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 know that
$\cot x = \frac{\cos x}{\sin x}$ Thus, rewriting the integral:

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

Step 2: Substituting $u = \sin x$

Let $u = \sin x$, so that $du = \cos x \, dx$. Then, the integral transforms into:

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

Step 3: Solve the Integral Using Integration by Parts

Using integration by parts, where we set:

• $v = \ln u \Rightarrow dv = \frac{du}{u}$
• $w = u \Rightarrow dw = du$

Applying the formula:

$\int v \, dw = vw - \int w \, dv$

we get:

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

Since $\int du = u$, we simplify:

$I = u \ln u - u + C$

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

$I = \sin x \ln (\sin x) - \sin x + C$

Thus, the final answer is:

$\int \cot x \ln(\sin x) \, dx = \sin x \ln (\sin x) - \sin x + C.$

---

Would you like a deeper explanation of any step?

Related Problems:

1. Evaluate $\int \cot x \ln(\cos x) \, dx$.
2. Compute $\int \tan x \ln(\sin x) \, dx$.
3. Solve $\int x \ln x \, dx$ using integration by parts.
4. Find $\int \ln(\sin x) \, dx$.
5. Compute $\int \cos x \ln(\sin x) \, dx$.

Tip:

When dealing with logarithms inside an integral, substitution and integration by parts are often useful techniques.