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gerryliyanaBest ResponseYou've already chosen the best response.1
Find the solution of gamma and beta function for \[\int\limits_{0}^{\pi/2} \frac{ d \theta }{ \sqrt{\sin \theta} }\]
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.1
@ajprincess @amistre64 @BAdhi
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
i dont have enough experience with beta and gamma
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
I am sorry @gerryliyana I havnt learnt this.:(
 one year ago

oldrin.batakuBest ResponseYou've already chosen the best response.0
I don't understand your original question, but this is an elliptic integral.
 one year ago

abb0tBest ResponseYou've already chosen the best response.1
If the root wasn't there I would have an idea but it's making it a problem with it there.
 one year ago

oldrin.batakuBest ResponseYou've already chosen the best response.0
@gerryliyana can you state more clearly what you wish to do?
 one year ago

abb0tBest ResponseYou've already chosen the best response.1
Isn't it \[\frac{ \Gamma (m) }{ 2 \times \Gamma (m) }\]
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.1
ok ok...,for eample i have \[\int\limits_{0}^{\pi/2} \frac{ d \theta }{ \sqrt{\sin \theta} } \] based on Beta function; \[B(p,q) = 2 \int\limits_{0}^{\pi/2} (\sin \theta)^{2p1} (\cos \theta)^{2q1} d \theta \] then, i have \[\int\limits_{0}^{\pi/2} \frac{ d \theta }{ \sqrt{\sin \theta} } \] \[\int\limits_{0}^{\pi/2} (\sin \theta)^{1/2} (\cos \theta)^{0} d \theta\] \[\int\limits\limits_{0}^{\pi/2} (\sin \theta)^{2 (1/4)1} (\cos \theta)^{2 (1/2)1} d \theta \] then i have \[B(p,q)=2 \int\limits\limits\limits_{0}^{\pi/2} (\sin \theta)^{2 (1/4)1} (\cos \theta)^{2 (1/2)1} d \theta \] \[\int\limits\limits\limits_{0}^{\pi/2} (\sin \theta)^{2 (1/4)1} (\cos \theta)^{2 (1/2)1} d \theta = \frac{ 1 }{ 2 } B(p,q)\] with p = 1/4 and q = 1/2; then (relation Beta and gamma) \[B(p,q) = \frac{ \Gamma (p) \Gamma (q) }{ \Gamma (p+q) }\] \[B(p,q) = \frac{ \Gamma (1/4) \Gamma (1/2) }{ \Gamma (1/4+1/2) }\] then i have \[\int\limits\limits\limits_{0}^{\pi/2} (\sin \theta)^{2 (1/4)1} (\cos \theta)^{2 (1/2)1} d \theta = \frac{ 1 }{ 2 } B(p,q)\] \[\int\limits\limits\limits\limits_{0}^{\pi/2} (\sin \theta)^{2 (1/4)1} (\cos \theta)^{2 (1/2)1} d \theta = \frac{ 1 }{ 2 } \frac{ \Gamma(1/4)\Gamma (1/2) }{ \Gamma (1/4+1/2 )}\] finally The solution of gamma and Beta function for \(\int\limits_{0}^{\pi/2} \frac{ d \theta }{ \sqrt{\sin \theta} } \) is \[\frac{ 1 }{ 2 } \frac{ \Gamma (1/4) \Gamma (1/2)}{ \Gamma (\frac{ 1 }{ 4 }+\frac{ 1 }{ 2 }) }\]
 one year ago

oldrin.batakuBest ResponseYou've already chosen the best response.0
Oh... okay. You didn't clarify that's what you wanted.
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.1
Ok guys actually, i wanna check my work.., hei how about these "find the solution of gamma and beta function for \[\int\limits_{0}^{\pi/2} (\tan^{3}\theta + \tan^{5} \theta) e^{\tan^{2}\theta} d \theta\]
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
let \(u=\tan^2 \theta\) see what happens, have u tried it yet?
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.1
yes., i've tried for this one, but it didn't work, oh my bad :(
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
it becomes\[\frac{1}{2}\int_{0}^{\infty} u e^{u} du\]right?
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.1
then, \[\frac{ 1 }{ 2 } \int\limits_{0}^{\infty} ue^{u} du = \frac{ 1 }{ 2 } \int\limits_{0}^{\infty} u^{21} e^{u} du\] then p=2
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.1
\[= \frac{ 1 }{ 2 } \Gamma(2)\]
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
and u know that\[\Gamma(n+1)=n!\]we are done :)
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.1
ah yeah.., thank you so much @mukushla and other woow we are done :)
 one year ago
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