1. **Problem statement:** Evaluate the integral $$I = \int \cot^3 x \csc^4 x \, dx.$$\n\n2. **Recall definitions and identities:** \n- $\cot x = \frac{\cos x}{\sin x}$\n- $\csc x = \frac{1}{\sin x}$\n- Useful derivative: $\frac{d}{dx} \csc x = -\csc x \cot x$\n\n3. **Rewrite the integral:** \n$$I = \int \cot^3 x \csc^4 x \, dx = \int \cot^3 x (\csc^2 x)^2 \, dx.$$\n\n4. **Express powers to use substitution:** \nWrite $\cot^3 x = \cot x \cdot \cot^2 x$ and use $\cot^2 x = \csc^2 x - 1$:\n$$I = \int \cot x (\csc^2 x - 1) \csc^4 x \, dx = \int \cot x \csc^6 x \, dx - \int \cot x \csc^4 x \, dx.$$\n\n5. **Use substitution:** Let $u = \csc x$, then $du = -\csc x \cot x \, dx = -u \cot x \, dx$, so $\cot x \, dx = -\frac{du}{u}$.\n\n6. **Rewrite integrals in terms of $u$:**\n- First integral: $\int \cot x \csc^6 x \, dx = \int \cot x u^6 \, dx = \int u^6 \cot x \, dx = \int u^6 \left(-\frac{du}{u}\right) = -\int u^5 \, du.$\n- Second integral: $\int \cot x \csc^4 x \, dx = \int u^4 \cot x \, dx = -\int u^3 \, du.$\n\n7. **Evaluate integrals:**\n$$-\int u^5 \, du = -\frac{u^6}{6} + C_1,$$\n$$-\int u^3 \, du = -\frac{u^4}{4} + C_2.$$\n\n8. **Combine results:**\n$$I = -\frac{u^6}{6} + \frac{u^4}{4} + C = -\frac{\csc^6 x}{6} + \frac{\csc^4 x}{4} + C.$$\n\n**Final answer:**\n$$\boxed{I = -\frac{\csc^6 x}{6} + \frac{\csc^4 x}{4} + C}.$$