Trigonometry Integral (Different cosine arguments)

Question

$$\int \frac{\cos\, 5x+\cos\, 4x}{1-2\cos\, 3x}dx$$

I tried using the $~\cos(A+B)~$ formula in the numerator and got $~2~\cos\left(\frac{x}{2}\right)\cos\left(\frac{9x}{2}\right)~$ but that doesn’t seem to take me anywhere. I know ultimately the arguments in the cos terms must be equal, but how do I get them to be?

Answer 1

Let's use a trig indenity to factor the numerator:

$$\int \frac{\cos5x+\cos4x}{1-2\cos3x}dx$$ $$= \int \frac{\cos 5x + \cos x + \cos4x + \cos2x - \cos2x - \cos x}{1-2\cos3x}dx$$ $$= \int \frac{2 \cos3x \cos2x + 2 \cos3x \cos x - \cos2x - \cos x}{1-2\cos3x}dx$$ $$= \int \frac{(2\cos3x - 1)(\cos2x + \cos x)}{1-2\cos3x}dx$$ $$= - \int \cos2x + \cos x dx$$

You can solve this by yourself, and get the answer:

$$ -\frac{\sin2x}{2} - \sin x + C$$

Answer 2

$$I=\int \frac{(\cos 5x +\cos 4x)}{(1-2\cos 3x)}dx$$ Multiplu up and down by $\sin 3x$ and simplify $$\implies I=\int \frac{2 \cos(9x/2) \cos(x/2) \sin 3x}{\sin 3x -\sin 6x} dx$$ $$I=-\int \frac{2 \cos(9x/2) \cos(x/2) \sin 3x} {2\cos(9x/2) \sin(3x/2)}dx=- \int 2\cos(x/2)\cos(3x/2)dx$$ $$\implies I=-\int (\cos 2x +\cos x) dx=-\frac{1}{2}\sin 2x- \sin x+C$$