Qual é a integral de # sin ^ 2 (x) cos ^ 4 (x) #?

Deixei #I=intsin^2xcos^4xdx#.

Usaremos as seguintes identidades para simplificar o Integrand: -

# [1] :2sin^2theta=1-cos2theta, [2] : 2cos^2theta=1+cos2theta#

# [3] : 2cosCcosD=cos(C+D)+cos(C-D)#

Agora, #sin^2xcos^4x=1/8(4sin^2xcos^2x)(2cos^2x)#

#=1/8(2sinxcosx)^2(1+cos2x)#

#=1/8(sin2x)^2(1+cos2x)#

#=1/8(sin^2 2x)(1+cos2x)#

#=1/16(2sin^2 2x)(1+cos2x)#

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

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

#=1/16{1-cos4x+cos2x-1/2(cos6x+cos2x)}#

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

#:. I=1/32int(2+cos2x-2cos4x-cos6x)dx#

#=1/32(2x+sin(2x)/2-(2sin(4x))/4-sin(6x)/6)#

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

Desfrute de matemática.!