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Hard Integration Question!!!

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    can anyone take me through this answer plz?

    use substitution x = sin& to integrate :

    1 / (1 - x²) ^ -3/2 dx.
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    Well you've been given the substitution so what have you done so far?
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    (Original post by Eager PPe ist)
    can anyone take me through this answer plz?

    use substitution x = sin& to integrate :

    1 / (1 - x²) ^ -3/2 dx.
    Well firstly you must know 1-sin^2x=cos^x
    then you must differentiate x = sin& to get dx/d&=cos&

    know what to do next?
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    The way you have set out the question is not very clear.

    If you mean:
    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    \int{\frac{1}{(1-x^2)^\frac{3}{2}}dx


    Then using a substitution of x = sinu think trig identities.
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    that's what i mean AEsp.

    i have used tirg identities and have integrated and have arrived here:

    I = -0.5sin2x ( 1 + cos2x) ^ -1/2
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    Ok then i believe your answer is wrong.

    

\displaystyle \int \frac{1}{(1-x^2)^{\frac{3}{2}}} dx



= \displaystyle \int (1-x^2)^{-\frac{3}{2}} dx

    Then using substitution x = sin u
    and trig identity  cos^2x + sin^2x = 1
    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    \frac{dx}{du} = cos u



    \displaystyle \int {(cos^2u)^{-\frac{3}{2}}cos u\, du



    \displaystyle = \int \frac{cos u}{\sqrt {(cos^2u)^3}}\, du



    \displaystyle = \int \frac{cosu}{cos^3u}\, du



    \displaystyle = \int \frac{1}{cos^2u}\, du



    \displaystyle = \int sec^2u\, du


    Im sure you recognise the integral of sec^2u
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    http://www.wolframalpha.com/input/?i=integrate+%281%2F%281+-+x%C2%B2%29%29^-%283%2F2%29+dx.+

