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Mathematics 20 Online
OpenStudy (anonymous):

integration of ln(sinx) ?

OpenStudy (anonymous):

\[\int\limits_{}^{}\ln(sinx)\]

OpenStudy (anonymous):

You must remember the by parts rule right?

OpenStudy (anonymous):

yes i know.

OpenStudy (anonymous):

1*ln(sinx)

OpenStudy (anonymous):

Take 1 as the function to integrate and ln(sinx) to differentiate

OpenStudy (anonymous):

\[\int\limits_{}^{}u.v=u \int\limits_{}^{}v-\int\limits_{}^{}(du/dx.\int\limits_{}^{}vdx)\]

OpenStudy (anonymous):

is it right ?

OpenStudy (anonymous):

ah Yes it is U = ln sinx V =1

OpenStudy (anonymous):

i got stuck up at \[x \ln sinx.\int\limits_{}^{}xcos x dx/sinx\] i cant make substitution of sinx =t as x remains there

OpenStudy (anonymous):

okay use by parts after substitution again

OpenStudy (anonymous):

\[x = sin^{-1}t\]

OpenStudy (anonymous):

trying ...

OpenStudy (anonymous):

\[x=\sin^{-1} t \] didnt work well ..it gets me more to intgrate

OpenStudy (anonymous):

yup okay I have to try it on my notebook

OpenStudy (anonymous):

thank you very much . sensei

OpenStudy (anonymous):

http://www.wolframalpha.com/input/?i=ln%28sinx%29

OpenStudy (anonymous):

ah I didn't get much ahead but see the integral scroll down a little

OpenStudy (anonymous):

This one was pretty tough. wondering why these questions are given to IIT aspirants

OpenStudy (anonymous):

lol

OpenStudy (anonymous):

Yea I know did you solve quadratic for IIT

OpenStudy (anonymous):

ah I still can't figure out how am I supposed to do them in 2 minutes I have posted a question go to my profile you will see

OpenStudy (anonymous):

question is quadratic related

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