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$$ \int \frac{(\cos 9x + \cos6x)}{2 \cos 5x - 1} dx $$
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?

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  • $\begingroup$ According to Maple that integral simplfies to $2{\cos(x)}^3\sin(x)-\cos(x)\sin(x)+\sin(x)$; you can check that it does not differ from $\cos(x)+\cos(4x)$ by a constant. $\endgroup$
    – Fimpellizzeri
    Jul 1, 2016 at 6:03
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    $\begingroup$ Moreover, while it is surely a lot of work, you can use the kinds of formulas you yourself suggested to reduce each of numerator and denominator to a polynomial in $\cos(x)$. In this particular case, the numerator ends up being a multiple of the denominator. It seems this example was carefully chosen, because changing some of those numbers around yield very ugly answers. $\endgroup$
    – Fimpellizzeri
    Jul 1, 2016 at 6:10
  • $\begingroup$ See math.stackexchange.com/questions/1817300/… $\endgroup$
    – lab bhattacharjee
    Jul 1, 2016 at 6:15

4 Answers 4

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Using Euler formula: $\cos x = 1/2 \times (e^{ix}+e^{-ix})$

$\cos 9x + \cos 6x = 1/2\times (e^{9ix}+e^{-9ix}+e^{6ix}+e^{-6ix}) \\ = 1/2 \times (e^{6ix}+e^{4ix}+e^{9ix}+e^{ix}+e^{-4ix}+e^{-6ix}+e^{-ix}+e^{-9ix}-e^{ix}-e^{-ix}-e^{4ix}-e^{-4ix} )\\= 1/2 \times (e^{5ix}+e^{-5ix}-1)(e^{ix}+e^{-ix}+e^{4ix}+e^{-4ix}) = (2\cos 5x-1)(\cos x + \cos 4x)$

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Using the "sums-to-products" formula $$\cos(A+B)+\cos(A-B)=2\cos A\cos B$$ we have $$\eqalign{ \cos9x+\cos6x &=\cos(5x+4x)+\cos(5x-4x)-\cos x\cr &\qquad\qquad\qquad+\cos(5x+x)+\cos(5x-x)-\cos4x\cr &=2\cos5x\cos4x-\cos4x+2\cos5x\cos x-\cos x\cr &=(2\cos5x-1)(\cos x+\cos4x)\ .\cr}$$

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  • $\begingroup$ That was extremely useful! Thanks! $\endgroup$
    – Anirudh Modi
    Jul 1, 2016 at 7:37
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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 $\cos 5x = \cos (2x+3x) .$

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$$cos(9x)+cos(6x)$$ $$=cos(10x-x)+ cos(5x+x) $$ $$=cos(x) cos(10x) + sin(10x) sin(x) + cos(5x)cos(x) - sin(x)sin(5x)$$ $$=cos(x)[2cos^2 (5x) - 1+cos(5x) ]+ sin(x) [2sin(5x)cos(5x) - sin(5x)]$$ $$=cos(x)(2cos(5x)-1)(cos(5x)+1) +sin(x)(sin(5x)(2cos(5x)-1)$$ The denominator cancels and you should be able to take it from here.

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