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

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

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0i dont have enough experience with beta and gamma

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

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

abb0t
 one year ago
Best ResponseYou've already chosen the best response.1If the root wasn't there I would have an idea but it's making it a problem with it there.

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

abb0t
 one year ago
Best ResponseYou've already chosen the best response.1Isn't it \[\frac{ \Gamma (m) }{ 2 \times \Gamma (m) }\]

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.1ok 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 \[\frac{ 1 }{ 2 } \frac{ \Gamma (1/4) \Gamma (1/2)}{ \Gamma (\frac{ 1 }{ 4 }+\frac{ 1 }{ 2 }) }\]

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

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

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

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

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

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

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

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

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