\cos x - \sin x)}{2(\sin x + \cos x)} \\ &= \frac{2\sin x}{2(\sin x + \cos x)} = \frac{\sin x}{\sin x + \cos x}. \end{aligned}$$

   Dividing top and bottom by $\sin x$ gives $\frac{1}{1 + \cot x}$⁠, the integrand. ✓

   Differentiating back is the test of an antiderivative (Property (I), Section 7.2.1). The last step reverses Step 2.
   Basics: integration undoes differentiation

Every fact used, and where it comes from