Question

Integration of `sin^(4)2xcos2xdx`?

This is one of my homework problems and though I have most of it, I'm confused about part of the answer that the textbook is wanting.

I've used 2x as my u and thus have 2dx as du for what I know results partially in `intsin^(4)ucosudu`

where I know I then need to do another u-sub to continue on, but my texbook is wanting that to also have a 1/2 in front of it like so:
`1/2intsin^(4)ucosudu`

and I'm unclear as to where that's coming from.

I've been looking back through my book in the hopes that I've missed a prior explanation or formula introduction that would explain this and I'm just not seeing anything that's making sense in this instance to me unless the textbook's solving method for this just skipped a few steps that I'm not recognizing. Can anyone please clarify where this 1/2 is coming from?

Answer 1

`I=1/10sin^5(2x)+C`

Explanation:

We want to to solve

`I=intsin^4(2x)cos(2x)dx`

Make a substitution `u=2x`

`I=intsin^4(u)cos(u)dx`

But we also have to change the "`dx`" to "`du`"

We know if `u=2x` then `(du)/dx=2` or `dx=1/2du`

Hence the "`1/2`"

`I=intsin^4(u)cos(u)1/2du=1/2intsin^4(u)cos(u)du`

To finish the integral, make the second substitution

Let `s=sin(u)` then `(ds)/(du)=cos(u)` or `du=1/cos(u)ds`

`I=1/2ints^4cos(u)1/cos(u)ds`

`=1/2ints^4ds`

`=1/2(1/5s^5)+C`

`=1/10s^5+C`

Substitute back `s=sin(u)` and `u=2x`

`I=1/10sin^5(2x)+C`