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    Hey,

    Can anyone give me a hint on how to integrate

     \int \frac{e^x}{\sqrt{1-e^{2x}}} . So far I've only come up with the fact that e^x and \sqrt{1-e^{2x}} are the legs of a right angled triangle with hypotenuse of length 1...
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    (Original post by FXLander)
    Hey,

    Can anyone give me a hint on how to integrate

     \int \frac{e^x}{\sqrt{1-e^{2x}}} . So far I've only come up with the fact that e^x and \sqrt{1-e^{2x}} are the legs of a right angled triangle with hypotenuse of length 1...
    Have you tried any substitutions?
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    i think a trig substitution for ex would work...
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    (Original post by Notnek)
    Have you tried any substitutions?
    I tried substituting \tan \theta = \frac{e^x}{\sqrt{1-e^{2x}}} since I drew the right triangle I was referring to, but I am not sure how to get rid of the dx when I do this...
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    (Original post by FXLander)
    I tried substituting \tan \theta = \frac{e^x}{\sqrt{1-e^{2x}}} since I drew the right triangle I was referring to, but I am not sure how to get rid of the dx when I do this...
    You're making it more complicated than it needs to be. Try the substitution that the bear gave you. With practice this will feel like an "obvious" substitution.

    EDIT : i.e. I mean u=e^x (this sub will lead to a standard/formula book integral)
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    (Original post by FXLander)
    I tried substituting \tan \theta = \frac{e^x}{\sqrt{1-e^{2x}}} since I drew the right triangle I was referring to, but I am not sure how to get rid of the dx when I do this...
    use a different trig function for e^x. consider trig identities.
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    (Original post by Notnek)
    You're making it more complicated than it needs to be. Try the substitution that the bear gave you. With practice this will feel like an "obvious" substitution.

    EDIT : i.e. I mean u=e^x
    After doing the substitution I got  \int \frac{du}{\sqrt{1-u^2}} which I then evaluated to \frac{\sqrt{1-e^{2x}}}{e^x} , however I am not sure whether this is correct...
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    (Original post by FXLander)
    After doing the substitution I got  \int \frac{du}{\sqrt{1-u^2}} which I then evaluated to \frac{\sqrt{1-e^{2x}}}{e^x} , however I am not sure whether this is correct...
    For:

     \displaystyle \int \dfrac{1}{ \sqrt{1-u^{2}} } \, du,

    look up a table of inverse trigonometric derrivatives.
 
 
 
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