Question

integrate log(sin^2x)

Answer 1

hmm any thoughts?

Answer 2

did you try anything

Answer 3

is that log base 10
or just ln?

Answer 4

Its ln. My friend tried using integration by parts but it goes on infinitely.

Answer 5

hmm integration by part! what do you mean it goes to infinity
do you there is no end the how many times you need to do integration by part

Answer 6

integration by part makes good start to me
since u sub would not work

Answer 7

true.

Answer 8

you might want to repeat that integration by part
call your initial integral I=int (ln(sin^2x))dx

Answer 9

ill try it out

Answer 10

yeah try and see if there is any cancellation
otherwise the integration by part cannot solve this

Answer 11

nope there isnt. it just gets more complicated.

Answer 12

hmm
from wolfram seems there is something you can do
http://www4b.wolframalpha.com/Calculate/MSP/MSP45861h30e7hbigg4ec5e00005972aaa8e0d8ac34?MSPStoreType=image/gif&s=56&w=521.&h=34.

Answer 13

that's a nasty solution by wolfram, but there should be a more interesting solution

Answer 14

wow that was so complicated :/

Answer 15

hehe that's a crazy one!
try you might find something interesting

Answer 16

i want to give it a try but i'm lazy hehehe

Answer 17

or the questions prolly wrong :P

Answer 18

where did u get the question! i don't think it's wrong lol

Answer 19

from my math book

Answer 20

what book?

Answer 21

You prolly wont know the book

Answer 22

did you find:
\[I=x\ln(\sin^2x)-2\int\limits x\cot x dx\]

Answer 23

oh wait yeah. that was so stupid

Answer 24

just say the name of the book! lol
is it stewart's book?

Answer 25

i think you can solve that second integral easily? why there is a problem

Answer 26

i should have just expanded\[xsin2x/(1-\cos2x)\]

Answer 27

lol i feel like an idiot.

Answer 28

expand what?
why did you make your life hard

Answer 29

cotx is something that you can integrate

Answer 30

yeah thanks

Answer 31

welcome

Answer 32

hmm i see you wrote 2sinxcosx=sin2x

Answer 33

darn the website is lagging again

Answer 34

a wise u v is
u=x , dv=cotx

Answer 35

we know integral of cotx already

Answer 36

yeah i did that and u end up with a bunch of stuff and integral log|sinx|

Answer 37

i found a nice solution

Answer 38

what's your result

Answer 39

ugh i give up it keeps going on..... or im not in the mood to solve this. Show me the solution

Answer 40

hold on...

Answer 41

i made a mistake
the result so far is \[\int\limits x\cot xdx=x\ln|\sin x|-\int\limits\ln(\sin x)dx\]

Answer 42

yeah i got that

Answer 43

we need another integration by part
this is a lot of fun carry the integration

Answer 44

i suppose that integral will be nice to deal with

Answer 45

you will probably go back to that cotx thingy

Answer 46

i told u it goes on infinitely

Answer 47

no it won't believe me

Answer 48

u ust said it urself

Answer 49

it goes back to the cot thingy

Answer 50

you will have
\[J=x\ln(\sin x)-\int\limits\ln(\sin x)dx=x\ln(\sin x)-x\ln(\sin x)+\int\limits x\cot xdx\]

Answer 51

where \[J=\int\limits x\cot xdx\]

Answer 52

nice cancellations in fact

Answer 53

what wait we still have \[\int\limits(xcotx)dx\]

Answer 54

hmm there is a problem hehe
i must of made some mistake somewhere

Answer 55

yes! i was hoping i would eliminate that j

Answer 56

but that doesnt happen instead u have an INFINITE loop

Answer 57

that's not the question hehee

Answer 58

bleh i dont think this problem is correct but if ur still determined that there is an answer then try solving it :P

Answer 59

yeah, I'm still sure that it ends somewhere
buy you still don't want to tell me what's name of the book lol

Answer 60

its from a sample paper for some competetive exam

Answer 61

well name them lol!
you don't have to anyway, hmm... i see that's why it a little of nasty
but it is not all complicated!

Answer 62

haha solve it and show me the final answer then

Answer 63

i still want to try one more trick hehe
but time is running out lol
thinking about substitution for \[\int\limits \ln(\sin x)dx\]

Answer 64

if i didn't solve it, i will come back with the solution later once i have time
don't close the question

Answer 65

k sure.

Answer 66

ideas are floating hehe
damned looping integral, i will solve it no matter what

Answer 67

seems my way was wrong hehe

Answer 68

@ganeshie8

Answer 69

hey forget it. My friend was being stupid and forgot to tell me that it was a definite integrals problem
That makes the whole thing much easier

Answer 70

yeah what are the limits

Answer 71

it should have some solution even if it's indefinite

Answer 72

i got stuck with \[\int\limits \ln(\sin x)dx\]
integration by part doesn't work

Answer 73

i had enough of this one lol, time to sleep hehe

Answer 74

http://math.stackexchange.com/questions/37829/computing-the-integral-of-log-sin-x?lq=1