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Math Help - Primitive of Reciprocal of sin^3 ax

  1. #1
    Super Member Matt Westwood's Avatar
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    Primitive of Reciprocal of sin^3 ax

    While I do some investigation of the complicated induction work on no. 335 ...

    No. 352 in my list (there were a load of easy sine formulas):

    \int \frac {\mathrm d x} {\sin^3 a x}

    which is supposed to work out to:

    -\frac {\cos a x} {2 a \sin^2 a x} + \frac 1 {2 a} \ln \left\vert{\tan \frac {a x} 2}\right\vert

    but is going a bit astray.

    I tried Parts, with u = \frac 1 {\sin a x} and got so far:

    -\frac {\cos a x} {a \sin^2 a x} - \int \frac {\cos^2 a x} {\sin^3 a x} \ \mathrm d x

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    Re: Primitive of Reciprocal of sin^3 ax

    your first answer is correct
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    Re: Primitive of Reciprocal of sin^3 ax

    oh you're having trouble getting there?
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    Re: Primitive of Reciprocal of sin^3 ax

    The last part is actually not as bad:
    S cos^2(ax)/sin^3(ax) dx right? Rewrite as
    S cot^2(ax)*csc(ax) dx
    Letting u=csc(ax), we have du=-a*cot^2(ax)dx. Then, cot^2(ax)dx = (-1/a)du Now the integral becomes
    S u*(-1/a)du = (-1/a) S u du = (-1/a) u^2/2 = (-1/a) csc^2(ax)/2 = (-1/2a) csc^2(ax) = -1/[2a*sin^2(ax)].

    Hmm... the natural logarithm doesn't show up anywhere. Maybe you should have started with S csc^3(ax) dx instead...
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    Super Member Matt Westwood's Avatar
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    Re: Primitive of Reciprocal of sin^3 ax

    Quote Originally Posted by dennydengler View Post
    The last part is actually not as bad:
    S cos^2(ax)/sin^3(ax) dx right? Rewrite as
    S cot^2(ax)*csc(ax) dx
    Letting u=csc(ax), we have du=-a*cot^2(ax)dx. Then, cot^2(ax)dx = (-1/a)du Now the integral becomes
    S u*(-1/a)du = (-1/a) S u du = (-1/a) u^2/2 = (-1/a) csc^2(ax)/2 = (-1/2a) csc^2(ax) = -1/[2a*sin^2(ax)].

    Hmm... the natural logarithm doesn't show up anywhere. Maybe you should have started with S csc^3(ax) dx instead...
    Yep, I'll try that properly -- except that \frac d{dx} \csc ax \ne a \cot^2 ax ... this appears to be where the ln comes in.
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    Re: Primitive of Reciprocal of sin^3 ax

    Oh oops, bad mistake.
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    Re: Primitive of Reciprocal of sin^3 ax

    Thanks from Matt Westwood and topsquark
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    Super Member Matt Westwood's Avatar
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    Re: Primitive of Reciprocal of sin^3 ax

    Quote Originally Posted by Matt Westwood View Post
    Yep, I'll try that properly -- except that \frac d{dx} \csc ax \ne a \cot^2 ax ... this appears to be where the ln comes in.
    I think the substitution to use is u = \sin^2 ax and then it turns into a simple integral \int \frac {1 - u^2}u \ \du unless I've blundered.
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    Super Member Matt Westwood's Avatar
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    Re: Primitive of Reciprocal of sin^3 ax

    I'm wishing I didn't know about this now, it looks like it might take all the fun away ...
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    Re: Primitive of Reciprocal of sin^3 ax

    sorry, just wanted to help. i though maybe you would be able to see where you went wrong in your steps.
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    Super Member Matt Westwood's Avatar
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    Re: Primitive of Reciprocal of sin^3 ax

    Quote Originally Posted by tinspire View Post
    sorry, just wanted to help. i though maybe you would be able to see where you went wrong in your steps.
    Yer alrite pal ... British sense of humour.
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    Re: Primitive of Reciprocal of sin^3 ax

    As mentioned earlier your new integral is

    - \int \csc ax \cot^2 ax \, dx =  - \int \csc ax (\csc^2 ax -1) \, dx
    = - \int \csc^3 ax \, dx  + \int \csc ax \, dx

    Now move over the csc^3 ax integral to the other side of the equation and evaluate the single csc ax integral.
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    Re: Primitive of Reciprocal of sin^3 ax

    Quote Originally Posted by Matt Westwood View Post
    Yer alrite pal ... British sense of humour.
    oh ok. nevermind, not sorry
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