How to perform the following integration $\int \frac{\cos 5x+5\cos 3x+10\cos x}{\cos 6x+ 6\cos 4x+ 15\cos 2x +10}dx$

Question

$$\int \frac{\cos 5x+5\cos 3x+10\cos x}{\cos 6x+ 6\cos 4x+ 15\cos 2x +10}dx$$

How to simplify the expression given? I tried using formulas for $\cos 2x$ and $\cos 3x$. I also tried the using $\cos x + \cos y = 2cos(\frac{x+y}{2})\cos(\frac{x+y}{2})$.

The options for answers are given are in form of half angles of trigonometric functions.

Answer 1

Using $$\cos 6x+6\cos 4x+15\cos 2x+10 = (\cos 6x+\cos 4x)+5(\cos 4x+\cos 2x)+10(1+\cos 2x)$$

So $$ = 2\cos 5x\cos x+5\cdot 2\cos 3x\cos x+20\cos^2 x = 2\cos x(\cos 5x+5\cos 3x+10\cos x)$$

So $$\int \frac{\cos 5x+5\cos 3x+10\cos x}{\cos 6x+ 6\cos 4x+ 15\cos 2x +10}dx=\frac{1}{2}\int \frac{1}{\cos x}dx$$

So $$ = \frac{1}{2}\int \sec xdx = \frac{1}{2}\ln |\sec x+\tan x|+\mathcal{C}$$

$$ = \frac{1}{2}\ln \left|\frac{1+\sin x}{\cos x}\right|+\mathcal{C} = \frac{1}{2}\ln\left|\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right|+\mathcal{C}$$

Answer 2

If you write $$\cos w = \frac{1}{2}\left(e^{iw}+e^{-iw}\right),$$ you find the numerator is $$\frac{1}{2}\left(e^{ix}+e^{-ix}\right)^5 = 16\cos^5(x)$$

The denominator is, likewise:

$$\frac{1}{2}\left(e^{ix}+e^{-ix}\right)^6=32\cos^6(x)$$

So the integrand is $\frac{1}{2\cos x}$.

This all comes because $1,5,10,10,5,1$ is the fifth row of Pascal's triangle, and $1,6,15,20,15,6,1$ is the sixth row.