integral of secx

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  1. bananarama2's Avatar
    1 15-02-2011  20:12
    • TSR Legend
    integral of secx
    \displaystyle\ resolve

    \displaystyle\int_0^{35^{\circ}} \ sec(x) \ dx \longrightarrow
    \displaystyle\int_0^{35^{\circ}} \ sec(x) \frac{sec(x)+tan(x)}{sec(x)+tan( x)} \ dx ....obviously ( I was told this was the first step)

    \displaystyle\ sec(x)+tan(x) = u

    \displaystyle\frac{du}{dx} = sec(x)\times tan(x) + sec^2(x) \ and \ then \ I'm \ stuck
  2. Farhan.Hanif93's Avatar
    2 15-02-2011  20:24
    • Section Moderator
    • PS Helper
    • TSR Idol
    • Location: Cambridge
    Re: integral of secx
    (Original post by wcp100)
    \displaystyle\ resolve

    \displaystyle\int_0^{35^{\circ}} \ sec(x) \ dx \longrightarrow
    \displaystyle\int_0^{35^{\circ}} \ sec(x) \frac{sec(x)+tan(x)}{sec(x)+tan( x)} \ dx ....obviously ( I was told this was the first step)

    \displaystyle\ sec(x)+tan(x) = u

    \displaystyle\frac{du}{dx} = sec(x)\times tan(x) + sec^2(x) \ and \ then \ I'm \ stuck
    Immediately there's a problem as your limits do not make sense. They must either be in radians or as a fraction of 360.

    What you need to note is that your integrand can be rewritten as:
    \dfrac{\sec^2 x + \sec x \tan x}{\tan x + \sec x}

    What is the derivative of \tan x + \sec x? Does it look familiar? What do you do with integrals with integrands of the form \dfrac{f'(x)}{f(x)}? *cough* natural logarithm *cough*.
    Last edited by Farhan.Hanif93; 15-02-2011 at 20:26.
  3. kfkle's Avatar
    3 15-02-2011  20:25
    • Respected Member
    Re: integral of secx
    Recall the integral of f'(x)/f(x).
  4. a²+b² = c²'s Avatar
    4 16-02-2011  15:19
    • New Member
    • Posts: 20
    Re: integral of secx
    I don't like that way of finding the integral of the secant.
    It assumes that we already know the answer, doesn't it?
    Why the hell would we multiply by that otherwise? :dontknow:
  5. nosignal456's Avatar
    5 16-02-2011  15:21
    • Exalted Member
    • Posts: 320
    Re: integral of secx
    this needs to be moved to the health and relationships forum.
    Post rating:
    2
  6. bananarama2's Avatar
    6 16-02-2011  19:17
    • TSR Legend
    Re: integral of secx
    (Original post by a²+b² = c²)
    I don't like that way of finding the integral of the secant.
    It assumes that we already know the answer, doesn't it?
    Why the hell would we multiply by that otherwise? :dontknow:
    You were thinking exactly what I was. Is there another way of doing it ?
  7. Farhan.Hanif93's Avatar
    7 16-02-2011  19:30
    • Section Moderator
    • PS Helper
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    • Location: Cambridge
    Re: integral of secx
    (Original post by a²+b² = c²)
    I don't like that way of finding the integral of the secant.
    It assumes that we already know the answer, doesn't it?
    Why the hell would we multiply by that otherwise? :dontknow:
    (Original post by wcp100)
    You were thinking exactly what I was. Is there another way of doing it ?
    Yes, there is.

    Note that \sec x = \dfrac{1}{\cos x} = \dfrac{\cos x}{1-\sin ^2x} = \dfrac{\cos x}{(1+\sin x)(1-\sin x)}.

    Then use a substitution of u =\sin x and use partial fractions.
  8. roar558's Avatar
    8 16-02-2011  19:33
    • Overlord in Training
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    Re: integral of secx
    (Original post by wcp100)
    \displaystyle\ resolve

    \displaystyle\int_0^{35^{\circ}} \ sec(x) \ dx \longrightarrow
    \displaystyle\int_0^{35^{\circ}} \ sec(x) \frac{sec(x)+tan(x)}{sec(x)+tan( x)} \ dx ....obviously ( I was told this was the first step)

    \displaystyle\ sec(x)+tan(x) = u

    \displaystyle\frac{du}{dx} = sec(x)\times tan(x) + sec^2(x) \ and \ then \ I'm \ stuck
    well you could just substitute in du for
    sec(x)\times tan(x) + sec^2(x) and u for

    \displaystyle\ sec(x)+tan(x) to get

    \displaystyle\int_0^{35^{\circ}} \frac{1}{u} \ du
  9. Get me off the £\?%!^@ computer's Avatar
    9 16-02-2011  22:47
    • Benevolent Member
    • Location: .
    Re: integral of secx
    (Original post by wcp100)
    You were thinking exactly what I was. Is there another way of doing it ?

    You can also use the well known t substitution.

    t=tan(x/2)

    cos x = (1-t^2)/(1+t^2)

    dx/dt=2/(1+t^2)

    It's all very straightforward and requires no cleverness at all.
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