    integral 1/(1/(1-x^2))^(3/2) dx
    Expanding the integrand 1/(1/(1-x^2))^(3/2) gives -2 sqrt(1/(1-x^2)) x^2+sqrt(1/(1-x^2))+sqrt(1/(1-x^2)) x^4:
    = integral (-2 sqrt(1/(1-x^2)) x^2+sqrt(1/(1-x^2))+sqrt(1/(1-x^2)) x^4) dx
    Integrate the sum term by term and factor out constants:
    = integral sqrt(1/(1-x^2)) dx-2 integral x^2 sqrt(1/(1-x^2)) dx+ integral x^4 sqrt(1/(1-x^2)) dx
    For the integrand x^2 sqrt(1/(1-x^2)), simplify powers:
    = integral sqrt(1/(1-x^2)) dx-2 integral x^2/sqrt(1-x^2) dx+ integral x^4 sqrt(1/(1-x^2)) dx
    For the integrand, x^2/sqrt(1-x^2) substitute x = sin(u) and dx = cos(u) du. Then sqrt(1-x^2) = sqrt(1-sin^2(u)) = cos(u) and u = sin^(-1)(x):
    = -2 integral sin^2(u) du+ integral sqrt(1/(1-x^2)) dx+ integral x^4 sqrt(1/(1-x^2)) dx
    For the integrand x^4 sqrt(1/(1-x^2)), simplify powers:
    = -2 integral sin^2(u) du+ integral sqrt(1/(1-x^2)) dx+ integral x^4/sqrt(1-x^2) dx
    For the integrand, x^4/sqrt(1-x^2) substitute x = sin(s) and dx = cos(s) ds. Then sqrt(1-x^2) = sqrt(1-sin^2(s)) = cos(s) and s = sin^(-1)(x):
    = integral sin^4(s) ds-2 integral sin^2(u) du+ integral sqrt(1/(1-x^2)) dx
    For the integrand sqrt(1/(1-x^2)), simplify powers:
    = integral sin^4(s) ds-2 integral sin^2(u) du+ integral 1/sqrt(1-x^2) dx
    The integral of 1/sqrt(1-x^2) is sin^(-1)(x):
    = integral sin^4(s) ds-2 integral sin^2(u) du+sin^(-1)(x)
    Write sin^2(u) as 1/2-1/2 cos(2 u):
    = integral sin^4(s) ds-2 integral (1/2-1/2 cos(2 u)) du+sin^(-1)(x)
    Integrate the sum term by term and factor out constants:
    = integral sin^4(s) ds-2 integral 1/2 du+ integral cos(2 u) du+sin^(-1)(x)
    The integral of 1/2 is u/2:
    = integral sin^4(s) ds-u+ integral cos(2 u) du+sin^(-1)(x)
    For the integrand cos(2 u), substitute p = 2 u and dp = 2 du:
    = 1/2 integral cos(p) dp+ integral sin^4(s) ds-u+sin^(-1)(x)
    The integral of cos(p) is sin(p):
    = (sin(p))/2+ integral sin^4(s) ds-u+sin^(-1)(x)
    Use the reduction formula, integral sin^m(s) ds = -(cos(s) sin^(m-1)(s))/m + (m-1)/m integral sin^(-2+m)(s) ds, where m = 4:
    = (sin(p))/2-1/4 sin^3(s) cos(s)+3/4 integral sin^2(s) ds-u+sin^(-1)(x)
    Write sin^2(s) as 1/2-1/2 cos(2 s):
    = (sin(p))/2-1/4 sin^3(s) cos(s)+3/4 integral (1/2-1/2 cos(2 s)) ds-u+sin^(-1)(x)
    Integrate the sum term by term and factor out constants:
    = (sin(p))/2-1/4 sin^3(s) cos(s)+3/4 integral 1/2 ds-3/8 integral cos(2 s) ds-u+sin^(-1)(x)
    For the integrand cos(2 s), substitute w = 2 s and dw = 2 ds:
    = (sin(p))/2-1/4 sin^3(s) cos(s)+3/4 integral 1/2 ds-u-3/16 integral cos(w) dw+sin^(-1)(x)
    The integral of 1/2 is s/2:
    = (sin(p))/2+(3 s)/8-1/4 sin^3(s) cos(s)-u-3/16 integral cos(w) dw+sin^(-1)(x)
    The integral of cos(w) is sin(w):
    = (sin(p))/2+(3 s)/8-1/4 sin^3(s) cos(s)-u-(3 sin(w))/16+sin^(-1)(x)+constant
    Substitute back for w = 2 s:
    = (sin(p))/2+(3 s)/8-1/4 sin^3(s) cos(s)-3/8 sin(s) cos(s)-u+sin^(-1)(x)+constant
    Substitute back for p = 2 u:
    = (3 s)/8-1/4 sin^3(s) cos(s)-3/8 sin(s) cos(s)-u+sin(u) cos(u)+sin^(-1)(x)+constant
    Substitute back for s = sin^(-1)(x):
    = -u+sin(u) cos(u)-3/8 sqrt(1-x^2) x-1/4 sqrt(1-x^2) x^3+11/8 sin^(-1)(x)+constant
    Substitute back for u = sin^(-1)(x):
    = 5/8 sqrt(1-x^2) x-1/4 sqrt(1-x^2) x^3+3/8 sin^(-1)(x)+constant
    Factor the answer a different way:
    = 1/8 (x sqrt(1-x^2) (5-2 x^2)+3 sin^(-1)(x))+constant
    Which is equivalent for restricted x values to:
    = 1/8 sqrt(1/(1-x^2)) (2 x^5-7 x^3+3 sqrt(1-x^2) sin^(-1)(x)+5 x)+constant
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    (Original post by g118)
    ..
    There's no need to post more than the link.

    See also the Guide to Posting and what it has to say about posting full solutions.


    If you read the thread, you'd also see that we don't think that's what the OP was actually trying to solve. (Although it is what he said he was trying to solve).
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    (Original post by g118)
    http://www.wolframalpha.com/input/?i=integrate+%281%2F%281+-+x%C2%B2%29%29^-%283%2F2%29+dx.+

