integrate

i got (-Î /2 ) ln 2
i didn't apply by parts ..i just used the properties of definite integrals and it's easy to solve the question that way

put 2x= y

2 dx=dy

let i be the value of the required integration

then i= (1/2)∫ log sin y dy ( LL 0 AND UL π)
i= ∫ log sin y dy ( LL 0 AND UL π/2)
i= ∫ log cos y dy ( LL 0 AND UL π/2)
ADDING the above two

2i=∫ log (sin y cos y ) dy ( LL 0 AND UL π/2)
multiply and divide by 2 inside the log
2i=∫ log (sin2y) dy - ∫ log2 dy ( LL 0 AND UL π/2)
2i=i- (Ï€/2) log2
hence i= -(Ï€/2) log2

sorry i should have used latex but i am not able to copy the document ....