1. **State the problem:** We need to evaluate the integral $$\int \cot^7(3\theta) \, d\theta$$. 2. **Rewrite the integral:** Recall that $$\cot x = \frac{\cos x}{\sin x}$$, so $$\cot^7(3\theta) = \cot^6(3\theta) \cdot \cot(3\theta)$$. 3. **Express in terms of powers of sine and cosine:** $$\cot^6(3\theta) = (\cot^2(3\theta))^3 = \left(\frac{\cos^2(3\theta)}{\sin^2(3\theta)}\right)^3 = \frac{\cos^6(3\theta)}{\sin^6(3\theta)}$$ So the integral becomes: $$\int \cot^7(3\theta) \, d\theta = \int \frac{\cos^6(3\theta)}{\sin^6(3\theta)} \cdot \cot(3\theta) \, d\theta = \int \frac{\cos^7(3\theta)}{\sin^7(3\theta)} \, d\theta$$ 4. **Use substitution:** Let $$u = \sin(3\theta)$$, then $$du = 3\cos(3\theta) d\theta$$ or $$d\theta = \frac{du}{3\cos(3\theta)}$$. Rewrite the integral in terms of $$u$$: $$\int \frac{\cos^7(3\theta)}{\sin^7(3\theta)} d\theta = \int \frac{\cos^7(3\theta)}{u^7} \cdot \frac{du}{3\cos(3\theta)} = \frac{1}{3} \int \frac{\cos^6(3\theta)}{u^7} du$$ 5. **Express $$\cos^6(3\theta)$$ in terms of $$u$$:** Since $$u = \sin(3\theta)$$, we have $$\cos^2(3\theta) = 1 - u^2$$, so $$\cos^6(3\theta) = (\cos^2(3\theta))^3 = (1 - u^2)^3$$. 6. **Substitute back:** $$\frac{1}{3} \int \frac{(1 - u^2)^3}{u^7} du = \frac{1}{3} \int (1 - u^2)^3 u^{-7} du$$ 7. **Expand the binomial:** $$(1 - u^2)^3 = 1 - 3u^2 + 3u^4 - u^6$$ So the integral becomes: $$\frac{1}{3} \int (u^{-7} - 3u^{-5} + 3u^{-3} - u^{-1}) du$$ 8. **Integrate term-by-term:** $$\int u^{-7} du = \frac{u^{-6}}{-6} = -\frac{1}{6u^6}$$ $$\int u^{-5} du = \frac{u^{-4}}{-4} = -\frac{1}{4u^4}$$ $$\int u^{-3} du = \frac{u^{-2}}{-2} = -\frac{1}{2u^2}$$ $$\int u^{-1} du = \ln|u|$$ 9. **Putting it all together:** $$\frac{1}{3} \left(-\frac{1}{6u^6} - 3 \left(-\frac{1}{4u^4}\right) + 3 \left(-\frac{1}{2u^2}\right) - \ln|u| \right) + C$$ Simplify inside the parentheses: $$-\frac{1}{6u^6} + \frac{3}{4u^4} - \frac{3}{2u^2} - \ln|u|$$ Multiply by $$\frac{1}{3}$$: $$-\frac{1}{18u^6} + \frac{1}{4u^4} - \frac{1}{2u^2} - \frac{1}{3} \ln|u| + C$$ 10. **Substitute back $$u = \sin(3\theta)$$:** $$\boxed{\int \cot^7(3\theta) d\theta = -\frac{1}{18 \sin^6(3\theta)} + \frac{1}{4 \sin^4(3\theta)} - \frac{1}{2 \sin^2(3\theta)} - \frac{1}{3} \ln|\sin(3\theta)| + C}$$