Question

Integrate cos^6 x using complex numbers?

Answer 1

like..
z=cosx+isinx
\[\frac{1}{z}=\cos x-i \sin x\]
\[(z+\frac{1}{z})^6=(2cosx)^6\]
now we use binomial..dont know how to do this part..

Answer 2

\[\LARGE C(6,0) (z^6+\frac{1}{z^6})+C(6,1)..\]
i basically want to know what to put in place of z^6+1/z^6 etc

Answer 3

\[\cos^{6} x = \frac{1}{2^{6}}(z+z^{-1})^{6}\]
\[(z+z^{-1})^{6} = z^{6}+6z^{4}+15z^{2}+20+15z^{-2}+6z^{-4}+z^{-6}\]
also
\[z = e^{ix}\]

Answer 4

z^6 = cos 6x + i sin6x
1/z^6 = cos 6x - i sin 6x

z^6 +1/z^6 = 2 cos 6x

Answer 5

what now?

Answer 6

z^4+1/z^4 = 2 cos 4x

Answer 7

z^2+1/z^2 = 2 cos 2x

Answer 8

you have the form
A cos 6x +Bcos4x+Ccos 2x
so easy to integrate

Answer 9

oh! thanks!

Answer 10

+20 ofcourse
welcome ^_^

Answer 11

@hartnn +2cosx i sinx ?

Answer 12

in z^6 = cos 6x + i sin6x