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integral of secx

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    \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
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    (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*.
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    Recall the integral of f'(x)/f(x).
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    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:
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    this needs to be moved to the health and relationships forum.
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    (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 ?
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    (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.
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    (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
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    (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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