The easiest way possible to evaluate $\int_{}^{}\frac{\ln{(\cos{x})}}{1-\cos{x}}dx$

Question

I want to solve this integral the easiest way possible, how should I do it? $$\int_{}^{}\frac{\ln{(\cos{x})}}{1-\cos{x}}dx$$

I tried with substitution, but it didn't work out. And I think that it isn't possible to use integration-by-parts method - at least I couldn't find a way to do it.

Answer 1

Using $1-\cos x= 2\sin^2 \frac x2$, $$I=\int \ln(\cos x) \cdot \frac 12 \csc^2 \frac x2 dx $$ Substitute $t=\cot \frac x2 \implies dt =-\frac 12\csc^2 \frac x2 dx$ : $$I=-\int\ln(\cos x) dt $$ Now, there’s a golden chance to use $$\cos x = \frac{1-\tan^2 (x/2)}{1+\tan^2(x/2)} = \frac{\cot^2 (x/2) -1}{\cot^2(x/2)+1} $$ $$I=\int \ln\left(\frac{t^2+1}{t^2-1} \right) dt =\int \ln(t^2+1)dt-\int \ln(t^2-1) dt$$ And then these can be integrated by parts easily.

Answer 2

Hint. Since $D\left(\frac{1}{\tan(x/2)}\right)=-\frac{1}{1-\cos(x)}$, by integrating by parts we have that $$\int\frac{\ln(\cos(x))}{1-\cos(x)}\,dx=-\frac{\ln(\cos(x))}{\tan(x/2)} -\int \frac{\tan(x)}{\tan(x/2)}\, dx.$$ Then use the Weierstrass substitution. Can you take it from here?

Answer 3

Integrating by parts is maybe the easiest

\begin{align} \int_{}^{}\frac{\ln{(\cos{x})}}{1-\cos{x}}dx &=-\int \ln(\cos x)d(\cot \frac x2)\\ &=-\cot \frac x2\ln(\cos x) -\int (1+\sec x)dx\\ &= -\cot \frac x2\ln(\cos x) -x -\ln(\tan x+\sec x) +C \end{align}