Question #8e0d6

1 Answer
Narad T.
Oct 17, 2016

#=16(cos^7(x/2)/7-cos^5(x/2)/5)+C#

Explanation:

Here are the steps
#intsin^3xcos(x/2)dx#
So we start with the substitution #u=x/2# so #du=dx/2#
#intsin^3xcos(x/2)dx=2intsin^3(2u)cosudu#
#=2int(2sinucosu)^3cosudu#
as #sin2u=2sinucosu#
#=16intsin^2usinucos^4udu#
#=16int(1-cos^2u)cos^4usinudu#
Now we use #v=cosu# #dv=-sinudu#
Integral #=16int(1-v^2)v^4-dv#
#=16int(v^6-v^4)dv#
#=16(v^7/7-v^5/5)#
#=16(cos^7u/7-cos^5u/5)#
#=16(cos^7(x/2)/7-cos^5(x/2)/5)+C#

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