Question

trig integral 1/sqrt. (sin2x) dx

Answer 1

hmmm \(\large \int\frac{1}{\sqrt{sin2x}}dx\) is it this?

Answer 2

yup

Answer 3

I don't think there's an elementary result...

Answer 4

looks like a dirty integral!

Answer 5

sorry my mistake..... its integral of sqrt. (1-cos4x) dx

Answer 6

\(1-cos4x ) right?

Answer 7

\(1-cos4x \)

Answer 8

with a square root

Answer 9

yeah! im just making sure it is not a power!
use double angle identity first

Answer 10

cos(2x)=2cos^2x-1
so cos4x=2cos^2(2x)-1

Answer 11

with this you should get rid of square root

Answer 12

will it be sqrt 2 -sqrt (2 cos^2 (2x))?

Answer 13

hmmm
\(1-cos4x=1-(2cos^22x-1)=2-2cos^22x=2(1-cos^22x)=2sin^22x\)

Answer 14

putting this back into our integral
\(\large \int\frac{1}{\sqrt{2sin^22x}}dx\)

Answer 15

since we achieved this
\(\large \int \frac{1}{\sqrt{2}sin(2x)}dx=\frac{1}{\sqrt{2}}\int\frac{1}{sin(2x)}dx\)

Answer 16

now you can integrate that one

Answer 17

thanks a lot

Answer 18

welcome!

Answer 19

sqrt of (1 +sinx)

Answer 20

you mean the result?

Answer 21

the result after integrating this

Answer 22

doesn't look to me a good one!
what did you do?

Answer 23

i haven't done the integral so i don't really know what it would look like

Answer 24

but i bet there should be an ln(...)

Answer 25

\[\int\limits_{0}^{\pi/2} \sqrt{(1 +\sin \theta)}\]

Answer 26

okiee thanks again

Answer 27

where did you come up with the interval? was it not indefinite integral?

Answer 28

no...this is the main question

Answer 29

i don't got what you are telling hehe
what does that have to do with our integral?

Answer 30

its a new question...where i have problem...and this one is not a indefinite integral...but i just wanna know that how to do the integration

Answer 31

how about u substitution?

Answer 32

i tried...but couldn't continue

Answer 33

what u sub did you choose?

Answer 34

u =sin theta

Answer 35

hmmm... i tried u sub for \(1+sin\theta\) the same problem
you have to consider by part

Answer 36

wolfram result
http://www.wolframalpha.com/input/?i=%5Cint+sqrt%281%2Bsinx%29dx
haha, but i don't think it is that way

Answer 37

i didn't got a bit :/

Answer 38

haha, you sure that's the real expression?

Answer 39

yeshh its the real expression...got more weirder integration in stock :/

Answer 40

well, let's stick with u sub haha
u=sintheta