Integral...












2












$begingroup$


I was told that there is a closed form to the integral $$int_{pi/4}^{pi/2}frac{cos(2theta)e^{cottheta}}{sin^3(2theta)left(e^{cottheta}-e^{tantheta}right)}dtheta$$



The given answer is




$$frac{ pi^2 }{48}$$




and I have verified it numerically. However, I don't know the methods used.



My question is: how to obtain the answer analytically?






integration definite-integrals







share|cite|improve this question











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closed as off-topic by RRL, Lord Shark the Unknown, Saad, Cesareo, Zacky Dec 21 '18 at 10:54


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Lord Shark the Unknown, Saad, Cesareo, Zacky

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 1




    $begingroup$
    So, why don't you share the closed form here?
    $endgroup$
    – Zacky
    Dec 20 '18 at 22:51










  • $begingroup$
    As not to spoil. For those that are interested, it is $pi^2/48$
    $endgroup$
    – william122
    Dec 20 '18 at 22:52






  • 1




    $begingroup$
    This is a recreational problem? I mean you share it for others, or you actually need help?
    $endgroup$
    – Zacky
    Dec 20 '18 at 22:55










  • $begingroup$
    I know the answer, interested in the method. But people can have fun while solving it as well!
    $endgroup$
    – william122
    Dec 20 '18 at 22:56






  • 1




    $begingroup$
    Probably $tantheta=u$ may work
    $endgroup$
    – Tito Eliatron
    Dec 20 '18 at 22:56


















2












$begingroup$


I was told that there is a closed form to the integral $$int_{pi/4}^{pi/2}frac{cos(2theta)e^{cottheta}}{sin^3(2theta)left(e^{cottheta}-e^{tantheta}right)}dtheta$$



The given answer is




$$frac{ pi^2 }{48}$$




and I have verified it numerically. However, I don't know the methods used.



My question is: how to obtain the answer analytically?






integration definite-integrals







share|cite|improve this question











$endgroup$



closed as off-topic by RRL, Lord Shark the Unknown, Saad, Cesareo, Zacky Dec 21 '18 at 10:54


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Lord Shark the Unknown, Saad, Cesareo, Zacky

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 1




    $begingroup$
    So, why don't you share the closed form here?
    $endgroup$
    – Zacky
    Dec 20 '18 at 22:51










  • $begingroup$
    As not to spoil. For those that are interested, it is $pi^2/48$
    $endgroup$
    – william122
    Dec 20 '18 at 22:52






  • 1




    $begingroup$
    This is a recreational problem? I mean you share it for others, or you actually need help?
    $endgroup$
    – Zacky
    Dec 20 '18 at 22:55










  • $begingroup$
    I know the answer, interested in the method. But people can have fun while solving it as well!
    $endgroup$
    – william122
    Dec 20 '18 at 22:56






  • 1




    $begingroup$
    Probably $tantheta=u$ may work
    $endgroup$
    – Tito Eliatron
    Dec 20 '18 at 22:56
















2












2








2


1



$begingroup$


I was told that there is a closed form to the integral $$int_{pi/4}^{pi/2}frac{cos(2theta)e^{cottheta}}{sin^3(2theta)left(e^{cottheta}-e^{tantheta}right)}dtheta$$



The given answer is




$$frac{ pi^2 }{48}$$




and I have verified it numerically. However, I don't know the methods used.



My question is: how to obtain the answer analytically?






integration definite-integrals







share|cite|improve this question











$endgroup$




I was told that there is a closed form to the integral $$int_{pi/4}^{pi/2}frac{cos(2theta)e^{cottheta}}{sin^3(2theta)left(e^{cottheta}-e^{tantheta}right)}dtheta$$



The given answer is




$$frac{ pi^2 }{48}$$




and I have verified it numerically. However, I don't know the methods used.



