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Help solving a fraction integral

  1. Sep 25, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex]\int[/tex](5sin xcosxdx)/(2+52sinx)


    3. The attempt at a solution

    I set u = sinx and du = cosxdx which gives me...

    [tex]\int[/tex](5udu)/(2+52u). I just need a little push as where to go from here, not the entire solution, just a push as what to do next. I have a feeling I need to use ln and know that ar = r ln(a) but don't know if I am supposed to use that on the 5u or not, any help is appreciated. Thanks in advance. :)
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 25, 2008 #2
    U could take this subs, at the very beginning

    [tex]\int\frac{5^{sin(x)}cosxdx}{2+(5^{sinx})^2}[/tex] so taking another substitution here will work quite nicely..

    [tex] 5^{sinx}=\sqrt2 u=>\sqrt2 du=5^{sinx}cos(x)ln(5)dx[/tex] now i guess you know how things turn out to be, right?
     
  4. Sep 25, 2008 #3
    That helped a lot, would the final answer then be (1/ln(5)[tex]\sqrt{}2[/tex])tan-1(5sinx/[tex]\sqrt{}2[/tex]) ? The only thing I was not sure about was the ln(5) as far as how to deal with that.
     
  5. Sep 26, 2008 #4
    take the derivative of your final answer and see if you get the integrand, then its okay, cuz i won't bother to do the whole thing. The general format of it looks okay, but i didn't check the details.
     
  6. Sep 26, 2008 #5

    HallsofIvy

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    52u= (5u)2 so the next step would be the substitution v= 5u, dv= ln(5) 5udu.
     
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