∫x(adsbygoogle = window.adsbygoogle || []).push({}); ^{2}√(3+2x-x^{2}) dx

Here's what I've already done:

completed the square

∫x^{2}√(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-sin^{2}θ)cosθ dθ

trig identity

4∫(2sinθ+1)^{2}√(cos^{2}θ)cosθ dθ

4∫(2sinθ+1)^{2}(cos^{2}θ)dθ

expanded + trig identity (cosθ = √(1-sin^{2}θ)

4∫(4sin^{2}θ+4sinθ+1)√(1-sin^{2}θ) cosθ dθ

u-sub

u = sinθ

du = cosθ dθ

4∫(4u^{2}+4u+1)√(1-u^{2}) 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∫u^{2}√(1-u^{2})

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?

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# I Need a little push on this integral using trig substitution.

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