2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Fimpellizieri Jul 1 '16 at 6:03
  • 1
    $\begingroup$ 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. $\endgroup$ – Fimpellizieri Jul 1 '16 at 6:10
  • $\begingroup$ See math.stackexchange.com/questions/1817300/… $\endgroup$ – lab bhattacharjee Jul 1 '16 at 6:15
2
$\begingroup$

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

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ That was extremely useful! Thanks! $\endgroup$ – Anirudh Modi Jul 1 '16 at 7:37
0
$\begingroup$

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 $\cos 5x = \cos (2x+3x) .$

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.