1. **State the problem:** Evaluate the integral $$\int \frac{\csc^2(x)}{\cot(x)} \, dx$$. 2. **Recall the trigonometric identities:** - $$\csc(x) = \frac{1}{\sin(x)}$$ - $$\cot(x) = \frac{\cos(x)}{\sin(x)}$$ - Also, $$\csc^2(x) = 1 + \cot^2(x)$$ but here we will use the definitions. 3. **Rewrite the integral using definitions:** $$\int \frac{\csc^2(x)}{\cot(x)} \, dx = \int \frac{\frac{1}{\sin^2(x)}}{\frac{\cos(x)}{\sin(x)}} \, dx = \int \frac{1}{\sin^2(x)} \cdot \frac{\sin(x)}{\cos(x)} \, dx = \int \frac{1}{\sin(x) \cos(x)} \, dx$$ 4. **Simplify the integral:** $$\int \frac{1}{\sin(x) \cos(x)} \, dx$$ 5. **Use substitution:** Let $$u = \sin(x)$$, then $$du = \cos(x) \, dx$$. Rewrite the integral in terms of $$u$$: $$\int \frac{1}{u \cos(x)} \, dx = \int \frac{1}{u} \cdot \frac{1}{\cos(x)} \, dx$$ But since $$du = \cos(x) \, dx$$, then $$\frac{dx}{\cos(x)} = \frac{du}{\cos^2(x)}$$ which complicates substitution. 6. **Alternative substitution:** Rewrite the integral as: $$\int \frac{1}{\sin(x) \cos(x)} \, dx = \int \frac{\sin(x) + \cos(x)}{\sin(x) \cos(x) (\sin(x) + \cos(x))} \, dx$$ which is not simpler. 7. **Use the identity:** Recall that $$\frac{1}{\sin(x) \cos(x)} = \frac{2}{\sin(2x)}$$. So the integral becomes: $$\int \frac{1}{\sin(x) \cos(x)} \, dx = \int \frac{2}{\sin(2x)} \, dx = 2 \int \csc(2x) \, dx$$ 8. **Integral of $$\csc(kx)$$:** $$\int \csc(kx) \, dx = -\frac{1}{k} \ln \left| \csc(kx) + \cot(kx) \right| + C$$ 9. **Apply the formula:** $$2 \int \csc(2x) \, dx = 2 \left(-\frac{1}{2} \ln \left| \csc(2x) + \cot(2x) \right| \right) + C = - \ln \left| \csc(2x) + \cot(2x) \right| + C$$ **Final answer:** $$\boxed{- \ln \left| \csc(2x) + \cot(2x) \right| + C}$$