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Math Help - Trigo Integrals

  1. #1
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    Trigo Integrals

    kindly check... this is an integral without using the formula of trig integrals:

    \int \cot(5x) + \csc(5x) dx

    solution:
    \int \frac{\cos(5x)}{\sin(5x)} + \csc(5x) dx
    from the first integral
    let u = sin(5x)
    du = cos(5x)dx

    from the right integral u multiply both denominator and numerator by (cot(5x) + csc(5x))
    \frac{1}{5}\ln{(\sin{5x})} + \int \frac{\csc(5x)\cot(5x) + \csc^2(5x)}{\cot(5x)+\csc(5x)} dx

    let u = \cot(5x)+\csc(5x)
    du = -5(\csc^2(5x) + \csc(5x)\cot(5x)) dx

    integrated to :
    \frac{1}{5}\ln{(\sin{5x})} - \frac{1}{5}\ln(\cot(5x) + \csc(5x))

    i ended up having
    \frac{1}{5} \ln(\frac{\sin^2(5x)}{\cos(5x) + 1}) + C


    the answer at back of book is supposed to be:
     \frac{1}{5} \ln (1-\cos(5x)) + C

    and you shud not use a formula of trig integral
    Last edited by ^_^Engineer_Adam^_^; June 27th 2007 at 05:32 AM.
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  2. #2
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    You could use the identity: cot(5x)-csc(5x)=-tan(\frac{5x}{2})
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  3. #3
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    is that half angled identity?
    im confused where the 2 appear
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  4. #4
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    Yep, it's a half-angle.

    tan(\frac{x}{2})=\frac{1-cos(x)}{sin(x)}=csc(x)-cot(x)

    Therefore, cot(x)-csc(x)=-tan(\frac{x}{2})
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  5. #5
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    but the trig formulas are not allowed:
    example : integral of tan(x) = |ln sec(x)| + C
    the formula of integrals tan, cot and sec and csc are not allowed becuz it is not yet discsed...
    is my solution right? i think the only problem i hav is on logarithm & trig identity
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  6. #6
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    waiiiitttt

    the question is supposed to be  \int{\cot(5x) +\csc(5x)dx}
    but it has still the same solution ive written
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  7. #7
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    After looking at your answer, you have it correct. It just needs hammered into the book form. Sometimes the books can simplify beyond all recognition(to paraphrase Soroban).
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  8. #8
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    kindly check... this is an integral without using the formula of trig integrals:

    \int \cot(5x) + \csc(5x) dx

    solution:
    \int \frac{\cos(5x)}{\sin(5x)} + \csc(5x) dx
    from the first integral
    let u = sin(5x)
    du = cos(5x)dx

    from the right integral u multiply both denominator and numerator by (cot(5x) + csc(5x))
    \frac{1}{5}\ln{(\sin{5x})} + \int \frac{\csc(5x)\cot(5x) + \csc^2(5x)}{\cot(5x)+\csc(5x)} dx

    let u = \cot(5x)+\csc(5x)
    du = -5(\csc^2(5x) + \csc(5x)\cot(5x)) dx

    integrated to :
    \frac{1}{5}\ln{(\sin{5x})} - \frac{1}{5}\ln(\cot(5x) + \csc(5x))

    i ended up having
    \frac{1}{5} \ln(\frac{\sin^2(5x)}{\cos(5x) + 1}) + C


    the answer at back of book is supposed to be:
     \frac{1}{5} \ln (1-\cos(5x)) + C

    and you shud not use a formula of trig integral
    Three things.
    1) I can see nothing wrong with your work.

    2) My calculator comes up with your answer.

    3) As far as I can tell the two answers do not differ by a constant.

    I think your book is wrong.

    Well, the answer is actually:
    \frac{1}{5} \ln(\frac{\sin^2(5x)}{|\cos(5x) + 1|}) + C

    But you have the idea.

    -Dan
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  9. #9
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    When I run it through my TI I get \frac{1}{5}ln(cos(5x)-1)
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  10. #10
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by topsquark View Post
    Three things.
    1) I can see nothing wrong with your work.

    2) My calculator comes up with your answer.

    3) As far as I can tell the two answers do not differ by a constant.

    I think your book is wrong.

    Well, the answer is actually:
    \frac{1}{5} \ln(\frac{\sin^2(5x)}{|\cos(5x) + 1|}) + C

    But you have the idea.

    -Dan
    Quote Originally Posted by galactus View Post
    When I run it through my TI I get \frac{1}{5}ln(cos(5x)-1)
    Okay I see my mistake now, but I still can't convince my calculator they are the same for some reason.

    \frac{1}{5} \ln(\frac{\sin^2(5x)}{|\cos(5x) + 1|}) = \frac{1}{5} \ln(\frac{1 - \cos^2(5x)}{|\cos(5x) + 1|})

    = \frac{1}{5} \ln(\frac{(1 -  cos(5x))(1 + cos(5x))}{|\cos(5x) + 1|})

    = \frac{1}{5}ln|1 - cos(5x)| = \frac{1}{5}ln|cos(5x) - 1|

    So the two answers are the same. But why can't I plug an x in my calculator to see that the two expressions are the same??

    -Dan

    Edit: No problems with the calculator difficulty. I still had my calculator on "exact." The differences between calculating the two expressions are infinitesimal, as expected.
    Last edited by topsquark; June 27th 2007 at 07:04 AM.
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  11. #11
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    We can use well-known results.
    Attached Thumbnails Attached Thumbnails Trigo Integrals-june31.gif  
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  12. #12
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    Quote Originally Posted by galactus View Post
    When I run it through my TI I get \frac{1}{5}ln(cos(5x)-1)
    However cos(5x)-1 is negative, which is not allowed for ln(). So you must take the absolute value or 1-cos(5x) instead.
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  13. #13
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    Smile

    thank u
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