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

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

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

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

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

question is quadratic related