1ya, put x=sin θ,
now
I=∫sin2θ ln(sinθ) dθ......which can be solved by considering sinθ=u and ln(sinθ) =v [1][1][1][1][1]
3 Answers
1you will get ,
x= sin a
so dx = cos a da
so i= sin2a ln (sin a ) da
write sin2a=(1- cos 2a)/2 and split both the terms solve them individually by parts .......i have given a hint (rather half of the solution) now proceed
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