Evaluate the integral sin(12x)cos(3x)dx?

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Answer 1 · 2018-05-09T12:04:19

The answer is #=4/9cos^3(3x)-8/15cos^5(3x) +C#

Let #u=3x#, #=>#, #du=3dx#

The integral is

#I=intsin(12x)cos(3x)dx=1/3intcos(u)sin(4u)du#

#sin(4u)=4cos^3usinu-4cosusin^3u#

#sin^2u=1-cos^2u#

#cosusin(4u)=cosu(4cos^3usinu-4cosusin^3u)#

#=sinu(4cos^4u-4cos^2usin^2u)#

#=sinu(4cos^4u-4cos^2u(1-cos^2u))#

#=sinu(8cos^4u-4cos^2u)#
Therefore,

#I=1/3intsinu(8cos^4u-4cos^2u)du#

#=4/3int(2cos^4u-cos^2u)sinudu#

Let #v=cosu#, #=>#, #dv=-sinudu#

Therefore,

#I=4/3int(v^2-2v^4)dv#

#=4/3*v^3/3-4/3*2*v^5/5#

#=4/9v^3-8/15v^5#

#=4/9cos^3u-8/15cos^5u#

#=4/9cos^3(3x)-8/15cos^5(3x) +C#