My question is: how to obtain the answer analytically?






integration definite-integrals




integration definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 21 '18 at 9:52









Lee David Chung Lin

4,29031241




4,29031241










asked Dec 20 '18 at 22:47









william122william122

54912




54912




closed as off-topic by RRL, Lord Shark the Unknown, Saad, Cesareo, Zacky Dec 21 '18 at 10:54


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Lord Shark the Unknown, Saad, Cesareo, Zacky

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by RRL, Lord Shark the Unknown, Saad, Cesareo, Zacky Dec 21 '18 at 10:54


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Lord Shark the Unknown, Saad, Cesareo, Zacky

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    $begingroup$
    So, why don't you share the closed form here?
    $endgroup$
    – Zacky
    Dec 20 '18 at 22:51










  • $begingroup$
    As not to spoil. For those that are interested, it is $pi^2/48$
    $endgroup$
    – william122
    Dec 20 '18 at 22:52






  • 1




    $begingroup$
    This is a recreational problem? I mean you share it for others, or you actually need help?
    $endgroup$
    – Zacky
    Dec 20 '18 at 22:55










  • $begingroup$
    I know the answer, interested in the method. But people can have fun while solving it as well!
    $endgroup$
    – william122
    Dec 20 '18 at 22:56






  • 1




    $begingroup$
    Probably $tantheta=u$ may work
    $endgroup$
    – Tito Eliatron
    Dec 20 '18 at 22:56
















  • 1




    $begingroup$
    So, why don't you share the closed form here?
    $endgroup$
    – Zacky
    Dec 20 '18 at 22:51










  • $begingroup$
    As not to spoil. For those that are interested, it is $pi^2/48$
    $endgroup$
    – william122
    Dec 20 '18 at 22:52






  • 1




    $begingroup$
    This is a recreational problem? I mean you share it for others, or you actually need help?
    $endgroup$
    – Zacky
    Dec 20 '18 at 22:55










  • $begingroup$
    I know the answer, interested in the method. But people can have fun while solving it as well!
    $endgroup$
    – william122
    Dec 20 '18 at 22:56






  • 1




    $begingroup$
    Probably $tantheta=u$ may work
    $endgroup$
    – Tito Eliatron
    Dec 20 '18 at 22:56










1




1




$begingroup$
So, why don't you share the closed form here?
$endgroup$
– Zacky
Dec 20 '18 at 22:51




$begingroup$
So, why don't you share the closed form here?
$endgroup$
– Zacky
Dec 20 '18 at 22:51












$begingroup$
As not to spoil. For those that are interested, it is $pi^2/48$
$endgroup$
– william122
Dec 20 '18 at 22:52




$begingroup$
As not to spoil. For those that are interested, it is $pi^2/48$
$endgroup$
– william122
Dec 20 '18 at 22:52




1




1




$begingroup$
This is a recreational problem? I mean you share it for others, or you actually need help?
$endgroup$
– Zacky
Dec 20 '18 at 22:55




$begingroup$
This is a recreational problem? I mean you share it for others, or you actually need help?
$endgroup$
– Zacky
Dec 20 '18 at 22:55












$begingroup$
I know the answer, interested in the method. But people can have fun while solving it as well!
$endgroup$
– william122
Dec 20 '18 at 22:56




$begingroup$
I know the answer, interested in the method. But people can have fun while solving it as well!
$endgroup$
– william122
Dec 20 '18 at 22:56




1




1




$begingroup$
Probably $tantheta=u$ may work
$endgroup$
– Tito Eliatron
Dec 20 '18 at 22:56






$begingroup$
Probably $tantheta=u$ may work
$endgroup$
– Tito Eliatron
Dec 20 '18 at 22:56












1 Answer
1






active

oldest

votes


















4












$begingroup$

HINT



First of all, notice that



begin{align*}
frac{cos(2theta)e^{cot(theta)}}{sin^{3}(2theta)(e^{cot(theta)}-e^{tan(theta)})} = frac{cot(2theta)csc^{2}(2theta)}{1 - e^{tan(theta)-cot(theta)}} = frac{cot(2theta)csc^{2}(2theta)}{1 - e^{-2cot(2theta)}}
end{align*}



Hence, according to the substitution $w = cot(2theta)$, where $mathrm{d}w = -2csc^{2}(2theta)mathrm{d}theta $, we obtain



begin{align*}
intfrac{cos(2theta)e^{cot(theta)}}{sin^{3}(2theta)(e^{cot(theta)}-e^{tan(theta)})}mathrm{d}theta = -frac{1}{2}intfrac{w}{1-e^{-2w}}mathrm{d}w
end{align*}



Can you proceed from here?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yes, thank you.
    $endgroup$
    – william122
    Dec 20 '18 at 23:12


