1. **State the problem:** Evaluate the integral $$\int 13 \cot^4 x \, dx$$. 2. **Recall the formula and identities:** We know that $$\cot^2 x = \csc^2 x - 1$$ and the integral of powers of cotangent can be simplified using this identity. 3. **Rewrite the integral:** $$\int 13 \cot^4 x \, dx = 13 \int \cot^4 x \, dx$$ 4. **Express $$\cot^4 x$$ in terms of $$\cot^2 x$$:** $$\cot^4 x = (\cot^2 x)^2 = (\csc^2 x - 1)^2 = \csc^4 x - 2 \csc^2 x + 1$$ 5. **Substitute back into the integral:** $$13 \int (\csc^4 x - 2 \csc^2 x + 1) \, dx = 13 \left( \int \csc^4 x \, dx - 2 \int \csc^2 x \, dx + \int 1 \, dx \right)$$ 6. **Use known integrals:** - $$\int \csc^2 x \, dx = -\cot x + C$$ - $$\int 1 \, dx = x + C$$ - For $$\int \csc^4 x \, dx$$, use the reduction formula or rewrite as: $$\int \csc^4 x \, dx = \int (\csc^2 x)^2 \, dx = \int (1 + \cot^2 x)^2 \, dx$$ 7. **Expand:** $$\int (1 + 2 \cot^2 x + \cot^4 x) \, dx$$ but this leads to a recursive integral. Instead, use the identity: $$\int \csc^4 x \, dx = -\cot x - \frac{\cot^3 x}{3} + C$$ 8. **Putting it all together:** $$13 \left( -\cot x - \frac{\cot^3 x}{3} - 2(-\cot x) + x \right) + C = 13 \left( -\cot x - \frac{\cot^3 x}{3} + 2 \cot x + x \right) + C$$ 9. **Simplify inside the parentheses:** $$-\cot x + 2 \cot x = \cot x$$ 10. **Final answer:** $$13 \left( \cot x - \frac{\cot^3 x}{3} + x \right) + C = 13 x + 13 \cot x - \frac{13}{3} \cot^3 x + C$$