Miscellaneous · Q7: $\int\frac{\sin x}{\sin(x-a)}\,dx$

Basics you need

Solution

1. Why this method

   The top and bottom involve the different angles $x$ and $x - a$⁠, so nothing cancels as it stands. Expressing $x$ as $(x - a) + a$ puts everything in terms of the one angle $x - a$ (and the constant $a$⁠).

   No standard formula has $\frac{\sin x}{\sin(x-a)}$⁠, and $\sin x$ is not the derivative of $\sin(x-a)$⁠. But once both are written in the same angle, the fraction separates into simple terms. Example 6(ii) of the book does exactly this for $\frac{\sin x}{\sin(x+a)}$⁠.
   Example 6(ii), Section 7.3.1 (Example 6).

2. Substitute $t = x - a$

   Then $x = t + a$ and $dx = dt$ (since $a$ is constant):

   $$\int\frac{\sin x}{\sin(x-a)}\,dx = \int\frac{\sin(t + a)}{\sin t}\,dt.$$
   The shift $t = x - a$ is one-one and differentiable for all $x$⁠, and $\sin t \ne 0$ exactly where the original integrand is defined.
   Substitution, Section 7.3.1. Basics: substitution

3. Expand and split

   $$\begin{aligned} \frac{\sin(t+a)}{\sin t} &= \frac{\sin t\cos a + \cos t\sin a}{\sin t} \\ &= \cos a + \sin a\,\frac{\cos t}{\sin t} \\ &= \cos a + \sin a\cot t. \end{aligned}$$
   The addition formula with $A = t$⁠, $B = a$⁠. Dividing each term of the numerator by $\sin t$ splits the fraction; $\cos a$ and $\sin a$ are constants because $a$ is.
   Addition formula (Basic 1). Basics: trigonometric identities

4. Integrate

   $$\begin{aligned} &\int(\cos a + \sin a\cot t)\,dt \\ &= t\cos a + \sin a\log|\sin t| + C_1. \end{aligned}$$
   A constant integrates to itself times $t$⁠; $\int\cot t\,dt = \log|\sin t|$ (Basic 2). The constant $\sin a$ comes outside the integral by property (IV).
   Standard formula (i) with $n = 0$⁠, Section 7.2; $\int\cot t\,dt$⁠, Section 7.3.1.

5. Return to $x$

   Put $t = x - a$⁠:

   $$\begin{aligned} &(x - a)\cos a + \sin a\log|\sin(x-a)| + C_1 \\ &= x\cos a + \sin a\log|\sin(x-a)| + C, \end{aligned}$$

   where $C = C_1 - a\cos a$⁠.

   The term $-a\cos a$ is a constant, so it is absorbed into the constant of integration. That is why the textbook's answer has $x\cos a$ rather than $(x - a)\cos a$⁠; the two differ only by a constant.
   Any constant may be absorbed into $C$ (Section 7.2).

6. Check: differentiate the answer

   $$\begin{aligned} &\frac{d}{dx}\big(x\cos a + \sin a\log|\sin(x-a)|\big) \\ &= \cos a + \sin a\cdot\frac{\cos(x-a)}{\sin(x-a)} \\ &= \frac{\sin(x-a)\cos a + \cos(x-a)\sin a}{\sin(x-a)} \\ &= \frac{\sin\big((x-a) + a\big)}{\sin(x-a)} = \frac{\sin x}{\sin(x-a)}. \ \checkmark \end{aligned}$$
   The derivative of $\log|\sin u|$ is $\cot u$ times $u'$⁠, with $u = x - a$⁠, $u' = 1$⁠. Recombining with the addition formula returns the integrand on every interval where $\sin(x - a) \ne 0$⁠.
   Chain rule, Chapter 5; addition formula. Basics: integration undoes differentiation

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