How do you find the integral of #int (cosx)^4 dx#?

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Answer 1 · 2018-03-16T07:02:59

The answer is #=1/32sin(4x)+1/4sin(2x)+3/8x+C#

First, linearize #cos^4x# by applying Euler's Identity

#cosx=(e^(ix)+e^(-ix))/2#

Therefore,

#cos^4x=((e^(ix)+e^(-ix))/2)^4#

#=1/16(e^(4ix)+e^(-4ix)+4e^(2ix)+4e^(-2ix)+6)#

#=1/8(cos(4x)+4cos(2x)+3)#

Therefore,

#intcos^4xdx=1/8intcos(4x)dx+1/2intcos(2x)+3/8int1dx#

#=1/32sin(4x)+1/4sin(2x)+3/8x+C#