Evaluate: `int(cosx)/(cos3x)dx`

Explanation

We have `int(cosx)/(cos 3x)dx = int(cosx)/((4cos^(3)x-3 cosx))dx`
`=int(dx)/(4cos^(2)x-3)=int(dx)/(4cosx^(2)x-3(sin^(2)x+cos^(2)x))`
`=int(dx)/(cos^(2)x-3sin^(2)x)=int(sec^(2)x)/(1-tan^(2)x)dx` [ on dividing the num . And denom . by `cos^(2)X`]
`=int(dt)/(1-3t^(2))` where tan x t and `sec^(2)xdx=dt`
`=(1)/(3)*int(dt)/(((1)/(3)-t^(2)))=(1)/(3)*int(dt)/([((1)/(sqrt(3)))^(2)-t^(2)])=(1)/(2)*(1)/(2*(1)/(sqrt(3)))log|((1)/(sqrt(3))+t)/((1)/(sqrt(3))-t)|+C`
`=(1)/(2sqrt(3))log|(1+sqrt(3)t)/(1-sqrt(3)t)|+C=(1)/(2sqrt(3))log|(1+sqrt(3)tanx)/(1-sqrt(3)tanx)|+C`.