Integrating $int_{0}^{a}sqrt{{tanh}(a)-{tanh}(x)};dx$












4












$begingroup$



What is the integral



$$int_{0}^{a}sqrt{{tanh}(a)-{tanh}(x)};dx$$




I have tried various substitutions but couldn't get the answer.



The substitution $x = {tanh}^{-1}y$ has simplified into rational function in y but I couldn't proceed further.



I got



$$int_{0}^{tanh(a)}frac{1}{sqrt{tanh(a)-y}}frac{1}{1-y^{2}} ; .$$






integration definite-integrals indefinite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    How about trying with a substitution $sqrt{operatorname{tanh}(a) - operatorname{tanh}{x}} = y$, instead? Might it work?
    $endgroup$
    – Matteo
    Jan 23 at 16:14










  • $begingroup$
    Typing tips: we use {} to indicate groups, {} to indicate a pair of braces. Your code has lots of uncompleted groups, hence the compiler cannot process them.
    $endgroup$
    – xbh
    Jan 23 at 16:29










  • $begingroup$
    Where did you get this integral?
    $endgroup$
    – Frank W.
    Jan 23 at 16:47










  • $begingroup$
    @FrankW. In quantum mechanics while using WKB method
    $endgroup$
    – Aierel
    Jan 23 at 16:48
















4












$begingroup$



What is the integral



$$int_{0}^{a}sqrt{{tanh}(a)-{tanh}(x)};dx$$




I have tried various substitutions but couldn't get the answer.



The substitution $x = {tanh}^{-1}y$ has simplified into rational function in y but I couldn't proceed further.



I got



$$int_{0}^{tanh(a)}frac{1}{sqrt{tanh(a)-y}}frac{1}{1-y^{2}} ; .$$






integration definite-integrals indefinite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    How about trying with a substitution $sqrt{operatorname{tanh}(a) - operatorname{tanh}{x}} = y$, instead? Might it work?
    $endgroup$
    – Matteo
    Jan 23 at 16:14










  • $begingroup$
    Typing tips: we use {} to indicate groups, {} to indicate a pair of braces. Your code has lots of uncompleted groups, hence the compiler cannot process them.
    $endgroup$
    – xbh
    Jan 23 at 16:29










  • $begingroup$
    Where did you get this integral?
    $endgroup$
    – Frank W.
    Jan 23 at 16:47










  • $begingroup$
    @FrankW. In quantum mechanics while using WKB method
    $endgroup$
    – Aierel
    Jan 23 at 16:48














4












4








4


0



$begingroup$



What is the integral



$$int_{0}^{a}sqrt{{tanh}(a)-{tanh}(x)};dx$$




I have tried various substitutions but couldn't get the answer.



The substitution $x = {tanh}^{-1}y$ has simplified into rational function in y but I couldn't proceed further.



I got



$$int_{0}^{tanh(a)}frac{1}{sqrt{tanh(a)-y}}frac{1}{1-y^{2}} ; .$$






integration definite-integrals indefinite-integrals







share|cite|improve this question











$endgroup$





What is the integral



$$int_{0}^{a}sqrt{{tanh}(a)-{tanh}(x)};dx$$




I have tried various substitutions but couldn't get the answer.



The substitution $x = {tanh}^{-1}y$ has simplified into rational function in y but I couldn't proceed further.



I got



$$int_{0}^{tanh(a)}frac{1}{sqrt{tanh(a)-y}}frac{1}{1-y^{2}} ; .$$






integration definite-integrals indefinite-integrals




integration definite-integrals indefinite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 23 at 20:09









Blue

48.7k870156




48.7k870156










asked Jan 23 at 15:46









AierelAierel

285




285












  • $begingroup$
    How about trying with a substitution $sqrt{operatorname{tanh}(a) - operatorname{tanh}{x}} = y$, instead? Might it work?
    $endgroup$
    – Matteo
    Jan 23 at 16:14










  • $begingroup$
    Typing tips: we use {} to indicate groups, {} to indicate a pair of braces. Your code has lots of uncompleted groups, hence the compiler cannot process them.
    $endgroup$
    – xbh
    Jan 23 at 16:29










  • $begingroup$
    Where did you get this integral?
    $endgroup$
    – Frank W.
    Jan 23 at 16:47










  • $begingroup$
    @FrankW. In quantum mechanics while using WKB method
    $endgroup$
    – Aierel
    Jan 23 at 16:48


















