- #1
uchuu-man chi
- 5
- 0
∫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?
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?