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Find the error

  1. Nov 17, 2013 #1
    ∫√(16-9χ²) dx

    this is what I tried

    let 9/16 χ² = sin²

    then x=4/3(sinu) and u=arcsin(3/4x) and dx=4/3(cosu) du

    ∫√(16(1-sin²u) )x 4/3(cosu) du

    ∫16/3 cos²u du

    ∫8/3(cos2u + 1) du

    4/3(sin2u) +8/3 u +k

    4/3(2sinucosu) +8/3 u +k

    8/3 (sin(arcsin(3/4χ))xcos(arcsin(3/4χ))) +8/3(arcsin(3/4χ))+k

    2χ√(1-16/9x²) +8/3arcsin(3/4χ) +k
     
  2. jcsd
  3. Nov 18, 2013 #2
    From the way you wrote it, it's a little hard for me to tell exactly what you did. Suppose I start the same way you did:

    ##\int \sqrt{16-9x^2} dx = 4\int \sqrt{1-9x/16}dx \hspace{50px}## (1)

    Let sin u = 3x/4; cos u du =3/4 dx; So (1) becomes

    ##3\int \sqrt {1-sin^2}(u) cos(u) du = 3\int cos^2(u) du##

    You can finish this up by using the half angle formula on the cos.
     
  4. Nov 18, 2013 #3
    can u solve it for me I wanna check plus how did that 4 outta the integral turned 3
     
    Last edited: Nov 18, 2013
  5. Nov 18, 2013 #4
    We do not solve anything for you. Moreover, you've failed to apply the standard PF format.
     
  6. Nov 18, 2013 #5
    Hi Saq,

    du = 3/4 dx.

    As Dirk says, we don't do your homework -- we provide hints, clues, get you started etc. I think I did enough that you should be able to finish up.
     
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