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Strange indefinite integral

  1. Dec 30, 2014 #1
    Hello, everyone

    I've trying to solve this integral but it seems like the methods I know are not enogh to solve it. So I'd be glad if you could give me some trick to get into the answer.
    Here it is:

    33wtvdw.jpg

    Thanks in advance!
     
  2. jcsd
  3. Dec 30, 2014 #2

    OldEngr63

    User Avatar
    Gold Member

    Maple knocked it out right away (I have no idea what the details are):

    Integral = num/den
    where
    num =(x*e^x*(tan((1/2)*e^x))^2-x*e^x+2*tan((1/2)*e^x))
    den=(1+tan((1/2)*e^x))*e^x

    Perhaps having the result in hand will enable you to go back and fill in the gaps.
     
  4. Dec 30, 2014 #3
    I found it to be e-xsin(ex) - xcos(ex) + c
    This was through Wolfram and I would guess integration by parts somehow
     
  5. Dec 30, 2014 #4
    Hey guys!

    I've just solved this thing. In fact, like Charles wrote, it is integration by parts. First you distribute the [itex]\sin(e^x)[/itex] into the parenthesis, then you do the substitution [itex]u=e^x[/itex]. After some steps you shall get something like this:

    [itex]I=\int \ln(u) \sin(u)du-\int \frac{\sin(u)}{u^2}du[/itex]

    Applying integration by parts:

    [itex]I=-\cos(u) ln(u)+\int \frac{\cos(u)}{u}du+\frac{\sin(u)}{u}-\int \frac{\cos(u)}{u}du[/itex]

    And there you are... As you see the non-primitive function integral gets cancelled.
     
  6. Dec 30, 2014 #5
    Eureka! And substituting back reduces it down. Aren't integrals just a blast?
     
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