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    Question: Use the substitution x = ln u to find \displaystyle\int \dfrac{e^x}{1+ e^{2x}}\ dx

    I'm stuck in the end; here's how I tried to do it:
    \dfrac{dx}{du} = \dfrac{1}{u} so  dx = \dfrac{du}{u}
    \Rightarrow \displaystyle\int \dfrac{e^{\ln u}}{1+ e^{2\ln u}}\ \dfrac{du}{u}
    \Rightarrow \displaystyle\int \dfrac{1}{1+ u^2}\ du

    I don't know how to do it after that last step . . .
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    you have the numerator wrong in your last line
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    (Original post by TenOfThem)
    you have the numerator wrong in your last line

    (Original post by Fing4)
    e^(lnu) is not 1
    Please see my post carefully, the numerator u is cancelled by 1/u when we replace dx by du/u.
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    (Original post by Zishi)
    Please see my post carefully, the numerator u is cancelled by 1/u when we replace dx by du/u.
    ooops

    reading whilst eating
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    In that case .... use the standard integral for

    \frac{1}{1+x^2}
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    (Original post by Zishi)
    Question: Use the substitution x = ln u to find \displaystyle\int \dfrac{e^x}{1+ e^{2x}}\ dx

    I'm stuck in the end; here's how I tried to do it:
    \dfrac{dx}{du} = \dfrac{1}{u} so  dx = \dfrac{du}{u}
    \Rightarrow \displaystyle\int \dfrac{e^{\ln u}}{1+ e^{2\ln u}}\ \dfrac{du}{u}
    \Rightarrow \displaystyle\int \dfrac{1}{1+ u^2}\ du

    I don't know how to do it after that last step . . .
    Your working is correct so far, ignore the above posts.

    Do you know the integrals/derivatives of the inverse trig funtions? You should be able to find one which matches this integral exactly.
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    (Original post by EEngWillow)
    Your working is correct so far, ignore the above posts.

    Do you know the integrals/derivatives of the inverse trig funtions? You should be able to find one which matches this integral exactly.
    Nope, but I guess I can workout derivatives/integrals of their inverse by working out from scratch. To solve this question do I need to learn some of 'em?
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    (Original post by TenOfThem)
    In that case .... use the standard integral for

    \frac{1}{1+x^2}
    What's the standard integral?
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    tan^{-1}x
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    (Original post by Zishi)
    Nope, but I guess I can workout derivatives/integrals of their inverse by working out from scratch. To solve this question do I need to learn some of 'em?
    Pretty much. I think they're in one of the FP modules?

    Anyway,

    \dfrac{d}{dx}arcsin(x) = \dfrac{1}{\sqrt{1-x^2}}

    \dfrac{d}{dx}arccos(x) = -\dfrac{1}{\sqrt{1-x^2}}

    \dfrac{d}{dx}arctan(x) = \dfrac{1}{x^2+1}
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    As EEngWillow said ... you need to know the differentials of the inverse trig functions


    http://www.cliffsnotes.com/study_gui...eId-39884.html

    I think you only use tan and sin in C4
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    (Original post by EEngWillow)
    Pretty much. I think they're in one of the FP modules?

    Anyway,

    \dfrac{d}{dx}arcsin(x) = \dfrac{1}{\sqrt{1-x^2}}

    \dfrac{d}{dx}arccos(x) = -\dfrac{1}{\sqrt{1-x^2}}

    \dfrac{d}{dx}arctan(x) = \dfrac{1}{x^2+1}

    (Original post by TenOfThem)
    As EEngWillow said ... you need to know the differentials of the inverse trig functions


    http://www.cliffsnotes.com/study_gui...eId-39884.html

    I think you only use tan and sin in C4
    Thanks.
 
 
 
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