Six times of $\displaystyle{x}$ and cosine raised to eight...

$1$ ..Integration of an $cos kx$ where $k$ is an integer gives $\frac {sin kx}{k}$ , and the sine of any integral multiple of $\pi$ gives $0$ .

$2$ ..Also summation and difference of two integers yields an integer.

In the given expression use $1 + cos 2x = 2{cos}^{2}{x}$ until you come to this : $\frac{1}{64}\int_{0}^{\pi}{cos 6x{(3+cos 4x + 4cos 2x)}^{2}} dx$ .

Now because of $1$ and $2$ almost all terms will yield zero after integration and substitution.Now,instead of going down the warpath and expanding the whole expression and then applying multiple angle formulas(like I foolishly did -_- ),let's pause a bit.

We need to find a term that is independent of $x$ .Everything is being multiplied by $cos 6x$ .So,what integral multiple of $x$ when combined with $6x$ will give that?Its $-6x$ obviously.

Now,in the expansion of the inner part we get one of the terms as $4 cos 2x cos 4x$ .Now , $2 cos 2x cos 4x = cos 6x + cos 2x$ .The $cos 2x$ does not matter.The $cos 6x$ does.Then the integral should be $4 cos 6x cos 6x$ and many other terms which are not important.Now, $4 cos 6x cos 6x = 2(1 +cos 12x )$ .Once again the $cos 12x$ is not important. Therefore the integral is $\frac{1}{64}\int_{0}^{\pi}{2+ other..unimportant..terms}dx$ ,which yields $\frac{\pi}{32}$ .

The cosine function obeys the following identity:

$\begin{aligned} cos (n\theta) & = 2\cos \theta \cos [(n-1)\theta] - \cos[(n-2)\theta] \\ \implies \cos \theta \cos (n\theta) & = \frac {\cos[(n+1)\theta] + \cos [(n-1)\theta]}2 \end{aligned}$

Therefore,

$\begin{aligned} \cos^8 x \cos 6x & = \frac {\cos^7 x}2 (\cos 7x + \cos 5x) \\ & = \frac {\cos^6 x}4 (\cos 8x + 2\cos 6x + \cos 4x) \\ & = \frac {\cos^5 x}8 (\cos 9x + 3\cos 7x + 3\cos 5x + \cos 3x) \\ & = \frac {\cos^2 x}{64} (\cos 12x + 6\cos 10x + 15\cos 8x + 20\cos 6x + 15\cos 4x + 6\cos 2x + 1) \\ & = \frac {1+\cos 2x}{128} (\cos 12x + 6\cos 10x + 15\cos 8x + 20\cos 6x + 15\cos 4x + 6\cos 2x + 1) \\ & = \frac {\cos 14x + 8\cos 12 x + 28\cos 10x + 56\cos 8x + 70 \cos 6x + 56\cos 4x + 28\cos 2x +8}{256} \end{aligned}$

Therefore,

$\begin{aligned} \int_0^\pi \cos 6x \cos^8 x \ dx & = \frac 1{256} \int_0^\pi (\cos 14x + 8\cos 12 x + \cdots + 28\cos 2x +8)\ dx \\ & = \frac 1{256} \left[\frac {\sin 14x}{14} + \frac {8\sin 12x}{12} + \cdots + \frac {28\sin 2x}2 +8x \right]_0^\pi \\ & = \frac 8{256}\pi = \frac \pi{32} \approx \boxed{0.0982} \end{aligned}$