Exercise 7.3 · Q13: $\int\frac{\cos 2x-\cos 2\alpha}{\cos x-\cos\alpha}\,dx$

Basics you need

Solution

1. Where is the integrand defined?

   The only danger is the denominator. We need

   $$\cos x - \cos\alpha \ne 0, \quad\text{that is,}\quad \cos x \ne \cos\alpha.$$

   Since $\cos x = \cos\alpha$ exactly when $x = 2n\pi \pm \alpha$ ($n$ an integer), the integrand is defined for all real $x$ except these points. We work on any interval that avoids them.

   An indefinite integral is an antiderivative on an interval where the integrand exists. Recording the excluded points first also tells us when the cancellation in Step 3 is allowed: we may divide by $\cos x - \cos\alpha$ only because it is not zero there.
   General solution of $\cos x = \cos\alpha$⁠: $x = 2n\pi \pm \alpha$ (Class XI, trigonometric equations).

2. Rewrite both double angles

   Apply $\cos 2\theta = 2\cos^2\theta - 1$ once with $\theta = x$ and once with $\theta = \alpha$⁠:

   $$\begin{aligned} \cos 2x - \cos 2\alpha &= \left(2\cos^2 x - 1\right) - \left(2\cos^2\alpha - 1\right) \\ &= 2\cos^2 x - 2\cos^2\alpha \\ &= 2\left(\cos^2 x - \cos^2\alpha\right). \end{aligned}$$
   The denominator is written in $\cos x$ and $\cos\alpha$⁠, so we want the numerator in the same "language". Of the three forms of $\cos 2\theta$⁠, the one using only $\cos\theta$ is $2\cos^2\theta - 1$⁠. The two $-1$s cancel, which is what makes this form so convenient here.
   $\cos 2\theta = 2\cos^2\theta - 1$ (Example 7(i), Section 7.3.2). Basics: trigonometric identities for integrals

3. Factorise and cancel

   Use $A^2 - B^2 = (A - B)(A + B)$ with $A = \cos x$⁠, $B = \cos\alpha$⁠:

   $$\begin{aligned} \cos 2x - \cos 2\alpha &= 2(\cos^2 x - \cos^2\alpha) \\ &= 2(\cos x - \cos\alpha)(\cos x + \cos\alpha). \end{aligned}$$

   Therefore, for every $x$ with $\cos x \ne \cos\alpha$⁠,

   $$\begin{aligned} f(x) &= \frac{2(\cos x - \cos\alpha)(\cos x + \cos\alpha)}{\cos x - \cos\alpha} \\ &= 2(\cos x + \cos\alpha) \\ &= 2\cos x + 2\cos\alpha. \end{aligned}$$
   The factor $\cos x - \cos\alpha$ appears on top and bottom, and it is non-zero on our domain (Step 1), so it cancels. A fraction we could not integrate directly has become a sum of a cosine and a constant.
   Cancelling a common factor is valid only where that factor is non-zero.

4. Integrate term by term

   $$\begin{aligned} \int (2\cos x + 2\cos\alpha)\,dx &= 2\int\cos x\,dx + 2\cos\alpha\int dx \\ &= 2\sin x + 2x\cos\alpha + C \\ &= 2(\sin x + x\cos\alpha) + C. \end{aligned}$$
   Property (V) splits the integral of a linear combination into the same combination of integrals. $2\cos\alpha$ is a constant multiplier (it does not depend on $x$⁠), so it comes outside, and $\int dx = x$⁠. Formula (ii) gives $\int\cos x\,dx = \sin x$⁠.
   Property (V), Section 7.2.1; standard formulae (i) and (ii), Section 7.2. Lesson: integration as anti-differentiation

5. Check: differentiate the answer back

   $\alpha$ is a constant, so $\frac{d}{dx}(x\cos\alpha) = \cos\alpha$⁠. Then

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

   The second line multiplies and divides by $\cos x - \cos\alpha$ (allowed, it is non-zero), and the third line is Step 2 read backwards. We are back at the integrand, so the answer is correct on every interval where $\cos x \ne \cos\alpha$⁠.

   $\frac{d}{dx}\sin x = \cos x$ (list in Section 7.2, (ii)); Property (I), Section 7.2.1. Basics: integration undoes differentiation

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