  • $begingroup$
    How about trying with a substitution $sqrt{operatorname{tanh}(a) - operatorname{tanh}{x}} = y$, instead? Might it work?
    $endgroup$
    – Matteo
    Jan 23 at 16:14










  • $begingroup$
    Typing tips: we use {} to indicate groups, {} to indicate a pair of braces. Your code has lots of uncompleted groups, hence the compiler cannot process them.
    $endgroup$
    – xbh
    Jan 23 at 16:29










  • $begingroup$
    Where did you get this integral?
    $endgroup$
    – Frank W.
    Jan 23 at 16:47










  • $begingroup$
    @FrankW. In quantum mechanics while using WKB method
    $endgroup$
    – Aierel
    Jan 23 at 16:48
















$begingroup$
How about trying with a substitution $sqrt{operatorname{tanh}(a) - operatorname{tanh}{x}} = y$, instead? Might it work?
$endgroup$
– Matteo
Jan 23 at 16:14




$begingroup$
How about trying with a substitution $sqrt{operatorname{tanh}(a) - operatorname{tanh}{x}} = y$, instead? Might it work?
$endgroup$
– Matteo
Jan 23 at 16:14












$begingroup$
Typing tips: we use {} to indicate groups, {} to indicate a pair of braces. Your code has lots of uncompleted groups, hence the compiler cannot process them.
$endgroup$
– xbh
Jan 23 at 16:29




$begingroup$
Typing tips: we use {} to indicate groups, {} to indicate a pair of braces. Your code has lots of uncompleted groups, hence the compiler cannot process them.
$endgroup$
– xbh
Jan 23 at 16:29












$begingroup$
Where did you get this integral?
$endgroup$
– Frank W.
Jan 23 at 16:47




$begingroup$
Where did you get this integral?
$endgroup$
– Frank W.
Jan 23 at 16:47












$begingroup$
@FrankW. In quantum mechanics while using WKB method
$endgroup$
– Aierel
Jan 23 at 16:48




$begingroup$
@FrankW. In quantum mechanics while using WKB method
$endgroup$
– Aierel
Jan 23 at 16:48










2 Answers
2






active

oldest

votes


















4












$begingroup$

$$I=int_0^a sqrt{tanh a-tanh x}dxoverset{tanh x=y}=int_0^{tanh a} sqrt{tanh a-y} frac{dy}{1-y^2}$$
Let us denote $tanh a=n$ for now. And our first goal must be to get rid of the square root, thus a substitution of $n-y =t^2$ would fit great.
$$I=int_0^n frac{sqrt{n-y}}{1-y^2}dy=int^0_{sqrt n} frac{t}{1-(n-t^2)^2}(-2t,)dt=int_0^{sqrt n} frac{2t^2}{1-(n-t^2)^2}dt$$
$$=int_0^sqrt n frac{1+n}{1+n-t^2}dt-int_0^sqrt n frac{1-n}{1-n+t^2}dt$$
$$=frac{1+n}{sqrt{1+n}}operatorname{arctanh}left(frac{t}{sqrt{1+n}}right)bigg|_0^sqrt n-frac{1-n}{sqrt{1-n}}arctanleft(frac{t}{sqrt{1-n}}right)bigg|_0^sqrt n$$
$$={sqrt{1+tanh a}}cdot operatorname{arctanh}left(sqrt{frac{{tanh a}}{1+tanh a}}right)-{sqrt{1-tanh a}}cdot arctanleft(sqrt{frac{{tanh a}}{1-tanh a}}right)$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This is the same as directly substituting the entire integrand with the new variable, is it not?
    $endgroup$
    – Matteo
    Jan 23 at 20:37










  • $begingroup$
    Pretty much yes, but it leaves a room to get rid of that $tanh a$, also this was OP's approach so I continued.
    $endgroup$
    – Zacky
    Jan 23 at 20:45



















1












$begingroup$

Hint



Check if this approach is useful. The substitution
$$ y = sqrt{tanh(a)-tanh(x)}$$
gives you
$$y^2 = tanh(a)-tanh(x),$$
and, therefore,
begin{eqnarray}
&&2y dy = -(1-tanh^2(x)) dx\
&&2ydy = -left[1-(tanh(a)-y^2)^2right] dx\
&&frac{2y dy}{(y^2-tanh(a))^2-1} = dx.
end{eqnarray}
Then the integral becomes rational, i.e.
begin{eqnarray}mathcal I &=& int sqrt{tanh(a)-tanh(x)}dx= \
&=&2int frac{y^2dy}{{(y^2-tanh(a))^2-1}}=\
&=&2intfrac{y^2dy}{(y^2-tanh(a)-1)(y^2+1-tanh(a))}=\
&=& 2int frac{y^2dy}{(y-sqrt{tanh(a)+1})(y+sqrt{tanh(a)+1})(y^2+1-tanh(a))},
end{eqnarray}
recalling that $-1leq tanh(x) leq 1$. You can proceed from here with partial fractions.






