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

    votes








    2 Answers
    2






    active

    oldest

    votes









    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

    7,49011061




    7,49011061












    • $begingroup$
      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

        868312




        868312






























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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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            The Binding of Isaac: Rebirth/Afterbirth

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            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me
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