    integral 1/(1/(1-x^2))^(3/2) dx
    Expanding the integrand 1/(1/(1-x^2))^(3/2) gives -2 sqrt(1/(1-x^2)) x^2+sqrt(1/(1-x^2))+sqrt(1/(1-x^2)) x^4:
    = integral (-2 sqrt(1/(1-x^2)) x^2+sqrt(1/(1-x^2))+sqrt(1/(1-x^2)) x^4) dx
    Integrate the sum term by term and factor out constants:
    = integral sqrt(1/(1-x^2)) dx-2 integral x^2 sqrt(1/(1-x^2)) dx+ integral x^4 sqrt(1/(1-x^2)) dx
    For the integrand x^2 sqrt(1/(1-x^2)), simplify powers:
    = integral sqrt(1/(1-x^2)) dx-2 integral x^2/sqrt(1-x^2) dx+ integral x^4 sqrt(1/(1-x^2)) dx
    For the integrand, x^2/sqrt(1-x^2) substitute x = sin(u) and dx = cos(u) du. Then sqrt(1-x^2) = sqrt(1-sin^2(u)) = cos(u) and u = sin^(-1)(x):
    = -2 integral sin^2(u) du+ integral sqrt(1/(1-x^2)) dx+ integral x^4 sqrt(1/(1-x^2)) dx
    For the integrand x^4 sqrt(1/(1-x^2)), simplify powers:
    = -2 integral sin^2(u) du+ integral sqrt(1/(1-x^2)) dx+ integral x^4/sqrt(1-x^2) dx
    For the integrand, x^4/sqrt(1-x^2) substitute x = sin(s) and dx = cos(s) ds. Then sqrt(1-x^2) = sqrt(1-sin^2(s)) = cos(s) and s = sin^(-1)(x):
    = integral sin^4(s) ds-2 integral sin^2(u) du+ integral sqrt(1/(1-x^2)) dx
    For the integrand sqrt(1/(1-x^2)), simplify powers:
    = integral sin^4(s) ds-2 integral sin^2(u) du+ integral 1/sqrt(1-x^2) dx
    The integral of 1/sqrt(1-x^2) is sin^(-1)(x):
    = integral sin^4(s) ds-2 integral sin^2(u) du+sin^(-1)(x)
    Write sin^2(u) as 1/2-1/2 cos(2 u):
    = integral sin^4(s) ds-2 integral (1/2-1/2 cos(2 u)) du+sin^(-1)(x)
    Integrate the sum term by term and factor out constants:
    = integral sin^4(s) ds-2 integral 1/2 du+ integral cos(2 u) du+sin^(-1)(x)
    The integral of 1/2 is u/2:
    = integral sin^4(s) ds-u+ integral cos(2 u) du+sin^(-1)(x)
    For the integrand cos(2 u), substitute p = 2 u and dp = 2 du:
    = 1/2 integral cos(p) dp+ integral sin^4(s) ds-u+sin^(-1)(x)
    The integral of cos(p) is sin(p):
    = (sin(p))/2+ integral sin^4(s) ds-u+sin^(-1)(x)
    Use the reduction formula, integral sin^m(s) ds = -(cos(s) sin^(m-1)(s))/m + (m-1)/m integral sin^(-2+m)(s) ds, where m = 4:
    = (sin(p))/2-1/4 sin^3(s) cos(s)+3/4 integral sin^2(s) ds-u+sin^(-1)(x)
    Write sin^2(s) as 1/2-1/2 cos(2 s):
    = (sin(p))/2-1/4 sin^3(s) cos(s)+3/4 integral (1/2-1/2 cos(2 s)) ds-u+sin^(-1)(x)
    Integrate the sum term by term and factor out constants:
    = (sin(p))/2-1/4 sin^3(s) cos(s)+3/4 integral 1/2 ds-3/8 integral cos(2 s) ds-u+sin^(-1)(x)
    For the integrand cos(2 s), substitute w = 2 s and dw = 2 ds:
    = (sin(p))/2-1/4 sin^3(s) cos(s)+3/4 integral 1/2 ds-u-3/16 integral cos(w) dw+sin^(-1)(x)
    The integral of 1/2 is s/2:
    = (sin(p))/2+(3 s)/8-1/4 sin^3(s) cos(s)-u-3/16 integral cos(w) dw+sin^(-1)(x)
    The integral of cos(w) is sin(w):
    = (sin(p))/2+(3 s)/8-1/4 sin^3(s) cos(s)-u-(3 sin(w))/16+sin^(-1)(x)+constant
    Substitute back for w = 2 s:
    = (sin(p))/2+(3 s)/8-1/4 sin^3(s) cos(s)-3/8 sin(s) cos(s)-u+sin^(-1)(x)+constant
    Substitute back for p = 2 u:
    = (3 s)/8-1/4 sin^3(s) cos(s)-3/8 sin(s) cos(s)-u+sin(u) cos(u)+sin^(-1)(x)+constant
    Substitute back for s = sin^(-1)(x):
    = -u+sin(u) cos(u)-3/8 sqrt(1-x^2) x-1/4 sqrt(1-x^2) x^3+11/8 sin^(-1)(x)+constant
    Substitute back for u = sin^(-1)(x):
    = 5/8 sqrt(1-x^2) x-1/4 sqrt(1-x^2) x^3+3/8 sin^(-1)(x)+constant
    Factor the answer a different way:
    = 1/8 (x sqrt(1-x^2) (5-2 x^2)+3 sin^(-1)(x))+constant
    Which is equivalent for restricted x values to:
    = 1/8 sqrt(1/(1-x^2)) (2 x^5-7 x^3+3 sqrt(1-x^2) sin^(-1)(x)+5 x)+constant
    3 years on, I think they may have worked it out
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    (Original post by DFranklin)
    There's no need to post more than the link.

    See also the Guide to Posting and what it has to say about posting full solutions.


    If you read the thread, you'd also see that we don't think that's what the OP was actually trying to solve. (Although it is what he said he was trying to solve).
    The thread is from April 2008, too. :lol:
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    (Original post by arob752)
    3 years on, I think they may have worked it out
    Well they did say it was a hard question, so maybe they are still working on it? :P

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