Evaluating $\int \frac{\cos ax+\cos bx}{1-2\cos cx}dx$

Question

I got this question on seeing the solution for evaluating the integral $$\int \frac{\cos 5x+\cos 4x}{1-2\cos 3x}dx$$ in my textbook. I searched this site, and found the following questions:

Another way to evaluate $\int\frac{\cos5x+\cos4x}{1-2\cos3x}{dx}$? (Exactly the same integral)
How to integrate $\frac {\cos (7x)-\cos (8x)}{1+2\cos (5x)} $ ?

In both of these questions and in my book, the first step involves multiplying the numerator and the denominator by $\sin px$ where $p=3$ in the first integral and $p=5$ in the second integral.

I wondered, why we must multiply both numerator and denominator by sine of "something" and how to determine that "something"? I was unable to answer the first question. But I was able to make some progress in solving the second question which I've discussed below:

Let us consider the following integral where $a,b,$ and $c$ are constants,

$$\int \frac{\cos ax+\cos bx}{1-2\cos cx}dx$$

In order to evaluate this integral we need to multiply both numerator and denominator by $\sin px$ where $p$ is some constant which we need to figure out. I guessed two possibilities:

$p=c$
$p=\frac{(a+b)}{3}$

Unfortunately, I was unable to tell which of the above two possibilities is the reason for the choice of $p$, because in both integrals (linked questions) the above two conditions are satisfied simultaneously.

In short, I'm confused why most of the sources multiply $\sin px$ in both numerator and denominator to solve this kind of integral. Is this some kind of a general rule or totally a guess? What are the constrains for the variable $p$ in $\sin px$? Or how do we determine $p$ in case of any integral of this form? Or is that also a guess?

Kindly explain the above two questions.

Thank you in advance.

Using Euler‘s identity, i.e. $$\cos(x)=\frac12(\exp(ix)+\exp(-ix))$$ seems most straightforward — Maximilian Janisch, Dec 13 '19 at 08:57
Have you considered that $c=\frac{a+b}3$ might be a condition for the method to work ? — , Dec 13 '19 at 08:57
@MaximilianJanisch, I'm sorry. I didn't read about "Euler's identity". — Vishnu, Dec 13 '19 at 08:59
It is very useful for integrals like this https://en.m.wikipedia.org/wiki/Integration_using_Euler%27s_formula — Maximilian Janisch, Dec 13 '19 at 09:00
@YvesDaoust, No. I didn't notice that. Thanks. Is that the reason for using $p=c$? — Vishnu, Dec 13 '19 at 09:01
It would be a reason for both $p=c$ and $p=\frac{a+b}3$ to work. — , Dec 13 '19 at 09:02
@YvesDaoust, Fine. So for a random value of $c$ this doesn't work. Am I right? BTW, how did you find this fact? (I think you could write an answer explaining both) — Vishnu, Dec 13 '19 at 09:05
I just questioned your affirmation that there are two cases :-) — , Dec 13 '19 at 09:05

score 2 · Accepted Answer · edited Dec 13 '19 at 13:44

In the integral $\int \dfrac{\cos 5x+\cos 4x}{1-2\cos 3x}dx$ , the numbers $5$, $4$ and $3$ are very carefully chosen such that the integrand can be simplified. It is not possible to solve the integral with any $a$, $b$ and $c$ ; instead of some specific carefully choosen $a$, $b$ and $c$.

Now, $$\int \dfrac{\cos 5x+\cos 4x}{1-2\cos 3x}dx = \int \dfrac{(\sin 3x)(\cos 5x+\cos 4x)}{(\sin 3x)(1-2\cos 3x)}dx$$ I have multiplied numerator and denominator by $\sin 3x$ to remove the coefficient "2" of $\cos 3x$ since $2\sin 3x\cos 3x$ would yield $\sin 6x$ (in general, $p$ should be equal to $c$ to remove that "2"). I am removing this "2" to apply the formula of $\sin A - \sin B$, in the hope that, doing similar thing (i.e. applying formula of $\cos A + \cos B$) in numerator would lead to cancellation of some common terms from numerator and denominator (that's exactly what will happen if you notice the solution further).

The cancellation is because of the choice of appropriate angles ($5x,4x$ and $3x$) of sine and cosine. This cancellation would not be possible if the angles are randomly chosen. What I mean to say is that besides having $p$ to be equal to $c$ , $a$ and $b$ should be well chosen so that the integrand can be simplified.

$$\int \dfrac{\cos 5x+\cos 4x}{1-2\cos 3x}dx= \int \dfrac{(\sin 3x)(\cos 5x+\cos 4x)}{\sin 3x-2\sin 3x.\cos 3x}dx$$

$$= \int \dfrac{(\sin 3x)(\cos 5x+\cos 4x)}{\sin 3x-\sin 6x}dx$$

$$= \int \dfrac{(\sin 3x)(\cos 5x+\cos 4x)}{2\cos \frac{9x}{2}.\sin\frac{-3x}{2}}dx$$

$$= \int \dfrac{(\sin 3x)(\require{cancel}\cancel{2} \cancel{\cos \frac{9x}{2}}.\cos \frac {x}{2})}{\cancel{2}\cancel{\cos \frac{9x}{2}}.\sin\frac{-3x}{2}}dx$$

$$= \int \dfrac{(2\cancel{\sin \frac{3x}{2}}.\cos \frac{3x}{2})(\cos \frac{x}{2})}{(-\cancel{\sin \frac{3x}{2}})}dx$$

$$= -\int (2\cos \frac{3x}{2}.\cos \frac{x}{2})dx$$

$$= -\int (\cos 2x+\cos x)dx$$

$$= \int (-\cos 2x-\cos x)dx$$ $$= -\dfrac {\sin 2x}{2} - \sin x + c$$

you dont need to put $$ in each line, you can use just one in the first line and separate the lines with \\ — Masacroso, Dec 13 '19 at 13:46

score 0 · Answer 2 · answered Aug 24 '23 at 13:52

In general \begin{align} \int\frac{\cos(2p+q)x \pm \cos(p+2q)x}{2\cos(p+q)x\mp 1}\ dx =\frac1p \sin px \pm \frac1q \sin qx \end{align} Examples:

$p=1$ and $q=2$ $$ \int\frac{\cos4x + \cos5x}{2\cos3x- 1}\ dx = \sin x + \frac12 \sin 2x $$
$p=2$ and $q=3$ $$ \int\frac{\cos7x- \cos8x}{2\cos5x+ 1}\ dx = \frac12\sin 2x - \frac13\sin 3x $$

Evaluating $\int \frac{\cos ax+\cos bx}{1-2\cos cx}dx$

2 Answers2