Question #33b66

1 Answer
Manasvini Nittala
Nov 25, 2017

the solution is #I=-cos(logx)+c#

Explanation:

let, #I=intsin(logx)/xdx->(1)#
let us use substitution method
let, #t=logx->(2)# then #dt=1/xdx->(3)# substitute #(2) and (3)# in #(1)#
we get,#I=intsintdtrArr-cost+crArr-cos(logx)+c#
#:.I=-cos(logx)+c#

Related questions
Impact of this question
749 views around the world
You can reuse this answer
Creative Commons License