# How do we evaluate this REALLY tricky integral? [closed]

$$\int{\frac{\cos 9x + \cos 6x}{2 \cos 5x-1} dx}$$ The objective is to find the answer in terms of $$\sin 4x$$ and $$\sin x$$

I would like to share my attempt and then ask a conceptual doubt as usual but this time I have no clue on how to even start off with this. I tried to apply the cosA + cos B formula but was in vain.

Any idea or hint would be appreciated. Also if you do know any sort of algorithm on approaching questions of these kind do let me know please.

• Try substituting $x\mapsto\pi-x$ Mar 31 at 18:02
• @user170231 Wait how did u do using that? I'm unlucky with that substitution Mar 31 at 18:06
• I cheated a bit. Mathematica fails to simplify the integrand as-is, but succeeds if you first replace $x$ with $\pi-x$: (Cos[9 x] + Cos[6 x])/(2 Cos[5 x] - 1) // Simplify // TrigReduce vs ((Cos[9 x] + Cos[6 x])/(2 Cos[5 x] - 1) /. x -> \[Pi] - x) // Simplify // TrigReduce. The latter gives a very similar expression to the one in Quanto's answer. Mar 31 at 18:30
• Apr 1 at 7:51

Note that

$$\cos6x =2\cos x\cos 5x-\cos 4x$$ $$\cos 9x=2\cos 4x\cos 5x-\cos x$$ Thus

$$\int{\frac{\cos 9x+\cos 6x}{2 \cos 5x-1} dx} =\int{\frac{(2 \cos 5x-1)(\cos x+\cos 4x)}{2 \cos 5x-1} dx}\\ =\int \cos x+\cos 4x\ dx=\sin x+\frac14\sin4x$$

Let $$E=e^{ix}$$. \begin{align} \int{\frac{\cos 9x + \cos 6x}{2 \cos 5x-1} dx}&=\int\frac{\frac{E^9+E^{-9}}2+\frac{E^6+E^{-6}}2}{2(\frac{E^5+E^{-5}}2)-1}dx\\ &=\int\frac{E^{18}+E^{15}+E^{3}+1}{2E^{4}(E^{10}-E^{5}+1)}dx\\ &=\int\frac{(E^{5}+1)(E^{3}+1)}{2E^{4}}dx\\ &=\int (\cos4x+\cos x)dx\\ &=\frac14\sin 4x+\sin x+c \end{align}

Answers : $$I＝\int\;\frac {\cos(9x)＋\cos(6x)}{－1＋2\cos(5x)}dx＝－\int\;\frac {2\cos(\textstyle\frac {15x}{2})\cos(\frac {3x}{2})}{1-2(2\cos^2(\frac {5x}{2})-1)}dx\\~\\＝-\int\;\frac {2\cos(\textstyle\frac {15x}{2})\cos(\frac {3x}{2})}{3-4\cos^2(\frac {5x}{2})}dx$$

Multiplying and dividing it by $$\cos (5x/2)$$ we get $$I＝-\int\;\frac {2\cos(\textstyle\frac {15x}{2})\cos(\frac {3x}{2})\cos(\frac {5x}{2})}{3\cos(\frac {5x}{2})-4\cos^3(\frac {5x}{2})}dx\\~\\＝－\int\frac {2\cos(\textstyle\frac {15x}{2})\cos(\frac {3x}{2})\cos(\frac {5x}{2})}{－\cos(\frac {15x}{2})}dx\\~\\＝\int\;\frac {2\cos(\textstyle\frac {15x}{2})\cos(\frac {3x}{2})\cos(\frac {5x}{2})}{\cos(\frac {15x}{2})}dx$$

using $$\cos(3\theta)=4\cos^3(\theta)-3\cos(\theta)$$

And therefore :

$$I＝\int2\cos(\frac {3x}{2})\cos(\frac {5x}{2})dx＝\int\;\cos(4x)＋\cos(x)\;dx\\~\\I＝\frac {\sin(4x)}{4}＋\sin(x)＋C$$

Since you want some more intuition, I use the path you started taking

So we are on the same page, I use the identities $$\cos(a)+\cos(b) = 2\cos((a+b)/2)\cos((a-b)/2)$$ $$\sin(a)-\sin(b) = 2\cos((a+b)/2)\sin((a-b)/2)$$ $$\sin(2x) = 2\sin x \cos x$$

First we have $$\cos(9x) + \cos(6x) = 2\cos(15x/2)\cos(3x/2)$$

We can also mess with the denominator like \begin{align*} &2\cos(5x)-1\\ =&\frac{\sin(10x)-\sin(5x)}{\sin(5x)}\\ =&\frac{2\cos(15x/2)\sin(5x/2)}{2\sin(5x/2)\cos(5x/2)}\\ =&\frac{\cos(15x/2)}{\cos(5x/2)} \end{align*}

Thus, we obtain \begin{align*} &\frac{\cos(9x)+\cos(6x)}{2\cos(5x)-1}\\ =& 2\cos(5x/2)\cos(3x/2)\\ =& \cos(4x) + \cos(x) \end{align*}