integrate: lim-.0-pi xtanx/secx.cosecx Share with your friends Share 22 Vipin Verma answered this ∫0πxtanxsecx×cosecxdx∫0πx×sinxcosx1cosx×1sinxdx= ∫0π xsin2xdx ∫0π x×(1-cos2x)2dx =∫0π (x2-xcos2x2)dx= x24 -(xsin2x4+ 18cos2x) = x24 -xsin2x4- 18cos2xSo putting the limits, we get [π24 -18]-[-18] =π24 (ans) 94 View Full Answer
