$$ \int \frac{(\cos 9x + \cos6x)}{2 \cos 5x - 1} dx $$
I know that it simplifies to $ \cos x + \cos 4x $ but I have no idea how to do that. I tried expanding $\cos 9x $ and $\cos 6x$ by using the formulas for $\cos 3x$ and $\cos 2x$. There is nothing i could think to simplify the $\cos 5x$ in the denominator
How to proceed while simplifying larger multiples to lower.
Is the any other way than simplifying the expression to calculate the integral?
On
If your simplification is correct then best option is to simplify down to $\cos x $ and $\sin x $.
$\cos 9x = \cos (3 \times (3x)) \; \& \cos 6x = \cos(3 \times (2x)) .$
And $\cos 5x = \cos (2x+3x) .$
On
Using Euler formula: $\cos x = 1/2 \times (e^{ix}+e^{-ix})$
$\cos 9x + \cos 6x = 1/2\times (e^{9ix}+e^{-9ix}+e^{6ix}+e^{-6ix}) \\ = 1/2 \times (e^{6ix}+e^{4ix}+e^{9ix}+e^{ix}+e^{-4ix}+e^{-6ix}+e^{-ix}+e^{-9ix}-e^{ix}-e^{-ix}-e^{4ix}-e^{-4ix} )\\= 1/2 \times (e^{5ix}+e^{-5ix}-1)(e^{ix}+e^{-ix}+e^{4ix}+e^{-4ix}) = (2\cos 5x-1)(\cos x + \cos 4x)$
On
$$cos(9x)+cos(6x)$$ $$=cos(10x-x)+ cos(5x+x) $$ $$=cos(x) cos(10x) + sin(10x) sin(x) + cos(5x)cos(x) - sin(x)sin(5x)$$ $$=cos(x)[2cos^2 (5x) - 1+cos(5x) ]+ sin(x) [2sin(5x)cos(5x) - sin(5x)]$$ $$=cos(x)(2cos(5x)-1)(cos(5x)+1) +sin(x)(sin(5x)(2cos(5x)-1)$$ The denominator cancels and you should be able to take it from here.
Using the "sums-to-products" formula $$\cos(A+B)+\cos(A-B)=2\cos A\cos B$$ we have $$\eqalign{ \cos9x+\cos6x &=\cos(5x+4x)+\cos(5x-4x)-\cos x\cr &\qquad\qquad\qquad+\cos(5x+x)+\cos(5x-x)-\cos4x\cr &=2\cos5x\cos4x-\cos4x+2\cos5x\cos x-\cos x\cr &=(2\cos5x-1)(\cos x+\cos4x)\ .\cr}$$