Integral of the power products, where m is an even number m = 2k, solve the following integral:#int_0^pisin^2x*cos^4xdx#?

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#int_0^pisin^2x*cos^4xdx#

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Answer 1 · 2018-04-06T06:29:18

#int_0^pisin^2x*cos^4xdx#

#=1/4int_0^pi(2sinxcosx)^2*cos^2xdx#

#=1/4int_0^pisin^2 2x*cos^2xdx#

#=1/4int_0^pi1/2(1-cos4x)*1/2(1+cos2x)dx#

#=1/16int_0^pi(1-cos4x+cos2x-cos4xcos2x)dx#

#=1/16int_0^pi(1-cos4x+cos2x-1/2(cos6x+cos2x))dx#

#=1/16[x-1/4sin4x+1/2sin2x-1/12sin6x-1/4sin2x]_0 ^pi#

#=1/16[(pi-1/4sin(4pi)+1/2sin(2pi)-1/12sin(6pi)-1/4sin(2pi))-(0-1/4sin(4*0)+1/2sin(2*0)-1/12sin(3*0)-1/4sin(2*0))]#

#=pi/16#