Would like to solve this problem. Did almost half of it but need to use a faster metod like Eulers rule? Can it work ?

Question

#int(sinx)^4*(cosx)^2#= (sinx)^4(1-sin²x) = #sin^4x-sin^6x# (then what formula can be used to solve it faster)

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Answer 1 · 2018-06-15T11:31:01

# 1/192(12x+sin6x-3sin4x-3sin2x)+C#.

To integrate even powers of #sin" &/or "cos# function/s, we

need to convert the power into multiple angle/s of

#sin" &/or "cos# function/s.

For this, the following Identities are used :

#sin^2x=(1-cos2x)/2, &, cos^2x=(1+cos2x)/2#.

We have, #sin^4xcos^2x=sin^2x*sin^2xcos^2x#

#=(1-cos2x)/2*1/4(4sin^2xcos^2x)=1/8(1-cos2x)(sin^2 2x)#,

#1/8(1-cos2x)*(1-cos4x)/2#,

#=1/16(1-cos4x-cos2x+cos4xcos2x)#,

#=1/32(2-2cos4x-2cos2x+2cos4xcos2x)#,

#=1/32{2-2cos4x-2cos2x+cos(4x+2x)+cos(4x-2x)}#,

#=1/32(2+cos6x-2cos4x-cos2x)#.

#:. intsin^4xcos^2xdx#,

#=1/32{2x+1/6sin6x-2*1/4sin4x-1/2sin2x}#,

#=1/192(12x+sin6x-3sin4x-3sin2x)+C#.