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    I'm quite stuck on this

    1
    ______________ dx

    (1-x^2)^1.5)

    Substitution is x=sinθ and the limits given are 0.5 and 0

    any help/advice would be appreciated!
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    (Original post by The Hedonist)
    I'm quite stuck on this

    1
    ______________ dx

    (1-x^2)^1.5)

    Substitution is x=sinθ and the limits given are 0.5 and 0

    any help/advice would be appreciated!
    1/(cos^2theta)^1.5 go from there . Also remember to change the limits.
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    Well what have you tried? Certainly that substitution looks like a good idea.
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    (Original post by Deep456)
    1/(cos^2theta)^1.5 go from there . Also remember to change the limits.
    but surely:

    \frac{dx}{d\theta} = cos\theta so  dx = cos\theta d\theta

    so the integral would be  \frac {cos\theta}{(cos^2\theta)^1^.^5} d\theta
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    (Original post by just george)
    but surely:

    dx/dtheta = cos(theta) so dx = cos(theta) dtheta

    so the integral would be cos(theta) / (cos^2theta)^1.5 dtheta
    All I have done is subbed in sin theta and used the identity. I haven't done anything else.

    What you have done is also correct ,yes.
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    (Original post by Deep456)
    All I have done is subbed in sin theta and used the identity. I haven't done anything else.
    sorry yeah it just looked a bit misleading to me, looked like you were saying the integral was  \frac{1}{(cos^2\theta)^1^.^5}

    The Hedonist - basically just use the substitution given, remember to change between dx and dtheta, the limits, and then hopefully you will be left with something you can integrate

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Updated: March 6, 2012
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