We need to evaluate the integral:

$I = \int \cot(x) \ln(\sin x) \, dx$

Step 1: Express $\cot(x)$ in terms of $\sin(x)$ and $\cos(x)$

We use the identity:

$\cot(x) = \frac{\cos(x)}{\sin(x)}$

Thus, the integral becomes:

$I = \int \frac{\cos(x)}{\sin(x)} \ln(\sin x) \, dx$

Step 2: Use Substitution

Let:

$u = \sin(x) \Rightarrow du = \cos(x) dx$

Rewriting the integral in terms of $u$:

$I = \int \frac{du}{u} \ln u$

Step 3: Solve the Integral

The integral:

$\int \frac{\ln u}{u} \, du$

is a standard result:

$\frac{(\ln u)^2}{2} + C$

Step 4: Substitute Back $u = \sin x$

$I = \frac{(\ln \sin x)^2}{2} + C$

Final Answer:

$\int \cot(x) \ln(\sin x) \, dx = \frac{(\ln \sin x)^2}{2} + C$

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Would you like a detailed explanation for any step?

Here are some related integrals you might find interesting:

1. $\int \tan(x) \ln(\cos x) \, dx$
2. $\int \ln(\sin x) \, dx$
3. $\int \ln(\cos x) \, dx$
4. $\int \cot^2(x) \ln(\sin x) \, dx$
5. $\int \cot(x) \, dx$

Tip: The substitution $u = \sin(x)$ is very useful for integrals involving trigonometric functions and logarithms!