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

HINT



First of all, notice that



begin{align*}
frac{cos(2theta)e^{cot(theta)}}{sin^{3}(2theta)(e^{cot(theta)}-e^{tan(theta)})} = frac{cot(2theta)csc^{2}(2theta)}{1 - e^{tan(theta)-cot(theta)}} = frac{cot(2theta)csc^{2}(2theta)}{1 - e^{-2cot(2theta)}}
end{align*}



Hence, according to the substitution $w = cot(2theta)$, where $mathrm{d}w = -2csc^{2}(2theta)mathrm{d}theta $, we obtain



begin{align*}
intfrac{cos(2theta)e^{cot(theta)}}{sin^{3}(2theta)(e^{cot(theta)}-e^{tan(theta)})}mathrm{d}theta = -frac{1}{2}intfrac{w}{1-e^{-2w}}mathrm{d}w
end{align*}



Can you proceed from here?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yes, thank you.
    $endgroup$
    – william122
    Dec 20 '18 at 23:12
















4












$begingroup$

HINT



First of all, notice that



begin{align*}
frac{cos(2theta)e^{cot(theta)}}{sin^{3}(2theta)(e^{cot(theta)}-e^{tan(theta)})} = frac{cot(2theta)csc^{2}(2theta)}{1 - e^{tan(theta)-cot(theta)}} = frac{cot(2theta)csc^{2}(2theta)}{1 - e^{-2cot(2theta)}}
end{align*}



Hence, according to the substitution $w = cot(2theta)$, where $mathrm{d}w = -2csc^{2}(2theta)mathrm{d}theta $, we obtain



begin{align*}
intfrac{cos(2theta)e^{cot(theta)}}{sin^{3}(2theta)(e^{cot(theta)}-e^{tan(theta)})}mathrm{d}theta = -frac{1}{2}intfrac{w}{1-e^{-2w}}mathrm{d}w
end{align*}



Can you proceed from here?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yes, thank you.
    $endgroup$
    – william122
    Dec 20 '18 at 23:12














4












4








4





$begingroup$

HINT



First of all, notice that



begin{align*}
frac{cos(2theta)e^{cot(theta)}}{sin^{3}(2theta)(e^{cot(theta)}-e^{tan(theta)})} = frac{cot(2theta)csc^{2}(2theta)}{1 - e^{tan(theta)-cot(theta)}} = frac{cot(2theta)csc^{2}(2theta)}{1 - e^{-2cot(2theta)}}
end{align*}



Hence, according to the substitution $w = cot(2theta)$, where $mathrm{d}w = -2csc^{2}(2theta)mathrm{d}theta $, we obtain



begin{align*}
intfrac{cos(2theta)e^{cot(theta)}}{sin^{3}(2theta)(e^{cot(theta)}-e^{tan(theta)})}mathrm{d}theta = -frac{1}{2}intfrac{w}{1-e^{-2w}}mathrm{d}w
end{align*}



Can you proceed from here?






share|cite|improve this answer









$endgroup$



HINT



First of all, notice that



begin{align*}
frac{cos(2theta)e^{cot(theta)}}{sin^{3}(2theta)(e^{cot(theta)}-e^{tan(theta)})} = frac{cot(2theta)csc^{2}(2theta)}{1 - e^{tan(theta)-cot(theta)}} = frac{cot(2theta)csc^{2}(2theta)}{1 - e^{-2cot(2theta)}}
end{align*}



Hence, according to the substitution $w = cot(2theta)$, where $mathrm{d}w = -2csc^{2}(2theta)mathrm{d}theta $, we obtain



begin{align*}
intfrac{cos(2theta)e^{cot(theta)}}{sin^{3}(2theta)(e^{cot(theta)}-e^{tan(theta)})}mathrm{d}theta = -frac{1}{2}intfrac{w}{1-e^{-2w}}mathrm{d}w
end{align*}



Can you proceed from here?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 20 '18 at 23:02









APC89APC89

2,413420




2,413420












  • $begingroup$
    Yes, thank you.
    $endgroup$
    – william122
    Dec 20 '18 at 23:12


















  • $begingroup$
    Yes, thank you.
    $endgroup$
    – william122
    Dec 20 '18 at 23:12
















$begingroup$
Yes, thank you.
$endgroup$
– william122
Dec 20 '18 at 23:12




$begingroup$
Yes, thank you.
$endgroup$
– william122
Dec 20 '18 at 23:12



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