Question

find the integral of
lnsinx

Answer 1

\[\int \ln (\sin x) dx\]
do integration by parts \[u = \ln (\sin x) \longrightarrow du = \frac{\cos x dx}{\sin x}\] \[dv = dx \longrightarrow v = x\] can you solve it now??

Answer 2

note the formula for integration by parts is \[uv - \int vdu\]

Answer 3

some feedbacks would be nice eh @KANNYTE ;)

Answer 4

i have tried to intergrate but it still comes back at lnsinx

Answer 5

GREAT!!!

Answer 6

is not v=1? @lgbasallote ?

Answer 7

you actually got it ;) can you write your equation here? @kannyte

Answer 8

v= 1? @Arnab09 ? what do you mean?

Answer 9

isn't the derivative of 1 = 0?

Answer 10

u said v=x above^^^
but if u=ln(sinx), then v=1

Answer 11

therefore \(\frac{d}{dx} (1) \ne dx\)

Answer 12

@Arnab09 http://www.wolframalpha.com/input/?i=integral+of+dx ;) hehehe

Answer 13

@KANNYTE pls write here what you got so far..the one with the \(\int \ln (\sin x)dx\) again...it's actually on the right track and im gonna show you the magic ^_^

Answer 14

http://www.wolframalpha.com/input/?i=integral+of+ln%28sinx%29
doesn't making sense.. :/

Answer 15

am having trouble with network, but am writting

Answer 16

oh wait... ln(sinx) o.O *facepalm* i took the derivative!!! wait sorry @KANNYTE i accidentally misled you...lemme recheck

Answer 17

oh wait i was right!

Answer 18

i was supposed to take the derivative -_-

Answer 19

continue writing @KANNYTE

Answer 20

@Arnab09 what was your question regarding my solution?

Answer 21

\[\int\limits uvdx=u \int\limits vdx-\int\limits [du/dx (\int\limits vdx)]dx\]
isn't it?

Answer 22

what is that formula?

Answer 23

for integration by parts..

Answer 24

the formula is \[\int udv = uv -\int vdu\]

Answer 25

and if we differenciate ln(sin x), it is not coming ln(sin x) again..!

Answer 26

derivative of \(\ln (\sin x) = \frac{1}{\sin x} (\cos xdx)\)

Answer 27

hm..

Answer 28

http://www.wolframalpha.com/input/?i=derivative+of+ln+%28sin+x%29

Answer 29

what about you @KANNYTE ? still typing?

Answer 30

i know that..^^
how to integrate that vdu here?

Answer 31

a second integration by parts

Answer 32

oh.. okay.. got it :)

Answer 33

this time you let \[u = x \longrightarrow du = dx\] \[dv = \cot x dx \longrightarrow \ln (\sin x)\]

Answer 34

actually ur formula and my formula of integration by parts method are same, @lgbasallote :P

Answer 35

lol you have to let \[u = \frac{\cos x}{\sin x}\] or else it becomes 0

Answer 36

im getting 0 even with that sub o.O can you check sir @eliassaab ?

Answer 37

@Arnab09,@lgbasallote. What you are getting from Wolfram Alpha indicates that this
integral does not have a closed form in terms of well know functions.

Answer 38

i see :/

Answer 39

true ;;; but this is quite interesting
\[ \int_0^{\pi/2} \ln \sin xdx = \frac{-\pi \ln 2 }{2}\]

Answer 40

http://www.wolframalpha.com/input/?i=integrate+ln%28sinx%29+from+0+to+pi%2F2

Answer 41

Also

\[
\int_{\frac{\pi }{4}}^{\frac{\pi
}{2}} \log (\sin (x)) \,
dx=\frac{1}{4} (2 C-\pi \log
(2))\\
C \text { is the Catalan Constant }\\
C=0.91596559417721901505460351493238411077414937428167
\]

Answer 42

This is getting more interesting ... !!
Can you give some link regarding this professor??

Answer 43

http://answers.yahoo.com/question/index?qid=20110926082734AAtEI4K

Answer 44

http://www.ams.org/journals/proc/2005-133-05/S0002-9939-04-07863-3/S0002-9939-04-07863-3.pdf

Answer 45

ty prof