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Thread: integral2

  1. #1
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    Question integral2

    integral

    du/(sqrt(1+u^2) )

    any help pls
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by alex83 View Post
    integral
    $\displaystyle
    \int \frac{du}{\sqrt{1+u^2}}$

    any help pls
    You can find this in a table of integrals. It's $\displaystyle arcsinh(u)$. (It's the "inverse hyperbolic sine of u".)
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  3. #3
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    Quote Originally Posted by alex83 View Post
    integral

    du/(sqrt(1+u^2) )

    any help pls
    As you'll see this is a standard trig integral, but it's better to do it by trig substitution.

    Construct a right triangle with a vertical leg of u, a horizontal leg of 1 and by Pythagoras the hypotenuse is $\displaystyle \sqrt{1+u^2}$. Let theta be the angle such that $\displaystyle \tan(\theta)=u$. You will also notice that $\displaystyle \cos(\theta)=\frac{1}{\sqrt{1+u^2}}$, which makes an easy substitution. Make sure to calculate $\displaystyle d\theta$ in terms of u, then make the full substitution, integrate, then back substitute.
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  4. #4
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    a classic problem, we can also put $\displaystyle v=u+\sqrt{1+u^2}.$
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