1. **Problem:** Evaluate the integral $$\int \cot^3(5x) \csc^5(5x) \, dx$$. 2. **Formula and rules:** Recall that $$\cot x = \frac{\cos x}{\sin x}$$ and $$\csc x = \frac{1}{\sin x}$$. Also, $$\frac{d}{dx} \csc x = -\csc x \cot x$$. 3. **Rewrite the integral:** $$\int \cot^3(5x) \csc^5(5x) \, dx = \int \cot^3(5x) \csc^4(5x) \csc(5x) \, dx$$. 4. **Express powers in terms of sine and cosine:** $$\cot^3(5x) = \frac{\cos^3(5x)}{\sin^3(5x)}, \quad \csc^4(5x) = \frac{1}{\sin^4(5x)}, \quad \csc(5x) = \frac{1}{\sin(5x)}$$. So the integral becomes: $$\int \frac{\cos^3(5x)}{\sin^3(5x)} \cdot \frac{1}{\sin^4(5x)} \cdot \frac{1}{\sin(5x)} \, dx = \int \frac{\cos^3(5x)}{\sin^8(5x)} \, dx$$. 5. **Use substitution:** Let $$u = \csc(5x)$$, then $$du = -5 \csc(5x) \cot(5x) \, dx$$. Rewrite $$\cot^3(5x) \csc^5(5x)$$ as $$u^5 \cot^3(5x)$$. 6. **Rewrite $$\cot^3(5x)$$ as $$\cot^2(5x) \cot(5x)$$ and use $$\cot^2(5x) = \csc^2(5x) - 1 = u^2 - 1$$. 7. **Express the integral in terms of $$u$$ and $$du$$:** $$\int \cot^3(5x) \csc^5(5x) \, dx = \int u^5 (u^2 - 1) \cot(5x) \, dx$$. From substitution, $$\cot(5x) \, dx = -\frac{du}{5u}$$. 8. **Substitute:** $$\int u^5 (u^2 - 1) \cot(5x) \, dx = \int u^5 (u^2 - 1) \left(-\frac{du}{5u}\right) = -\frac{1}{5} \int u^4 (u^2 - 1) \, du$$. 9. **Simplify the integral:** $$-\frac{1}{5} \int (u^6 - u^4) \, du = -\frac{1}{5} \left( \frac{u^7}{7} - \frac{u^5}{5} \right) + C = -\frac{1}{35} u^7 + \frac{1}{25} u^5 + C$$. 10. **Back-substitute $$u = \csc(5x)$$:** $$-\frac{1}{35} \csc^7(5x) + \frac{1}{25} \csc^5(5x) + C$$. **Final answer:** $$\int \cot^3(5x) \csc^5(5x) \, dx = -\frac{1}{35} \csc^7(5x) + \frac{1}{25} \csc^5(5x) + C$$.