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

  1. Mar 15, 2017 #1
    ∫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?
     
  2. jcsd
  3. Mar 15, 2017 #2

    haruspex

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    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?
     
  4. Mar 15, 2017 #3
    oo ok thank you I was able to get it.
     
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