You could use:
\cos^8 x = \frac{\cos 8 x + 8 \cos 6x + 28 \cos 4x + 56 \cos 2x + 35}{128}
This should not be too much of an issue if you use the same method as used in integrating cos2x
Here you will have to use the same thing 3 times....
can you?
You could use:
\cos^8 x = \frac{\cos 8 x + 8 \cos 6x + 28 \cos 4x + 56 \cos 2x + 35}{128}
No, start with 4\cos^4 x = (1+\cos 2x)^2 and without much sweat you get that expression
qustions is wrong :P...ders no dx......hence the function ∫cos8x is non-integrateble