What is the integral of #sin^6(x)#?

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Answer 1 · 2018-04-28T17:13:53

#I=1/192[60x-45sin2x+9sin4x-sin6x]+c#

Note:
#color(red)((1)2sin^2theta=1-cos2theta#
#color(blue)((2)cos^2theta=1+cos2theta#
#color(violet)((3)2cosAcosB=cos(A+B)+cos(A-B)#

Here,

#I=intsin^6xdx#

Now,

#sin^6x=(sin^2x)^2(sin^2x)#

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

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

#=1/8(1-2cos2x+(1+cos4x)/2)(1-cos2x)#

#=1/16(2-4cos2x+1+cos4x)(1-cos2x)#

#=1/16(3-4cos2x+cos4x)(1-cos2x)#

#=1/16[3-4cos2x+cos4x-3cos2x+4cos^2 2x-cos4xcos2x]#

#=1/16[3-7cos2x+cos4x+2xx2cos^2 2x-1/2*2cos4xcos2x]#

#=1/16[3-7cos2x+cos4x+2(1+cos4x)-1/2(cos6x+cos2x)]#

#=1/16[3-7cos2x+cos4x+2+2cos4x-1/2(cos6x+cos2x)]#

#=1/32[6-14cos2x+2cos4x+4+4cos4x-cos6x-cos2x]#

#=1/32[10-15cos2x+6cos4x-cos6x]#

So,

#I=1/32int[10-15cos2x+6cos4x-cos6x]dx#

#I=1/32[10x-(15sin2x)/2+(6sin4x)/4-(sin6x)/6]+c#

#I=1/32[10x-(15sin2x)/2+(3sin4x)/2-(sin6x)/6]+c#

#I=1/192[60x-45sin2x+9sin4x-sin6x]+c#