What is the evaluation of the following of Integral?: #int1/(cos3x-cosx)dx#

ans - #1/4|cosecx-log|secx+tanx||+C#

1 Answer
Mar 17, 2018

#I=1/4[cosecx-log|secx+tanx|]+c#

Explanation:

We have, #color(red)(cos3x=4cos^3x-3cosx#
#=>cos3x-cosx=4cos^3x-3cosx-cosx##=>cos3x-cosx=4cos^3x-4cosx=4cosx(cos^2x-1)#
#=>cos3x-cosx=-4cosx(1-cos^2x)=-4cosxsin^2x#
Hence,
#I=int1/(cos3x-cosx) dx=intcolor(red)(1)/(-4cosxsin^2x)dx#
#I=-1/4intcolor(red)((cos^2x+sin^2x))/(cosxsin^2x)dx#
#I=-1/4int[cos^2x/(cosxsin^2x)+sin^2x/(cosxsin^2x)]dx#
#I=-1/4int[cosx/sin^2x+1/cosx]dx#
#I=-1/4int[cosecxcotxdx+secx]dx#
#I=-1/4[-cosecx+log|secx+tanx|]+c#
#I=1/4[cosecx-log|secx+tanx|]+c#
Hint:,
#(1)intcosecxcotxdx=-cosecx+c#
#(2)intsecxdx=log|secx+tanx|+c#