What is the net area between #f(x) = cos^2xsin^2x # and the x-axis over #x in [0, 3pi ]#?

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Answer 1 · 2017-03-12T08:44:04

The answer is #=3/8pi=1.178u^2#

We start, by simplifying

#sin2x=2sinxcosx#

and

#cos2x=1-2sin^2x#

#sin^2x=(1-cos2x)/2#

#sin^2 2x=1/2(1-cos4x)#

Therefore,

#f(x)=cos^2xsin^2x#

#=1/4sin^2 2x#

#=1/4*1/2*(1-cos4x)#

#=1/8(1-cos4x)#

The area is

#=1/8int_0^(3pi)(1-cos4x)dx#

#=1/8[x-1/4sin4x]_0^(3pi)#

#=1/8(3pi-1/4sin12pi-0-0)#

#=3/8pi=1.178#