Question

ITSURGENT.......... Integrate (2logsinx-logsin2x) within the limit 90 to 0

Answer 1

I= ∫90-0 (2logsinx-logsin2x)  this would be the first equation
I= ∫90-0 (2 log sin(90-x)-log sin 2x) dx
I= ∫90-0 (2log cosx-log sin 2x) dx  this would be our second equation
adding first and second
2I=∫90-0(2log sinx-log sin2x)dx+ ∫90-0(log cosx- log sin2x)dx
2I=2∫90-0(log sinx+logcosx)dx - 2∫90-0 log sin2xdx
I= ∫90-0 log sinxcosxdx - ∫90-0 lod sin2x
I=∫90-0 log sinxcoxdx+ ∫90-0 log 2dx+ ∫90-0 lod 2dx -∫90-0 log sin2x
I=∫90-0 log sinxcosxdx-∫90-0log2dx-∫90-0 logsin2xdx
I=∫90-0 lod sin2xdx - ∫90-0 log 2xdx-∫90-0 log sin2xdx
I=-∫90-0 log 2dx
I=-log2∫90-0 dx
I=-log2 (90 minus 0)
I= -90 log2
I= - pi/2 log2