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Evaluating Definite integral

  1. May 14, 2013 #1
    1. The problem statement, all variables and given/known data
    If ##\displaystyle P=\int_0^{\pi} \frac{\cos x}{(x+4)^2}dx## and ##\displaystyle I=\int_0^{\pi/2} \frac{\sin (2x)}{2x+4}dx##, then the value of ##P+2I-\frac{1}{\pi+4}## is equal to


    2. Relevant equations



    3. The attempt at a solution
    By substituting 2x=t i.e 2dx=dt, and replacing t with x, I can be rewritten as
    [tex]I=\frac{1}{2}\int_0^{\pi} \frac{\sin x}{x+4}dx[/tex]
    [tex]P+2I=\int_0^{\pi} \frac{\cos x+x\sin x+4\sin x}{(x+4)^2}dx[/tex]

    How should I proceed from here? :confused:

    Any help is appreciated. Thanks!
     
  2. jcsd
  3. May 14, 2013 #2

    CompuChip

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    Try computing $$\frac{d}{dx} \frac{\cos x}{x + 4}$$.
     
  4. May 15, 2013 #3
    Ah yes, but how did you think of this?

    So ##P+2I=1/4+1/(4+\pi)##, hence the answer is ##1/4##.

    Thanks CompuChip! :smile:
     
  5. May 15, 2013 #4

    CompuChip

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    Having f/T² in one term and f'/T in other was a pretty big hint, which reminded me of the quotient rule
    [tex]\left( \frac{f}{g} \right)' = \frac{f'}{g} - \frac{f g'}{g^2}[/tex]
    So I tried that hoping it would get me somewhere, and luckily it worked out exactly (if you don't get the minus signs wrong, as I initially did - there the -1/(pi + 4) was the clue that something might need to cancel out).
     
  6. May 15, 2013 #5
    Thank you again! That is very helpful. :smile:
     
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