integrate 0 to pie/2 (2log|cosx|-log|sin2x|)dx Share with your friends Share 2 Manbar Singh answered this Let I = ∫0π/2 2 logcos x - logsin 2x dxNow,we know that,cos x = cos x, 0⩽x⩽π/2sin 2x = sin 2x, 0⩽2x⩽π So,I = ∫0π/2 2 log cos x - log sin 2x dxI = ∫0π/2logcos2x - logsin 2x dxI =∫0π/2logcos2xsin 2x dxI = ∫0π/2logcos x . cos x2 sin x . cos x dxI = ∫0π/2logcot x2 dxI = ∫0π/2logcot x - log 2 dxI = ∫0π/2logcot x dx - log 2∫0π/2dx I = ∫0π/2logcot x dx - log 2 x0π/2I =∫0π/2logcot x dx - log 2π2 - 0I = ∫0π/2logcot x dx - π2log 2 ..........1Let I1 = ∫0π/2logcot x dx .........2⇒ I1 =∫0π/2 logcotπ2 - x dx⇒ I1 = ∫0π/2logtan x dx .........3Adding 2 and 3, we get 2I1 = ∫0π/2logcot x dx + ∫0π/2logtan x dx⇒ 2I1 = ∫0π/2logcot x + logtan x dx⇒ 2I1 = ∫0π/2 logcot x . tan x dx⇒ 2I1 = ∫0π/2logcos xsin x×sin xcos x dx⇒ 2I1 = ∫0π/2 log 1 dx⇒ 2I1 = ∫0π/20 dx as, log 1 = 0⇒I1 = 0Now, from 1, we get I = 0 - π2 log 2⇒I =-π2 log 2 7 View Full Answer