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Math Help - Integral (secx)^-1

  1. #1
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    Integral (secx)^-1

    /int sec^{-1}xdx

    How do I start?
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  2. #2
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    Quote Originally Posted by circuscircus View Post
    /int sec^{-1}xdx

    How do I start?
    Start by expressing the question clearly. Presumably you want:

    integral arcsec(x) dx.

    Start by putting u=arcsec(x), then sec(u)=x, or 1/cos(u)=x.

    RonL
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  3. #3
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    u=arcsecx; sec u = x

    \int arcsecx dx

    \int sec u = \int x

    ln|sec u + tan u| + C_{1} = 1/2 x^2 + C_{2}

    ln|sec(arcsecx) + tan(arcsecx)| = 1/2 x^2 + C_{3}; c_{3}=C_{2}-C_{1}\

    ln|x + \sqrt{x^2-1}| = 1/2 x^2 + C_{3}; C=-C_{3}

    = ln|x + \sqrt{x^2-1}| - 1/2 x^2 + C

    correct?
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  4. #4
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    Should be:

    xsec^{-1}(x)-ln|x+\sqrt{x^{2}-1}|+C

    or some form thereof.
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  5. #5
    Senior Member polymerase's Avatar
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    Quote Originally Posted by circuscircus View Post
    u=arcsecx; sec u = x

    \int arcsecx dx

    \int sec u = \int x

    ln|sec u + tan u| + C_{1} = 1/2 x^2 + C_{2}

    ln|sec(arcsecx) + tan(arcsecx)| = 1/2 x^2 + C_{3}; c_{3}=C_{2}-C_{1}\

    ln|x + \sqrt{x^2-1}| = 1/2 x^2 + C_{3}; C=-C_{3}

    = ln|x + \sqrt{x^2-1}| - 1/2 x^2 + C

    correct?
    No, it should be,

    xarcsecx-ln|x + \sqrt{x^2-1}|+C
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  6. #6
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    Where does xarcsecx come from? Also what happens to 1/2 x^2?
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  7. #7
    Math Engineering Student
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    circus, you need to apply integration by parts
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  8. #8
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    Hello, circuscircus!

    polymerase is absolutely correct . . .


    \int \sec^{\text{-}1}\!(x)\,dx

    Integrate by parts . . .

    \begin{array}{ccccccc}u & = & \sec^{\text{-}1}(x) & \quad & dv & = & dx \\<br />
du & = & \frac{dx}{x\sqrt{x^2-1}} & \quad & v & = & x \end{array}

    Then we have: . x\!\cdot\!\sec^{\text{-}1}(x) - \int\frac{dx}{\sqrt{x^2-1}}

    Trig Substitution: . x\,=\,\sec\theta\quad\Rightarrow\quad dx \,=\,\sec\theta\tan\theta\,d\theta,\;\;\sqrt{x^2-1}\,=\,\tan\theta

    . . Substitute: . x\!\cdot\!\sec^{\text{-}1}(x) - \int\frac{\sec\theta\tan\theta\,d\theta}{\tan\thet  a} \;=\;x\!\cdot\!\sec^{\text{-}1}(x) - \int\sec\theta\,d\theta

    . . and we have: . x\!\cdot\!\sec^{\text{-}1}(x) - \ln|\sec\theta + \tan\theta| + C


    Back-substitute: . x\!\cdot\!\sec^{\text{-}1}(x) - \ln|x + \sqrt{x^2-1}| + C

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  9. #9
    Math Engineering Student
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    Quote Originally Posted by Soroban View Post
    \int\frac{dx}{\sqrt{x^2-1}}
    Here's another attempt

    Set u=x+\sqrt{x^2-1}\implies du=1+\frac x{\sqrt{x^2-1}}\,dx\,\therefore\,du=\frac u{\sqrt{x^2-1}}\,dx, the integral becomes to

    \int\frac1{\sqrt{x^2-1}}\,dx=\int\frac1u\,du=\ln|u|+k

    Back substitute and we are done.
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