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    I did this one in the exam:

    use the substitution x=sec(t) to show that

    (integral between upper:2 and lower:root 2) 1/(x^3*root(x^2 - 1)) dx = (root(3) - 2)/8 + (pi)/24

    I get

    1/8 - (pi)/24

    Anyone see how to do it? (Sorry about the notation - I hope they enable LaTeX here some time!)

    Thanks
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    (Original post by sat)
    I did this one in the exam:

    use the substitution x=sec(t) to show that

    (integral between upper:2 and lower:root 2) 1/(x^3*root(x^2 - 1)) dx = (root(3) - 2)/8 + (pi)/24

    I get

    1/8 - (pi)/24

    Anyone see how to do it? (Sorry about the notation - I hope they enable LaTeX here some time!)

    Thanks
    I get the answer, the integral should simplify to cos^2(t) with bounds as pi/3 and pi/4.
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    (Original post by sat)
    I did this one in the exam:

    use the substitution x=sec(t) to show that

    (integral between upper:2 and lower:root 2) 1/(x^3*root(x^2 - 1)) dx = (root(3) - 2)/8 + (pi)/24

    I get

    1/8 - (pi)/24

    Anyone see how to do it? (Sorry about the notation - I hope they enable LaTeX here some time!)

    Thanks
    x = sect , so dx = [sint/(cost)^2] dt
    x : sqrt(2) -> 2, so t : pi/4 -> pi/3
    x^3sqrt(x^2 - 1) = (sect)^3.tant = sint/(cost)^4
    I = integration { (cost)^4/sint * sint/(cost)^2 dt }
    = I{(cost)^2dt}
    = 1/2* I{cos2t - 1)dt}
    = 1/2* (sin2t/2 - t) (pi/4 -> pi/3)
    = 1/4*[sin(2pi/3) - sin(pi/2)] - pi/24
    = (sqrt(3) - 2)/8 - pi/24.
    :cool:
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    I had got down to the cos^2 bit previously, but I must have chosen the wrong set of trig. identities to go fro there. I did get there in the end!
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    I remember doing that question, is it from this year's paper?
 
 
 
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