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Math Help - integrating hyperbolic functions

  1. #1
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    integrating hyperbolic functions

    how do i integrate the following?

    1/sqrt(x^2+4*x+13)


    also,

    how do i find dy/dx for

    y=cosh^-1(secx)?

    Thanks!
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  2. #2
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    Re: integrating hyperbolic functions

    What ideas have you had so far?
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: integrating hyperbolic functions

    Quote Originally Posted by Rine198 View Post
    how do i integrate the following?

    1/sqrt(x^2+4*x+13)


    also,

    how do i find dy/dx for

    y=cosh^-1(secx)?

    Thanks!
    For the integral part:

    \int\frac{dx}{\sqrt{x^2+4x+13}}=\int\frac{dx}{\sqr  t{(x+2)^2+9}}\overset{x+2=t}{=}\int\frac{dt}{\sqrt  {t^2+9}}\overset{t=3tan(v)}{=}\int\frac{3sec^2(v)d  v}{\sqrt{9tan^2(v)+9}}=\int\frac{3sec^2(v)dv}{3sex  (v)}=\int sec(v)dv=\log{(tan(v)+sex(v))}+C \overset{v=arctan(\frac{t}{3})}{=}

    \log{\frac{1}{3}(\sqrt{t^2+9}+t)}+C \overset{t=x+2}{=} \log{\frac{1}{3}(\sqrt{x^2+4x+13}+x+2)}+C
    Last edited by Also sprach Zarathustra; June 13th 2011 at 07:42 PM.
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  4. #4
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    Re: integrating hyperbolic functions

    Quote Originally Posted by Rine198 View Post
    how do i integrate the following?

    1/sqrt(x^2+4*x+13)


    also,

    how do i find dy/dx for

    y=cosh^-1(secx)?

    Thanks!
    \displaystyle \begin{align*}y &= \cosh^{-1}{(\sec{x})} \\ \cosh{y} &= \sec{x} \\ \frac{d}{dx}(\cosh{y}) &= \frac{d}{dx}(\sec{x}) \\ \frac{d}{dy}(\cosh{y})\,\frac{dy}{dx} &= \frac{d}{dx}\left[(\cos{x})^{-1}\right] \\ \sinh{y}\,\frac{dy}{dx} &= \sin{x}\cos^{-2}{x} \\ \frac{dy}{dx} &= \frac{\sin{x}}{\cos^2{x}\sinh{y}} \\ \frac{dy}{dx} &= \frac{\sin{x}}{\cos^2{x}\sinh{\left[\cosh^{-1}{(\sec{x})}\right]}} \\ \frac{dy}{dx} &= \frac{\sin{x}}{\cos^2{x}\sqrt{\left\{\cosh{\left[\cosh^{-1}{\left(\sec{x}\right)}\right]}\right\}^2 - 1}} \\ \frac{dy}{dx} &= \frac{\sin{x}}{\cos^2{x}\sqrt{\sec^2{x} - 1}} \\ \frac{dy}{dx} &= \frac{\sin{x}}{\cos^2{x}\sqrt{\tan^2{x}}} \\ \frac{dy}{dx} &= \frac{\sin{x}}{\cos^2{x}\tan{x}} \\ \frac{dy}{dx} &= \frac{\sin{x}}{\frac{\cos{x}}{\sin{x}}} \\ \frac{dy}{dx} &= \frac{\sin^2{x}}{\cos{x}}\end{align*}
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  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: integrating hyperbolic functions

    y=cosh^{-1}(sec(x))


    (cosh^{-1}(x))'=\frac{1}{\sqrt{x+1}\sqrt{x-1}}


    (sec(x))'=tan(x)sec(x)


    Hence:

    {cosh^{-1}(sec(x))}'=(sec(x))'\frac{1}{\sqrt{sec(x)+1}\sqr  t{sec(x)-1}}=\frac{tan(x)sec(x)}{\sqrt{sec(x)+1}\sqrt{sec(x  )-1}}
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