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Writing arcsin in log form

  1. Oct 11, 2007 #1
    It seems to that i went wrong way.
    1. The problem statement, all variables and given/known data

    [tex]\int \frac{arcsin(e^x)dx}{e^x}[/tex]

    3. The attempt at a solution
    [tex]=-\frac{arcsin(e^x)}{e^x}+\int \frac{dx}{\sqrt{1-e^{2x}}}[/tex]

    [tex]\int \frac{dx}{\sqrt{1-e^{2x}}}=\frac{x}{\sqrt{1-e^{2x}}}+\int \frac{xe^{2x}dx}{(1-e^{2x})^{\frac{3}{2}}}[/tex]
    Last edited: Oct 11, 2007
  2. jcsd
  3. Oct 11, 2007 #2
    Ever thought about writing arcsin in log form (the inverse function of sin when sin is written in complex numbers), that way something might happen between the logs and the exp's. I haven't tried it, so I don't know, its just a suggestion.
  4. Oct 12, 2007 #3
    Here is the another way I tried.
    [tex]-\int arcsin(\frac{1}{u})du=-\int arccsc(u)du=-uarccsc(u)-\int\frac{du}{\sqrt{u^2-1}}[/tex]
  5. Oct 12, 2007 #4
    Than for
    [tex]u=sec\theta\rightarrow du=\frac{sin\theta d\theta}{cos^2\theta}[/tex]
    Last edited: Oct 12, 2007
  6. Oct 12, 2007 #5
    But I don't know,actually,how to integrate last one.:frown:
  7. Oct 12, 2007 #6

    trig identity

  8. Oct 12, 2007 #7
    Ok,than please check everything.
    [tex]\int \frac{arcsin(e^x)dx}{e^x}[/tex]


    [tex]-\int arcsin(\frac{1}{u})du=-\int arccsc(u)du=-uarccsc(u)-\int\frac{du}{\sqrt{u^2-1}}[/tex]

    [tex]u=sec\theta\rightarrow du=\frac{sin\theta d\theta}{cos^2\theta}[/tex]



  9. Oct 12, 2007 #8
    Sorry,I've posted answer to another question here.
    Last edited: Oct 13, 2007
  10. Oct 13, 2007 #9

    Gib Z

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    Homework Helper

    Post 7, you made a sign error. Suddenly the sec integral becomes positive...other than that, the answer is fine, but it would look nicer if you took the e^(-x) under the sqrt sign in the log, as the argument of the log is always positive anyway, no need for absolute value signs.
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