1. We are asked to find the integral $$\int \csc^{2} x \cot^{3} x \, dx$$. 2. Recall the identities and derivatives: - $\frac{d}{dx} \cot x = -\csc^{2} x$ - $\cot^{n} x$ can be integrated using substitution if we express powers properly. 3. Rewrite the integral: $$\int \csc^{2} x \cot^{3} x \, dx = \int \cot^{3} x \csc^{2} x \, dx$$ 4. Use substitution: let $u = \cot x$, then $du = -\csc^{2} x \, dx$ or $-du = \csc^{2} x \, dx$. 5. Substitute into the integral: $$\int \cot^{3} x \csc^{2} x \, dx = \int u^{3} (-du) = -\int u^{3} \, du$$ 6. Integrate: $$-\int u^{3} \, du = -\frac{u^{4}}{4} + C = -\frac{\cot^{4} x}{4} + C$$ 7. Therefore, the integral evaluates to: $$\boxed{-\frac{\cot^{4} x}{4} + C}$$ This matches the option with $-\frac{\cot^{4} x}{4} + C$ (not listed exactly but the closest is $\frac{\cot^{4} x}{4} + C$ with a sign difference). The correct integral includes the negative sign due to substitution. Slug: integral csc cot Subject: calculus Desmos: {"latex":"y=-\frac{\cot^{4} x}{4}+C","features":{"intercepts":true,"extrema":true}} q_count:1