Integration of sqrt(sinx)
interesting... \[\int \sqrt{\sin x}dx\] right?
Yeah!
@lgbasallote Can you plz give me the solution. Completely raked my brains and still no solution!!!
lol i dont wanna hurt mine =)))
Yeah right!!!!!!!!
@experimentX Can you plz give me the solution. Completely raked my brains and still no solution!!!
One was is \[ \sqrt{\sin x} = {1 \over 2i}(e^{ix}-e^{-ix})\]
@experimentX Can you tell me how to get the answer. I am interested in the process by which you solved.
oh, that follows from the definition \[ e^{ix} = \cos x + i\sin x\]
Try Integration by parts, perhaps?
@experimentX can you please elaborate.
do you know that Euler's identity?
No i dont know.
sorry .. that wouldn't work. I didn't realize it does not have any closed form http://www.wolframalpha.com/input/?i=integrate+sqrt%28sin+x%29
What the heck is this? http://reference.wolfram.com/mathematica/ref/EllipticE.html
elliptic integrals are non elementary integrals associated with arc length of ellipse. I don't know much about it.
@mukushla \[\int\limits \sqrt{\sin (\tan^{-1}{2t})} \ \ dx\] dx turns into dt here how exactly? o_O
is the integral solvable??
@.Sam. HELP!!
wolfram says it is not.
I'm just guessing \[\int\limits_{}^{}\sqrt{sinx}~dx\] \[\int\limits\limits_{}^{}\sin(x)^\frac{1}{2}~dx\] \[-\frac{2\sin(x)^\frac{3}{2}}{3\cos(x)}\] But you can't use this because this integral is not elementary. So, you have to reduce it to an elliptic integral like what they said.
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