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,

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

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

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