integration of ln(sinx) ?
\[\int\limits_{}^{}\ln(sinx)\]
You must remember the by parts rule right?
yes i know.
1*ln(sinx)
Take 1 as the function to integrate and ln(sinx) to differentiate
\[\int\limits_{}^{}u.v=u \int\limits_{}^{}v-\int\limits_{}^{}(du/dx.\int\limits_{}^{}vdx)\]
is it right ?
ah Yes it is U = ln sinx V =1
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
okay use by parts after substitution again
\[x = sin^{-1}t\]
trying ...
\[x=\sin^{-1} t \] didnt work well ..it gets me more to intgrate
yup okay I have to try it on my notebook
thank you very much . sensei
ah I didn't get much ahead but see the integral scroll down a little
This one was pretty tough. wondering why these questions are given to IIT aspirants
lol
Yea I know did you solve quadratic for IIT
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
question is quadratic related
http://openstudy.com/users/ishaan94#/users/ishaan94/updates/4e5be0050b8b1f45b4865084
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