Question

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

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

Answer 1

`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`