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Thread: Integration by parts of arcsec(x)

  1. #1
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    Integration by parts of arcsec(x)

    Having problems with the substitution. I don't see how it fits, although I feel like I am doing it correctly.

    int arcsec(x) dx
    =xarcsec(x) - int (x)/((x)(sqrt((x^2)-1)))
    =xarcsec(x) - int 1/(sqrt(x^2)-1)

    Here I try to use substitution but it doesn't seem to go well for me. Thanks in advance!
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  2. #2
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    Quote Originally Posted by Beeorz View Post
    Having problems with the substitution. I don't see how it fits, although I feel like I am doing it correctly.

    int arcsec(x) dx
    =xarcsec(x) - int (x)/((x)(sqrt((x^2)-1)))
    =xarcsec(x) - int 1/(sqrt(x^2)-1)

    Here I try to use substitution but it doesn't seem to go well for me. Thanks in advance!
    The easy way is to make a hyperbolic substitution.

    The hard way is to make the substitution $\displaystyle x = \sec \theta$.

    Show the details of your calculation.
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  3. #3
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    I've tried using:

    u=arcsec(x)
    du=1/(xsqrt((x^2)-1)))

    Seems like the correct choice but I don't see how it would work. By hyperbolic sub-ing do you mean putting in:

    arccosh(x) = 1/sqrt((x^2)-1)
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  4. #4
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    Hello, Beeorz!

    Your work is correct . . .


    $\displaystyle \int \text{arcsec}(x)\,dx $


    . . $\displaystyle \begin{array}{cccccccc}u &=& \text{arcsec}(x) & & dv &=& dx \\ \\[-3mm]
    du &=&\dfrac{dx}{x\sqrt{x^2-1}} & & v &=& x \end{array}$


    $\displaystyle x\!\cdot\!\text{arcsec}(x) - \int\frac{x}{x\sqrt{x^2-1}}\,dx \;=\;x\!\cdot\text{arcsec}(x) - \int \frac{dx}{\sqrt{x^2-1}}$ . . . . good work!

    Let: .$\displaystyle x \:=\:\sec\theta \quad\Rightarrow\quad dx \:=\:\sec\theta\tan\theta\,d\theta$


    $\displaystyle x\!\cdot\!\text{arcsec}(x) - \int\frac{\sec\theta\tan\theta\,d\theta}{\tan\thet a} \;=\;x\!\cdot\!\text{arcsec}(x) - \int\sec\theta\,d\theta \;=$ .$\displaystyle x\!\cdot\!\text{arcsec}(x) - \ln|\sec\theta + \tan\theta| + C$


    Back-substitute: .$\displaystyle \sec\theta \:=\:x\quad\Rightarrow\quad \tan\theta \:=\:\sqrt{x^2-1}$


    Therefore: .$\displaystyle \boxed{x\!\cdot\!\text{arcsec}(x) - \ln|x + \sqrt{x^2-1}| + C}$

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    errr still confused as to how you are getting tan on the bottom...
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  6. #6
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    $\displaystyle sec^2\theta - 1 = tan^2\theta$

    so with the substitution of $\displaystyle x = sec\theta$

    $\displaystyle \sqrt{sec^2\theta -1} = tan\theta$
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  7. #7
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    Quote Originally Posted by Beeorz View Post
    Seems like the correct choice but I don't see how it would work. By hyperbolic sub-ing do you mean putting in:
    arccosh(x) = 1/sqrt((x^2)-1)
    $\displaystyle \frac{d}{dx} \cosh^{-1}{x} = \frac{1}{\sqrt{x^2-1}}$
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