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Math Help - Hyperbolic Trig Proof

  1. #1
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    Hyperbolic Trig Proof

    Hey all,

    I'm supposed to show that Sec (Theta) = Cosh (Theta)

    Given:

    x = ln (Sec (theta) + tan (theta))

    I have NO idea where to start, any pointers?

    -B

    P.S.

    I'm supposed to compute the integral of x^5(Sin(x^3)). I should use parts to integrate this, correct?
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  2. #2
    Member Mentia's Avatar
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    But Sec(T) is not equal to Cosh(T), for instance:

    Sec(Pi) = -1,
    Cosh(Pi) = 11.59

    Are you sure you have written down the problem correctly?
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  3. #3
    Member Mentia's Avatar
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    With regards to your P.S.,

    use the substitution:

    u = x^3,

    then you need only integrate by parts once.
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by TyrsFromAbove37 View Post
    P.S.

    I'm supposed to compute the integral of x^5sin(x^3). I should use parts to integrate this, correct?
    Nope, at least not right away. Do a u-substitution, letting t=x^3. Thus dt=3x^2\,dx

    Now \int x^5sin(x^3)\,dx = \frac{1}{3}\int x^3sin(x^3)*3x^2\,dx = \frac{1}{3}\int tsin(t)\,dt..

    NOW you use integration by parts (letting u=t and dv = sin(t))to obtain \frac{1}{3}\int tsin(t)\,dt = \frac{1}{3}(sin(t)-tcos(t)).

    Plugging back in for t yields: \frac{1}{3}sin(x^3)-\frac{1}{3}x^3cos(x^3).

    Thus: \int x^5sin(x^3)\,dx = \frac{1}{3}sin(x^3)-\frac{1}{3}x^3cos(x^3) + C
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  5. #5
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    Quote Originally Posted by Mentia View Post
    But Sec(T) is not equal to Cosh(T), for instance:

    Sec(Pi) = -1,
    Cosh(Pi) = 11.59
    I thought it had something to do with the ln function, possibly dealing with the derivatives of the trigonometric functions as they related to a fraction, which would eliminate the trig function in and of itself and leave the fraction with only constants and x terms...(Hope that made sense)

    At any rate, I'll ask the prof tomorrow to clarify, maybe he miswrote the problem...

    Thanks,

    -B
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  6. #6
    Super Member redsoxfan325's Avatar
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    Regarding your actual question, I believe I have the solution.

    Take the derivative of both sides of your equation:

    \frac{d}{dx}[x] = \frac{d}{d\theta}[ln(sec(\theta)+tan(\theta))]

    Thus, 1 = sec(\theta). (Yeah, that complicated expression on the right is just \int sec(\theta)\,d\theta.)

    Since 1=sec(\theta), \theta = arcsec(1) = 0.

    Thus sec(0) = cosh(0) = 1.

    QED
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  7. #7
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    Thanks for all the help guys, it is greatly appreciated...

    -B
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  8. #8
    Member Mentia's Avatar
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    Did you simply want a value for which Sec(T) = Cosh(t)? There are an infinite number of them, you can show that

    T=ln(4- { e}^{ i+1} - {e }^{i-1 } - {e }^{1-i } - {e }^{1+i } )

    but remember that the complex natural log is a multivalued function. In putting it in this form I may have even gotten rid of some solutions... anyone have some pointers on finding all solutions?
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  9. #9
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    I was able to clarify what my prof wanted for the sec = cosh problem:

    He wanted us to take the value for cosh (having to do with the e^x terms)

    and relate it to the ln function (as it pertains to e^x terms)

    Hope that makes sense...

    At any rate, I tried plugging the values into the cosh function, but didn't get an answer or I didn't get an answer that made the problem look done...

    Any thoughts?

    -B
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  10. #10
    Member Mentia's Avatar
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    Ok i figured it out. You did write it down slightly wrong. What he wants is for you to show Cosh(x) = Sec(T).

    You can raise e by both sides of your equation. Then you know e^x. You also know e^-x. Then add them. You can solve it down to Cosh(x) = Sec(T).
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  11. #11
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    I'm a bit confused by your wording. Mind clarifying what you mean when you talk about raising both sides by e?

    We talked a bit about this one in class and I was still a bit lost...

    Thanks,

    -B
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  12. #12
    Member Mentia's Avatar
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    For example if we had:

    x = ln (y)

    Then raise e to the power of both sides:

    e^x = e ^ [ln (y)] = y

    This is the same as saying:

    If 2 = 2 then e^2 = e^2. Right?

    Doing this is often used if you have a natural log since it is the inverse natural log:

    e^[ln (x)] = x
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  13. #13
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    I applied that to the cosh function and ended up getting:

    (Sec(theta)+Tan(theta)) - (Sec(theta)+Tan(theta)) / 2

    This cancels out to 0...I don't know where to take it from there...

    -B
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  14. #14
    Member Mentia's Avatar
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    You should get:

     {e }^{ x} = ( \frac{ 1}{ cos( \theta )} )(1+sin( \theta ))

    Then this means that:

     {e }^{ -x} = (cos( \theta )) {(1+sin( \theta )) }^{ -1}

    Then
    <br />
{e }^{ x} + {e }^{ -x} = ( \frac{ 1}{ cos( \theta )} )(1+sin( \theta )) + (cos( \theta )) ( \frac{ 1}{ 1 + sin( \theta )} )

    Now get a common denominator for the right side of the equation and solve it down. You should find that the right side is equal to 2/cos(T). Then divide both sides by 2 and you find Cosh(x) = Sec(T)
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  15. #15
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    Alright, that makes sense, thanks for the help.

    -B
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