Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Find the solution of gamma and beta function!

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

Find the solution of gamma and beta function for \[\int\limits_{0}^{\pi/2} \frac{ d \theta }{ \sqrt{\sin \theta} }\]
i dont have enough experience with beta and gamma

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

me neither :(
I am sorry @gerryliyana I havnt learnt this.:(
I don't understand your original question, but this is an elliptic integral.
If the root wasn't there I would have an idea but it's making it a problem with it there.
@gerryliyana can you state more clearly what you wish to do?
Isn't it \[\frac{ \Gamma (m) }{ 2 \times \Gamma (m) }\]
For your beta
ok ok...,for eample i have \[\int\limits_{0}^{\pi/2} \frac{ d \theta }{ \sqrt{\sin \theta} } \] based on Beta function; \[B(p,q) = 2 \int\limits_{0}^{\pi/2} (\sin \theta)^{2p-1} (\cos \theta)^{2q-1} d \theta \] 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 \[\frac{ 1 }{ 2 } \frac{ \Gamma (1/4) \Gamma (1/2)}{ \Gamma (\frac{ 1 }{ 4 }+\frac{ 1 }{ 2 }) }\]
Oh... okay. You didn't clarify that's what you wanted.
Ok guys actually, i wanna check my work.., hei how about these "find the solution of gamma and beta function for \[\int\limits_{0}^{\pi/2} (\tan^{3}\theta + \tan^{5} \theta) e^{-\tan^{2}\theta} d \theta\]
let \(u=\tan^2 \theta\) see what happens, have u tried it yet?
yes., i've tried for this one, but it didn't work, oh my bad :(
it becomes\[\frac{1}{2}\int_{0}^{\infty} u e^{-u} du\]right?
wait a sec..,
then, \[\frac{ 1 }{ 2 } \int\limits_{0}^{\infty} ue^{-u} du = \frac{ 1 }{ 2 } \int\limits_{0}^{\infty} u^{2-1} e^{-u} du\] then p=2
\[= \frac{ 1 }{ 2 } \Gamma(2)\]
and u know that\[\Gamma(n+1)=n!\]we are done :)
ah yeah.., thank you so much @mukushla and other woow we are done :)
very welcome :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question