Exercise 7.2 · Q24: $\int\frac{2\cos x-3\sin x}{6\cos x+4\sin x}\,dx$

Basics you need

Solution

1. Take the common factor out of the denominator

   $$\frac{2\cos x - 3\sin x}{6\cos x + 4\sin x} = \frac12\cdot\frac{2\cos x - 3\sin x}{3\cos x + 2\sin x}.$$
   Both coefficients of the denominator, 6 and 4, are even. Removing the 2 leaves a denominator whose derivative we can compare directly with the numerator.

2. Differentiate the new denominator

   $$\begin{aligned} \frac{d}{dx}\left(3\cos x + 2\sin x\right) &= -3\sin x + 2\cos x \\ &= 2\cos x - 3\sin x. \end{aligned}$$

   This is exactly the numerator.

   $\frac{d}{dx}\cos x = -\sin x$⁠, $\frac{d}{dx}\sin x = \cos x$ (Chapter 5, Table 5.3). Basics: integration undoes differentiation

3. Substitute $t = 3\cos x + 2\sin x$

   Let $t = 3\cos x + 2\sin x$⁠. By Step 2, $dt = (2\cos x - 3\sin x)\,dx$⁠. Hence

   $$\int\frac{2\cos x - 3\sin x}{6\cos x + 4\sin x}\,dx = \frac12\int\frac{dt}{t}.$$
   The numerator times $dx$ is precisely $dt$⁠, and the denominator (after Step 1) is $t$⁠. No $x$ remains.
   Change of variable, Section 7.3.1; Property (IV), Section 7.2.1. Lesson: integration by substitution

4. Integrate and return to $x$

   $$\begin{aligned} \frac12\int\frac{dt}{t} &= \frac12\log|t| + C \\ &= \frac12\log\left|3\cos x + 2\sin x\right| + C. \end{aligned}$$

   Written in the book's order, this is $\frac12\log\left|2\sin x + 3\cos x\right| + C$⁠.

   The modulus stays: $3\cos x + 2\sin x$ is negative for some $x$ (for instance $x = \pi$ gives $-3$⁠).
   Standard formula (xii), Section 7.2. Basics: logarithms and 1/x

5. Check: differentiate the answer back

   With $u = 3\cos x + 2\sin x$ and $\frac{d}{dx}\log|u| = \frac{u'}{u}$⁠:

   $$\begin{aligned} \frac{d}{dx}\left[\frac12\log|u|\right] &= \frac12\cdot\frac{2\cos x - 3\sin x}{3\cos x + 2\sin x} \\ &= \frac{2\cos x - 3\sin x}{6\cos x + 4\sin x}. \end{aligned}$$

   This is the integrand, so the answer is correct.

   Chain rule, Theorem 4, Section 5.3.1; Property (I), Section 7.2.1.

6. Where the answer is valid

   We need $3\cos x + 2\sin x \ne 0$⁠. Where $\cos x = 0$⁠, the expression is $2\sin x = \pm 2 \ne 0$⁠, so those points are fine. Elsewhere, dividing by $\cos x$⁠, the condition is

   $$\tan x \ne -\frac32.$$

   So the excluded points are $x = k\pi - \tan^{-1}\frac32$⁠, $k \in \Z$⁠, and the answer holds on each open interval between them.

   These are exactly the zeros of the denominator $6\cos x + 4\sin x = 2(3\cos x + 2\sin x)$⁠, where the integrand itself is undefined.

7. A different substitution changes only the constant

   Suppose we skip Step 1 and put $s = 6\cos x + 4\sin x$⁠. Then $ds = 2(2\cos x - 3\sin x)\,dx$⁠, and

   $$\begin{aligned} \int f(x)\,dx &= \frac12\int\frac{ds}{s} \\ &= \frac12\log\left|6\cos x + 4\sin x\right| + C'. \end{aligned}$$

   Since $|6\cos x + 4\sin x| = 2\,|3\cos x + 2\sin x|$⁠,

   $$\begin{aligned} &\frac12\log\left|6\cos x + 4\sin x\right| \\ &\quad= \frac12\log 2 + \frac12\log\left|3\cos x + 2\sin x\right|. \end{aligned}$$

   The two answers differ by the constant $\frac12\log 2$⁠, which is absorbed into the arbitrary constant. Both are correct.

   Indefinite integrals are families $F(x) + C$⁠; two antiderivatives that differ by a constant describe the same family (Property (II), Section 7.2.1).
   $\log(ab) = \log a + \log b$ for $a, b > 0$ (Section 5.4).

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