share|cite|improve this answer











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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

    oldest

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    4












    $begingroup$

    $$I=int_0^a sqrt{tanh a-tanh x}dxoverset{tanh x=y}=int_0^{tanh a} sqrt{tanh a-y} frac{dy}{1-y^2}$$
    Let us denote $tanh a=n$ for now. And our first goal must be to get rid of the square root, thus a substitution of $n-y =t^2$ would fit great.
    $$I=int_0^n frac{sqrt{n-y}}{1-y^2}dy=int^0_{sqrt n} frac{t}{1-(n-t^2)^2}(-2t,)dt=int_0^{sqrt n} frac{2t^2}{1-(n-t^2)^2}dt$$
    $$=int_0^sqrt n frac{1+n}{1+n-t^2}dt-int_0^sqrt n frac{1-n}{1-n+t^2}dt$$
    $$=frac{1+n}{sqrt{1+n}}operatorname{arctanh}left(frac{t}{sqrt{1+n}}right)bigg|_0^sqrt n-frac{1-n}{sqrt{1-n}}arctanleft(frac{t}{sqrt{1-n}}right)bigg|_0^sqrt n$$
    $$={sqrt{1+tanh a}}cdot operatorname{arctanh}left(sqrt{frac{{tanh a}}{1+tanh a}}right)-{sqrt{1-tanh a}}cdot arctanleft(sqrt{frac{{tanh a}}{1-tanh a}}right)$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      This is the same as directly substituting the entire integrand with the new variable, is it not?
      $endgroup$
      – Matteo
      Jan 23 at 20:37










    • $begingroup$
      Pretty much yes, but it leaves a room to get rid of that $tanh a$, also this was OP's approach so I continued.
      $endgroup$
      – Zacky
      Jan 23 at 20:45
















    4












    $begingroup$

    $$I=int_0^a sqrt{tanh a-tanh x}dxoverset{tanh x=y}=int_0^{tanh a} sqrt{tanh a-y} frac{dy}{1-y^2}$$
    Let us denote $tanh a=n$ for now. And our first goal must be to get rid of the square root, thus a substitution of $n-y =t^2$ would fit great.
    $$I=int_0^n frac{sqrt{n-y}}{1-y^2}dy=int^0_{sqrt n} frac{t}{1-(n-t^2)^2}(-2t,)dt=int_0^{sqrt n} frac{2t^2}{1-(n-t^2)^2}dt$$
    $$=int_0^sqrt n frac{1+n}{1+n-t^2}dt-int_0^sqrt n frac{1-n}{1-n+t^2}dt$$
    $$=frac{1+n}{sqrt{1+n}}operatorname{arctanh}left(frac{t}{sqrt{1+n}}right)bigg|_0^sqrt n-frac{1-n}{sqrt{1-n}}arctanleft(frac{t}{sqrt{1-n}}right)bigg|_0^sqrt n$$
    $$={sqrt{1+tanh a}}cdot operatorname{arctanh}left(sqrt{frac{{tanh a}}{1+tanh a}}right)-{sqrt{1-tanh a}}cdot arctanleft(sqrt{frac{{tanh a}}{1-tanh a}}right)$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      This is the same as directly substituting the entire integrand with the new variable, is it not?
      $endgroup$
      – Matteo
      Jan 23 at 20:37










    • $begingroup$
      Pretty much yes, but it leaves a room to get rid of that $tanh a$, also this was OP's approach so I continued.
      $endgroup$
      – Zacky
      Jan 23 at 20:45














