Question

Question #8e0d6

Answer 1

`=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`