Ritika Das
Last Activity: 7 Years ago
Let 1/x=dv
After integrating, log x.dx = v
Also, let tan x = u
Therefore,sec2x.dx=du
Integral of u.dv= uv-integral vdu
=log x.tan x- integral log x.sec2x.dx
Integral of tanx/x dx= log x.tanx- integral logx sec2x.dx.........(1)
Right hand side 2nd term consideration, we get : –
Integral log x.sec2x dx,
By integration by parts method,
Let u= log x
Therefore du= (1/x).dx integral sec2x dx= dv ,
Putting value we get v= tan x
So thus log x.tan x- integral tan x/x.dx
Putting value in next part of the right hand side we get..... (1)
Integral tan x/x.dx= log x.tan x + log x.tan -integral tan x/x.dx
2integral tan x/x.dx= 2log x.tan x
Thus,
Integral tan x/x =log x.tan x+c
There you go!