    4












    4








    4





    $begingroup$

    $$I=int_0^a sqrt{tanh a-tanh x}dxoverset{tanh x=y}=int_0^{tanh a} sqrt{tanh a-y} frac{dy}{1-y^2}$$
    Let us denote $tanh a=n$ for now. And our first goal must be to get rid of the square root, thus a substitution of $n-y =t^2$ would fit great.
    $$I=int_0^n frac{sqrt{n-y}}{1-y^2}dy=int^0_{sqrt n} frac{t}{1-(n-t^2)^2}(-2t,)dt=int_0^{sqrt n} frac{2t^2}{1-(n-t^2)^2}dt$$
    $$=int_0^sqrt n frac{1+n}{1+n-t^2}dt-int_0^sqrt n frac{1-n}{1-n+t^2}dt$$
    $$=frac{1+n}{sqrt{1+n}}operatorname{arctanh}left(frac{t}{sqrt{1+n}}right)bigg|_0^sqrt n-frac{1-n}{sqrt{1-n}}arctanleft(frac{t}{sqrt{1-n}}right)bigg|_0^sqrt n$$
    $$={sqrt{1+tanh a}}cdot operatorname{arctanh}left(sqrt{frac{{tanh a}}{1+tanh a}}right)-{sqrt{1-tanh a}}cdot arctanleft(sqrt{frac{{tanh a}}{1-tanh a}}right)$$






    share|cite|improve this answer











    $endgroup$



    $$I=int_0^a sqrt{tanh a-tanh x}dxoverset{tanh x=y}=int_0^{tanh a} sqrt{tanh a-y} frac{dy}{1-y^2}$$
    Let us denote $tanh a=n$ for now. And our first goal must be to get rid of the square root, thus a substitution of $n-y =t^2$ would fit great.
    $$I=int_0^n frac{sqrt{n-y}}{1-y^2}dy=int^0_{sqrt n} frac{t}{1-(n-t^2)^2}(-2t,)dt=int_0^{sqrt n} frac{2t^2}{1-(n-t^2)^2}dt$$
    $$=int_0^sqrt n frac{1+n}{1+n-t^2}dt-int_0^sqrt n frac{1-n}{1-n+t^2}dt$$
    $$=frac{1+n}{sqrt{1+n}}operatorname{arctanh}left(frac{t}{sqrt{1+n}}right)bigg|_0^sqrt n-frac{1-n}{sqrt{1-n}}arctanleft(frac{t}{sqrt{1-n}}right)bigg|_0^sqrt n$$
    $$={sqrt{1+tanh a}}cdot operatorname{arctanh}left(sqrt{frac{{tanh a}}{1+tanh a}}right)-{sqrt{1-tanh a}}cdot arctanleft(sqrt{frac{{tanh a}}{1-tanh a}}right)$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 23 at 20:12

























    answered Jan 23 at 20:06









    ZackyZacky

    7,4301961




    7,4301961












    • $begingroup$
      This is the same as directly substituting the entire integrand with the new variable, is it not?
      $endgroup$
      – Matteo
      Jan 23 at 20:37










    • $begingroup$
      Pretty much yes, but it leaves a room to get rid of that $tanh a$, also this was OP's approach so I continued.
      $endgroup$
      – Zacky
      Jan 23 at 20:45


















    • $begingroup$
      This is the same as directly substituting the entire integrand with the new variable, is it not?
      $endgroup$
      – Matteo
      Jan 23 at 20:37










    • $begingroup$
      Pretty much yes, but it leaves a room to get rid of that $tanh a$, also this was OP's approach so I continued.
      $endgroup$
      – Zacky
      Jan 23 at 20:45
















    $begingroup$
    This is the same as directly substituting the entire integrand with the new variable, is it not?
    $endgroup$
    – Matteo
    Jan 23 at 20:37




    $begingroup$
    This is the same as directly substituting the entire integrand with the new variable, is it not?
    $endgroup$
    – Matteo
    Jan 23 at 20:37












    $begingroup$
    Pretty much yes, but it leaves a room to get rid of that $tanh a$, also this was OP's approach so I continued.
    $endgroup$
    – Zacky
    Jan 23 at 20:45




    $begingroup$
    Pretty much yes, but it leaves a room to get rid of that $tanh a$, also this was OP's approach so I continued.
    $endgroup$
    – Zacky
    Jan 23 at 20:45











