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Math Help - Intgration again - Trig Sub

  1. #1
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    Intgration again - Trig Sub

    solve:

    \int \frac{dx}{x^2 (\sqrt{x^2+1})}


    I use these trig subs:

    x=tan\theta
    dx=sec^2 \theta \\d\theta

    after substituting and simplifying I end up here:

    \int \frac{sec\theta}{tan^2\theta} \\d\theta


    Any help in getting "unstuck" would be appreciated!
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by Got5onIt View Post
    solve:

    \int \frac{dx}{x^2 (\sqrt{x^2+1})}
    Define x=\frac1u, so the integral becomes to

    -\int\frac u{\sqrt{1+u^2}}\,du=-\sqrt{1+u^2}+k.

    Back substitute and you're done.
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  3. #3
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    Thanks Kriz for the help!

    I can't understand how x=\frac{1}{u} substitutes into the denominator of the original integral and give you \sqrt{1+u^2}? Why is it not \sqrt{1+((\frac{1}{u})^2)} ?
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  4. #4
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    Trig sub works well with this one. You have the correct integral in terms of trig functions.

    I will use x instead of theta for less typing.

    \int\frac{sec{\theta}}{tan^{2}{\theta}}d{\theta}

    Rewrite as \int{csc(x)cot(x)}dx

    Let u=csc(x), \;\ du=-csc(x)cot(x)dx

    That gives the easiest of integrals:

    -\int{du}

    -u

    \boxed{-csc(x)}
    Last edited by galactus; October 31st 2007 at 04:13 PM.
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  5. #5
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    Thanks Galactus! I was hoping the fraction could be rewritten in some other trig form. I did not know it was cscXcotX. Where can I find a reference for trig subs?


    Quote Originally Posted by galactus View Post
    Trig sub works well with this one. You have the correct integral in terms of trig functions.

    I will use x instead of theta for less typing.

    \int\frac{sec{\theta}}{tan^{2}{\theta}}d{\theta}

    Rewrite as \int{csc(x)cot(x)}dx

    Let u=csc(x), \;\ du=-csc(x)cot(x)dx

    That gives the easiest of integrals:

    -\int{du}

    -u

    \boxed{-csc(x)}
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