Souradeep Majumder
Last Activity: 14 Years ago
yah answer is possible........
log sinx+ log cos=log(sinxcox)+logsin2x-log2
2x=z
2dx=dz
limit is 0 to pi...........
adjust 1/2 which is comes from 2dx=dz with pi then limit gonna be 0 to pi/2 integration log(sinz)dz which is log(sinx)dx=i
2i=i-log2integration o to pi/2 dx
then i think you can easily solve......... hobei amar
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