The correct option is
D −(π/2)log2We are given, I=∫π/20(2logsinx−logsin2x)dx
→ We know that
∫baf(x)dx=∫baf(a+b−x)dx.
So, I=∫π/20[2log(sin(π2−x)))−log(sin2(π2−2))]dx
∴I=∫π/20(2logcosx−logsin2x)dx........(2)
We add Equation (1) & (2). We get
∴2I=∫π/20(2logsinx−logsin2x+2logcosx−logsinx)dx
∴2I=∫π/20[logsinx+logcosx−logsin2x]dx
So, I=∫π/20[logsinx+logcosx−log(2sinxcosx)]dx
As we know, log(a.b)=loga+logb
I=∫π/20logsinx+logcosx−[log2+logsinx+logcosx]dx
I=∫π/20(logsinx+logcosx−logx−logsinx−logcosx)dx
I=∫π/20log2dx=−log2[x]π/20
I=−log2[π2−0]=−(π/2)log2