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Thread: Need help with Trig Integral

  1. #1
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    Feb 2010
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    Need help with Trig Integral

    Hello,

    I've been trying to work the following problem, but I am getting lost from the start. I'm pretty sure if I could get it started I wouldn't have a problem with it.

    $\displaystyle (1+4cot(x))/(4-cot(x))$

    Thanks for any help you can offer!
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  2. #2
    Super Member
    Joined
    Jan 2009
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    $\displaystyle \frac{1 + 4\cot(x)}{4 - \cot(x) }$

    $\displaystyle = \frac{ \tan(x) \cdot (1 + 4\cot(x) )}{ \tan(x) \cdot ( 4 - \cot(x))} $

    $\displaystyle = \frac{ \tan(x) + 4 }{ 4\tan(x) - 1 } $

    $\displaystyle = -\frac{ \tan(x) + 4}{ 1 - 4\tan(x) }$

    then use the formula

    $\displaystyle \tan(A + B) = \frac{ \tan(A) + \tan(B) }{ 1 - \tan(A)\tan(B) } $

    By substituting $\displaystyle A = x , B = \tan^{-1}(4) $

    We obtain $\displaystyle \frac{1 + 4\cot(x)}{4 - \cot(x) } = - \tan( x + \tan^{-1}(4) ) $


    Wow , it is actually a tagent function so the integral is log. function :

    $\displaystyle \ln{ \left [ \cos( x + \tan^{-1}(4) ) \right ]} + C $
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  3. #3
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    Thank you!

    Multiplying by 1 is something I would've never seen in my life. I always seem to forget that as an option.
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