Need a little push on this integral using trig substitution.

[uchuu-man chi]

∫x²√(3+2x-x²) dx

Here's what I've already done:

completed the square
∫x²√(4-(x-1)²) dx

(x-1) = 2sinθ
sinθ = (x-1)/2
x = 2sinθ+1
dx = 2cosθ dθ

trig sub + pulled out constants
4∫(2sinθ+1)²√(1-sin²θ)cosθ dθ

trig identity
4∫(2sinθ+1)²√(cos²θ)cosθ dθ

4∫(2sinθ+1)²(cos²θ)dθ

expanded + trig identity (cosθ = √(1-sin²θ)
4∫(4sin²θ+4sinθ+1)√(1-sin²θ) cosθ dθ

u-sub
u = sinθ
du = cosθ dθ

4∫(4u²+4u+1)√(1-u²) du

I proceeded to multiply them, and split them into 3 integrals. But, I still ended up with the one of the integrals being:

16∫u²√(1-u²)

which is exactly where I started. I feel like I'm missing something painfully obvious. Can someone give me a push into the right direction?

 

[haruspex]

expanded + trig identity (cosθ = √(1-sin2θ)
4∫(4sin2θ+4sinθ+1)√(1-sin2θ) cosθ dθ

This is a backwards step. You just got rid of the √, don't bring it back.
Expand the squared term instead.
How would you deal with ∫cos²?

 

[uchuu-man chi]

This is a backwards step. You just got rid of the √, don't bring it back.
Expand the squared term instead.
How would you deal with ∫cos²?

oo ok thank you I was able to get it.