    1












    $begingroup$

    Hint



    Check if this approach is useful. The substitution
    $$ y = sqrt{tanh(a)-tanh(x)}$$
    gives you
    $$y^2 = tanh(a)-tanh(x),$$
    and, therefore,
    begin{eqnarray}
    &&2y dy = -(1-tanh^2(x)) dx\
    &&2ydy = -left[1-(tanh(a)-y^2)^2right] dx\
    &&frac{2y dy}{(y^2-tanh(a))^2-1} = dx.
    end{eqnarray}
    Then the integral becomes rational, i.e.
    begin{eqnarray}mathcal I &=& int sqrt{tanh(a)-tanh(x)}dx= \
    &=&2int frac{y^2dy}{{(y^2-tanh(a))^2-1}}=\
    &=&2intfrac{y^2dy}{(y^2-tanh(a)-1)(y^2+1-tanh(a))}=\
    &=& 2int frac{y^2dy}{(y-sqrt{tanh(a)+1})(y+sqrt{tanh(a)+1})(y^2+1-tanh(a))},
    end{eqnarray}
    recalling that $-1leq tanh(x) leq 1$. You can proceed from here with partial fractions.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Hint



      Check if this approach is useful. The substitution
      $$ y = sqrt{tanh(a)-tanh(x)}$$
      gives you
      $$y^2 = tanh(a)-tanh(x),$$
      and, therefore,
      begin{eqnarray}
      &&2y dy = -(1-tanh^2(x)) dx\
      &&2ydy = -left[1-(tanh(a)-y^2)^2right] dx\
      &&frac{2y dy}{(y^2-tanh(a))^2-1} = dx.
      end{eqnarray}
      Then the integral becomes rational, i.e.
      begin{eqnarray}mathcal I &=& int sqrt{tanh(a)-tanh(x)}dx= \
      &=&2int frac{y^2dy}{{(y^2-tanh(a))^2-1}}=\
      &=&2intfrac{y^2dy}{(y^2-tanh(a)-1)(y^2+1-tanh(a))}=\
      &=& 2int frac{y^2dy}{(y-sqrt{tanh(a)+1})(y+sqrt{tanh(a)+1})(y^2+1-tanh(a))},
      end{eqnarray}
      recalling that $-1leq tanh(x) leq 1$. You can proceed from here with partial fractions.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Hint



        Check if this approach is useful. The substitution
        $$ y = sqrt{tanh(a)-tanh(x)}$$
        gives you
        $$y^2 = tanh(a)-tanh(x),$$
        and, therefore,
        begin{eqnarray}
        &&2y dy = -(1-tanh^2(x)) dx\
        &&2ydy = -left[1-(tanh(a)-y^2)^2right] dx\
        &&frac{2y dy}{(y^2-tanh(a))^2-1} = dx.
        end{eqnarray}
        Then the integral becomes rational, i.e.
        begin{eqnarray}mathcal I &=& int sqrt{tanh(a)-tanh(x)}dx= \
        &=&2int frac{y^2dy}{{(y^2-tanh(a))^2-1}}=\
        &=&2intfrac{y^2dy}{(y^2-tanh(a)-1)(y^2+1-tanh(a))}=\
        &=& 2int frac{y^2dy}{(y-sqrt{tanh(a)+1})(y+sqrt{tanh(a)+1})(y^2+1-tanh(a))},
        end{eqnarray}
        recalling that $-1leq tanh(x) leq 1$. You can proceed from here with partial fractions.






        share|cite|improve this answer











        $endgroup$



        Hint



        Check if this approach is useful. The substitution
        $$ y = sqrt{tanh(a)-tanh(x)}$$
        gives you
        $$y^2 = tanh(a)-tanh(x),$$
        and, therefore,
        begin{eqnarray}
        &&2y dy = -(1-tanh^2(x)) dx\
        &&2ydy = -left[1-(tanh(a)-y^2)^2right] dx\
        &&frac{2y dy}{(y^2-tanh(a))^2-1} = dx.
        end{eqnarray}
        Then the integral becomes rational, i.e.
        begin{eqnarray}mathcal I &=& int sqrt{tanh(a)-tanh(x)}dx= \
        &=&2int frac{y^2dy}{{(y^2-tanh(a))^2-1}}=\
        &=&2intfrac{y^2dy}{(y^2-tanh(a)-1)(y^2+1-tanh(a))}=\
        &=& 2int frac{y^2dy}{(y-sqrt{tanh(a)+1})(y+sqrt{tanh(a)+1})(y^2+1-tanh(a))},
        end{eqnarray}
        recalling that $-1leq tanh(x) leq 1$. You can proceed from here with partial fractions.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 23 at 17:11

























        answered Jan 23 at 16:50









        MatteoMatteo

        848311




        848311






























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            Mario Kart Wii

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            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
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            What does “Dominus providebit” mean?

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            5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
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            Antonio Litta Visconti Arese

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            Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
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