Evaluate: `intcos2x" ln "(1+tanx)dx`

Answer

`(1)/(2)[sin2xlog_(e)(1+tanx)-xlog|sinx+cosx|]+C`

Explanation

`int cos 2x log_(e)(1+tan x)dx`
`=(1)/(2)sin2x log_(e)(1+tanx)-(1)/(2)int(sec^(2)x)/((1+tanx))sin2xdx`
`=(1)/(2)sin2x log_(e)(1+tanx)-int(tanx+1-1)/(1+tanx)dx`
`=(1)/(2)sin2x log_(e)(1+tanx)-(1)/(2)int(2cos x+sinx-sinx)/(sinx+cosx)dx`
`=(1)/(2)sin2x log_(e)(1+tanx)x+(x)/(2)+(1)/(2)int(cos x-sinx)/(sinx+cosx)dx`
`=(1)/(2)[sin2xlog_(e)(1+tanx)-x+log|sinx+cosx|]+C`