Question

\[\int\frac{1}{\sqrt{sinx}}
dx\]

Answer 1

zomg

Answer 2

lol, whats wrong

Answer 3

algebraic sub....

Answer 4

try it then :P

Answer 5

\[u = \sqrt{\sin x}\]\[u^2 = \sin x\]\[x = \sin^{-1}(u^2)\]Is that how you start?

Answer 6

this looks like one of @Mimi_x3's old integrals

Answer 7

you will get into a dead end

Answer 8

Right, what do I have to do?

Answer 9

its dirty integral :P

Answer 10

I dislike the taste of ugly integrals :)

Answer 11

hehe uglier than ugly integrals :P

Answer 12

substitute sin x= t^2

Answer 13

it is an integral that does not have a closed form..

Answer 14

reduction formula

Answer 15

wallis formula

Answer 16

lol, do you know what it is igba

Answer 17

riemann sum

Answer 18

no idea. fancy words?

Answer 19

or use sqrt(cosec x) = 1/2[{ (sqrt(cosec x) + cot (x))} + { sqrt(cosec x) - sqrt cot (x)}]

Answer 20

last time i checked Riemann sum was used for definite integral :P

Answer 21

sqrt(cosec x) = 1/2[ {sqrt(cosec x) + sqrtcot (x))} + { sqrt(cosec x) - sqrt cot (x)}]

Answer 22

How about just saying the following? :P\[\ln|\sqrt{\sin x}|\]

Answer 23

lol, that is not integrating properly..

i iwsh it was that easy :p

Answer 24

sqrt(cosec x) = 1/2[{ (sqrt(cosec x) + cot (x))} + { sqrt(cosec x) - sqrt cot (x)}]..
use this and integrate the rhs ...i.e sum of two integrands so integrate them separately

Answer 25

but its an integral that does not have a closed form; can you do it like that?

Answer 26

There's actual theory dedicated to this... it's not definite:
http://en.wikipedia.org/wiki/Elliptic_integral

Answer 27

sinx = t^2
=>cos x = sqrt(1-t^4)
now dx = 2t dt / (sqrt(1-t^4))
making that substituion,
we have
2 integral ( dt / sqrt(1-t^4) )
does this help ?

Answer 28

(definite as in 'closed form', exact expression)

Answer 29

well does it mean it cant be solved analytically?