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Math Help - Integral secant and csc

  1. #1
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    Integral secant and csc

    Can someone explain to me the substiution which is made for these integrals, I never find out why, it is they just just do it, and have no reasoning, perhaps a little insight could help
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by RockHard View Post
    Can someone explain to me the substiution which is made for these integrals, I never find out why, it is they just just do it, and have no reasoning, perhaps a little insight could help
    The trick is to multiply secant by \frac{\sec x+\tan x}{\sec x+\tan x} and cosecant by \frac{\csc x+\cot x}{\csc x+\cot x}. The substitution for these integrals should then become obvious.
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  3. #3
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    Quote Originally Posted by RockHard View Post
    Can someone explain to me the substiution which is made for these integrals, I never find out why, it is they just just do it, and have no reasoning, perhaps a little insight could help
    can't tell you how someone came up with it, but is sure is elegant ...

    \int \sec{x} \, dx

    \int \sec{x} \cdot \frac{\sec{x}+\tan{x}}{\sec{x}+\tan{x}} \, dx

    \int \frac{\sec^2{x}+\sec{x}\tan{x}}{\sec{x}+\tan{x}} \, dx

    let u = \sec{x}+\tan{x}

    du = (\sec{x}\tan{x} + \sec^2{x}) \, dx

    substitute ...

    \int \frac{du}{u}

    \ln|u| + C

    back substitute ...

    \ln|\sec{x}+\tan{x}| + C
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    Yes, I know you use that to arrive by substituting the denominator to get the natural log, however, I am one of those that want to know why rather than this is how you do it and the hows its goes with no reason. sorry to be a pain thanks for the input though. I just wanted to know how do you manage or who decided this will work, as in the reasoning.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by skeeter View Post
    can't tell you how someone came up with it, but is sure is elegant ...

    \int \sec{x} \, dx

    \int \sec{x} \cdot \frac{\sec{x}+\tan{x}}{\sec{x}+\tan{x}} \, dx
    Indeed!

    Had I not been aware of this, I would have done it differently. The idea is similar, I would have multiplied by \frac {\cos x}{\cos x}. More intuitive, and a bit more work and certainly not as elegant. But you get to the same thing eventually.
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  6. #6
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    or


     \sec(x) = \sec(x) \cdot \frac{\sqrt{ \sec^2(x)-1}}{\sqrt{ \sec^2(x)-1}}

     = \frac{ \sec(x) \tan(x) }{\sqrt{ \sec^2(x)-1}}

     = \frac{d(\sec(x))}{dx}\cdot  \frac{1}{\sqrt{ \sec^2(x)-1}}

     \int \sec(x) ~dx = \int \frac{d(\sec(x))}{\sqrt{ \sec^2(x)-1}}

     =\cosh^{-1}(\sec(x)) + C
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by RockHard View Post
    .... I just wanted to know how do you manage or who decided this will work, as in the reasoning.
    who knows who came up with it? I don't.

    math is like that sometimes. sometimes someone recognizes a pattern and applies it backwards to get somewhere. being very familiar with trig derivatives no doubt was a motivating factor here. seeing how that would work backwards might have led to such a method.

    sometimes, you don't know where it came from or why someone would even think of that. yet you can't deny that it works, and that it is stunningly elegant.
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