Integral limit 0 toπ/4 log sinx Share with your friends Share 0 Mayank answered this Let K = ∫0π/4 log sin x dx⇒K = 12∫0π/2 log sin x dx⇒K = 12I ......1 now from 1, we getK = 12 × -π2 log 2 = -π4 log 2 0 View Full Answer
