What is the integral of #sin^2(x)cos^4(x) #?

Question

105034 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-04T16:45:07

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

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

We will use the following Identities to simplify the Integrand :-

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

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

Now, #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#.

Enjoy Maths.!