integrate

put x= sin θ

you 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

ya, put x=sin θ,

now
I=∫sin²Î¸ ln(sinθ) dθ......which can be solved by considering sinθ=u and ln(sinθ) =v [1][1][1][1][1]