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Homework Help: Integration By Parts - Another Problem

  1. Sep 21, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]
    \int {\frac{{\cos ydy}}
    {{\sin ^2 y + \sin y - 6}} }
    [/tex]


    3. The attempt at a solution
    [tex]
    \int {\frac{{\cos ydy}}
    {{\sin ^2 y + \sin y - 6}} = } \int {\frac{{\cos ydy}}
    {{(\sin y - 2)(\sin y + 3)}}}
    [/tex]

    Now I attempt to split this into partial fractions:

    [tex]
    \begin{gathered}
    \frac{{\cos y}}
    {{(\sin y - 2)(\sin y + 3)}} = \frac{A}
    {{\sin y - 2}} + \frac{B}
    {{\sin y + 3}} \hfill \\
    \cos y = A(\sin y + 3) + B(\sin y - 2) \hfill \\
    \cos y = A\sin y + 3A + B\sin y - 2B \hfill \\
    \cos y = (A + B)\sin y + 3A - 2B \hfill \\
    \end{gathered}
    [/tex]

    ....from here i'm not sure what to do to solve for the coefficients. I can't see any trig identities that would help either. How would you solve for these?

    Thank you.
     
  2. jcsd
  3. Sep 21, 2008 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Oh, RedBarchetta!

    This is screaming out for a substitution! :wink:
     
  4. Sep 21, 2008 #3
    You're right. :rofl: It just took me a while to notice that. It definitely decreases the difficulty level.

    Thanks.
     
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