Integral of logarithm of cotangent divided by squared sine.

Question

My task is: $\int\frac{\log(|\cot x|)}{\sin^2x}\,dx$

If i try using the formula: $\int f(x)*g'(x) dx = f(x)*g(x) - \int f'(x)*g(x)dx$

At some point it throws out

$\int \tan(x)\,dx$

which even Wolfram says is incalculable on my basic university level.

I can't come up with any good substitution method. Any ideas? Thanks.

Answer 1

Are you sure Wolfram says $\displaystyle\tan x\,\mathrm d x$ is incalculable? It is simply, by the substitution $u=\cos x$, $\color{red}{-\ln(\cos x)}$.

Here, again the substitution $$u=\cot x,\quad\mathrm d u=-\frac1{\sin^2x}\,\mathrm dx,$$ (dropping the absolute value, since the logarithm requires the argument be positive) leads to to a well-known integral $$\int\frac{\ln(\cot x)}{\sin^2x}\,\mathrm dx =-\int\ln u\,\mathrm du= -(u\ln u-u).$$

Answer 2

If you have gotten things down to $\int \tan x\mathrm \;dx$, this is certainly possible to do with undergrad techniques... It is a u substitution! $$ \int\frac{\sin x}{\cos x}\;\mathrm dx\stackrel{u=\cos x}{=}-\int\frac{1}{u}\;\mathrm du=\ln|\sec x|+C $$

Answer 3

Integrating by parts setting $$u=\ln (|\cot x|), v'= \frac {1}{\sin^2 x} $$ We have: $$I = \int \frac {\log |\cot x|}{\sin^2 x} \, dx = -\cot x \ln |\cot x| - \int \csc^2 x\, dx $$ $$= -\cot x \ln |\cot x| +\cot x $$