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











$endgroup$



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












$begingroup$


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



-

Popular posts from this blog

Ellipse (mathématiques)

Image
Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
Read more

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
Read more