$\cos^3(x) = \left(\dfrac{e^{ \imath x}+e^{ -\imath x}}{2}\right)^3=$
$\dfrac{3}{4}\dfrac{e^{\imath x}+e^{-\imath x}}{2}+\dfrac{1}{4}\dfrac{e^{\imath 3x}+e^{-\imath 3x}}{2}=$
$\dfrac{3}{4}\cos(x)+\dfrac{1}{4}\cos(3x)$
I know how to do this problem the easy way, where I have trouble though is when I apply Euler's formula.
integral Cos^3(x)dx= Sin(x) - (1/3) Sin^3(x)+c
however when I try to do this with complex numbers I get stuck at this point.
I can do the integral I am just not sure how to go back to trig functions.
Cos^3(x) = (1/4) ((e^3ix - e^-3ix)/2 - 3(e^ix - e^-ix)/2)
I think this part -3(e^ix - e^-ix)/2) becomes -3Cos(x)
but what about the rest?
ps- I'm on spring break or else I would ask my professor (calc 2)