Exercise 7.3 · Q22: $\int\frac{dx}{\cos(x-a)\cos(x-b)}$

Basics you need

Solution

1. Where is the integrand defined, and what we assume

   We need $\cos(x - a) \ne 0$ and $\cos(x - b) \ne 0$⁠, that is, $x \ne a + \frac{\pi}{2} + n\pi$ and $x \ne b + \frac{\pi}{2} + n\pi$ ($n$ an integer). We work on an interval avoiding all these points.

   The method divides by $\sin(a - b)$⁠, so in Steps 2 to 5 we assume

   $$\sin(a - b) \ne 0, \quad\text{that is,}\quad a - b \ne n\pi.$$

   The remaining case $a - b = n\pi$ is treated in Step 6.

   The textbook's answer has $\sin(a - b)$ in a denominator, so it silently assumes this condition. Stating it, and covering the other case, makes the solution complete.

2. Write $\sin(a - b)$ using the two angles

   Since $(x - b) - (x - a) = a - b$⁠,

   $$\begin{aligned} \sin(a - b) &= \sin\big[(x - b) - (x - a)\big] \\ &= \sin(x - b)\cos(x - a) \\ &\quad - \cos(x - b)\sin(x - a). \end{aligned}$$
   The denominator contains $\cos(x - a)$ and $\cos(x - b)$⁠. Expanding the sine of the difference of these two angles produces exactly those cosines in each term, which is what will let us cancel. Although $x$ appears inside, the left side is the constant $\sin(a - b)$⁠.
   $\sin(A - B) = \sin A\cos B - \cos A\sin B$ with $A = x - b$⁠, $B = x - a$⁠. Basics: trigonometric identities for integrals

3. Multiply and divide by $\sin(a - b)$⁠, then split

   Write $k = \sin(a - b)$⁠, a non-zero constant. Then

   $$f(x) = \frac1k\cdot\frac{\sin(a - b)}{\cos(x - a)\cos(x - b)}.$$

   Replace the top $\sin(a - b)$ by its expansion from Step 2 and split:

   $$\begin{aligned} \frac{\sin(a - b)}{\cos(x - a)\cos(x - b)} &= \frac{\sin(x - b)\cos(x - a)}{\cos(x - a)\cos(x - b)} \\ &\quad - \frac{\cos(x - b)\sin(x - a)}{\cos(x - a)\cos(x - b)} \\ &= \frac{\sin(x - b)}{\cos(x - b)} - \frac{\sin(x - a)}{\cos(x - a)} \\ &= \tan(x - b) - \tan(x - a). \end{aligned}$$

   So

   $$f(x) = \frac{1}{\sin(a - b)}\big[\tan(x - b) - \tan(x - a)\big].$$
   Multiplying by $\frac{\sin(a - b)}{\sin(a - b)} = 1$ changes nothing (allowed because $\sin(a - b) \ne 0$⁠). After the expansion, $\cos(x - a)$ cancels in the first fraction and $\cos(x - b)$ cancels in the second, which is valid because both are non-zero on our interval.
   $\frac{A - B}{D} = \frac AD - \frac BD$⁠; $\tan\theta = \frac{\sin\theta}{\cos\theta}$⁠. Basics: splitting fractions

4. Integrate the two tangents

   By Section 7.3.1 (i), with the substitution $u = x - b$ (so $du = dx$⁠), $\int\tan(x - b)\,dx = -\log|\cos(x - b)|$⁠, and likewise for $x - a$⁠. Hence

   $$\begin{aligned} \int f(x)\,dx &= \frac{1}{\sin(a - b)}\Big[-\log|\cos(x - b)| \\ &\qquad + \log|\cos(x - a)|\Big] + C \\ &= \frac{1}{\sin(a - b)}\log\left|\frac{\cos(x - a)}{\cos(x - b)}\right| + C. \end{aligned}$$
   $\frac{1}{\sin(a - b)}$ is a constant, so it stays outside (Property (IV)), and the difference is integrated term by term (Property (V)). The minus sign in front of $\tan(x - a)$ turns $-\log|\cos(x - a)|$ into $+\log|\cos(x - a)|$⁠. Finally the two logarithms combine into the logarithm of a quotient.
   $\int\tan x\,dx = \log|\sec x| + C$⁠, Section 7.3.1; Properties (IV), (V), Section 7.2.1. Lesson: integration by substitution

5. Check: differentiate the answer back

   By the chain rule,

   $$\begin{aligned} \frac{d}{dx}\log|\cos(x - a)| &= \frac{-\sin(x - a)}{\cos(x - a)} \\ &= -\tan(x - a), \end{aligned}$$

   and similarly for $b$⁠. Let $F(x)$ be our answer without $C$⁠, and again $k = \sin(a - b)$⁠. Then

   $$\begin{aligned} F'(x) &= \frac1k\big[-\tan(x - a) + \tan(x - b)\big] \\ &= \frac1k\cdot\frac{\sin(a - b)}{\cos(x - a)\cos(x - b)} \\ &= \frac{1}{\cos(x - a)\cos(x - b)}. \end{aligned}$$

   The second line is the split of Step 3 read backwards (put the two tangents over the common denominator $\cos(x - a)\cos(x - b)$ and use Step 2 for the numerator). This is the integrand, so the answer is correct whenever $\sin(a - b) \ne 0$⁠.

   Chain rule (Theorem 4, Section 5.3.1); Property (I), Section 7.2.1. Basics: integration undoes differentiation

6. The excluded case $a - b = n\pi$

   If $a - b = n\pi$⁠, then $x - b = (x - a) + n\pi$⁠, so $\cos(x - b) = (-1)^n\cos(x - a)$⁠. The integrand becomes $(-1)^n\sec^2(x - a)$⁠, and

   $$\int\frac{dx}{\cos(x - a)\cos(x - b)} = (-1)^n\tan(x - a) + C.$$
   Here the two cosines are the same function up to sign, so no splitting is needed: standard formula (iv) with $u = x - a$ does it. This case is not covered by the textbook's formula, which would have $\sin(a - b) = 0$ in a denominator.
   $\cos(\theta + n\pi) = (-1)^n\cos\theta$⁠; standard formula (iv), Section 7.2.

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