Integrate #int sin^4(x)dx#?

#int sin^4(x)dx#

How do you use the half-angle identities to reduce the function?

2 Answers
Øko
Feb 8, 2018

#I=1/32(12x+sin(4x)-8sin(2x))+C#

Explanation:

We want to solve

#I=intsin^4(x)dx#

Use double-angle identities (equivalent to half-angle):

  • #cos(2a)=1-2sin^2(a)#

  • #cos(2a)=2cos^2(a)-1#

Rewrite the integrand using double-angle identities

#I=intsin^2(x)sin^2(x)dx#

#=int1/2(1-cos(2x))1/2(1-cos(2x))dx#

#=1/4int(1-cos(2x))^2dx#

#=1/4int1+cos^2(2x)-2cos(2x)dx#

#=1/4int1+1/2(1+cos(4x))-2cos(2x)dx#

#=3/8intdx+1/8intcos(4x)dx-1/2intcos(2x)dx#

Integrate

#I=3/8x+1/32sin(4x)-1/4sin(2x)+C#

#=1/32(12x+sin(4x)-8sin(2x))+C#

Narad T.
Feb 8, 2018

The answer is #=1/32sin4x-1/4sin2x+3/8x+C#

Explanation:

Another way to perform this integral

Applying Euler's relation

#cosx+isinx=e^(ix)#

#cosx-isinx=e^(-ix)#

#2isinx=(e^(ix)-e^(-ix))#

#sinx=(e^(ix)-e^(-ix))/(2i)#

#cosx=(e^(ix)+e^(-ix))/2#

#i^2=-1#

#sin^4(x)=(e^(ix)-e^(-ix))^4/(2i)^4#

#=1/16(e^(4ix)-4e^(2ix)+6-4e^(-2ix)+e^(-4ix))#

#=1/16(e^(4ix)+e^(-4ix)-4(e^(2ix)+e^(-2ix))+6)#

#=1/8cos(4x)-1/2cos(2x)+3/8#

Therefore,

#intsin^4(x)dx=int(1/8cos(4x)-1/2cos(2x)+3/8)dx#

#=1/32sin4x-1/4sin2x+3/8x+C#

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