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Math Help - intergration with inverse trig

  1. #1
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    intergration with inverse trig

    Evaluate: \int_{\sqrt{2}}^{2} \dfrac{dx}{x\sqrt{x^{2}-1}}

    \dfrac{1}{x\sqrt{x^{2}-1}}dx is the same as  arcsec(x)

    im not able to evaluate it from  \sqrt{2} to 2
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  2. #2
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    Quote Originally Posted by viet View Post
    Evaluate: \int_{\sqrt{2}}^{2} \dfrac{dx}{x\sqrt{x^{2}-1}}

    \dfrac{1}{x\sqrt{x^{2}-1}}dx is the same as  arcsec(x)

    im not able to evaluate it from  \sqrt{2} to 2
    1 / [x*sqrt(x^2 -1)] dx is not the same as arcsec(x).
    Rather, arcsec(x) +C = INT.[1 / x*sqrt(x^2 +1)]dx.

    So, after integration,
    = arcsec(x) | (sqrt(2) to 2)
    = arcsec(2) -arcsec(sqrt(2))
    = pi/3 -pi/4
    = pi/12 -----------------answer.
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  3. #3
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    Quote Originally Posted by ticbol View Post
    1 / [x*sqrt(x^2 -1)] dx is not the same as arcsec(x).
    Rather, arcsec(x) +C = INT.[1 / x*sqrt(x^2 +1)]dx.

    So, after integration,
    = arcsec(x) | (sqrt(2) to 2)
    = arcsec(2) -arcsec(sqrt(2))
    = pi/3 -pi/4
    = pi/12 -----------------answer.
    It should be,
    \sec^{-1}|x|+C
    Note the absolute value.
    In this case it does not matter because of the integration limits are non-negative. But in general it must have an absolute value attached.
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  4. #4
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    Quote Originally Posted by ThePerfectHacker View Post
    It should be,
    \sec^{-1}|x|+C
    Note the absolute value.
    In this case it does not matter because of the integration limits are non-negative. But in general it must have an absolute value attached.
    Yeah?
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