Question

integration of log tanx

Answer 1

I have solved this integral by substitution

method

$Let\:I=\int{log\,tanx}\:dx$

Take

$t=log\,tanx$

$\frac{dt}{dx}=\frac{1}{tanx}sec^2x$

$\frac{dt}{dx}=\frac{1}{tanx}(1+tan^2x)$

$\frac{dt}{dx}=\frac{1}{t}(1+t^2)$

$\implies\:dx=\frac{t\:dt}{1+t^2}$

Now,

$I=\int{t\,\frac{t}{1+t^2}}dt$

$I=\int{\frac{t^2\:dt}{1+t^2}}dt$

$I=\int[\frac{(1+t^2)-1}{1+t^2}]dt$

$I=\int[\frac{1+t^2}{1+t^2}-\frac{1}{1+t^2}]dt$

$I=\int[1-\frac{1}{1+t^2}]dt$

$I=t-tan^{-1}t+C$

$\implies\boxed{\bf\,I=log\,tanx-tan^{-1}(log\,tanx)+C}$